Pandas 2.2 中文官方教程和指南（十三）
Pandas 2.2 中文官方教程和指南（十三）
ApacheCN_飞龙
发布于 2024-04-25 19:56:20
发布于 2024-04-25 19:56:20
写时复制（CoW）
注意
写时复制将成为 pandas 3.0 的默认设置。我们建议现在就启用它以从所有改进中受益。
写时复制首次引入于版本 1.5.0。从版本 2.0 开始，大部分通过 CoW 可能实现和支持的优化已经实现。从 pandas 2.1 开始，所有可能的优化都得到支持。
写时复制将在版本 3.0 中默认启用。
CoW 将导致更可预测的行为，因为不可能用一个语句更新多个对象，例如索引操作或方法不会产生副作用。此外，通过尽可能延迟复制，平均性能和内存使用将得到改善。
先前的行为
pandas 的索引行为很难理解。一些操作返回视图，而其他操作返回副本。根据操作的结果，改变一个对象可能会意外地改变另一个对象：
In [1]: df = pd.DataFrame({"foo": [1, 2, 3], "bar": [4, 5, 6]})
In [2]: subset = df["foo"]
In [3]: subset.iloc[0] = 100
In [4]: df
Out[4]:
foo bar
0 100 4
1 2 5
2 3 6 改变subset，例如更新其值，也会更新df。确切的行为很难预测。写时复制解决了意外修改多个对象的问题，它明确禁止这种情况。启用写时复制后，df保持不变：
In [5]: pd.options.mode.copy_on_write = True
In [6]: df = pd.DataFrame({"foo": [1, 2, 3], "bar": [4, 5, 6]})
In [7]: subset = df["foo"]
In [8]: subset.iloc[0] = 100
In [9]: df
Out[9]:
foo bar
0 1 4
1 2 5
2 3 6 接下来的部分将解释这意味着什么，以及它如何影响现有应用程序。
迁移到写时复制
写时复制将成为 pandas 3.0 的默认和唯一模式。这意味着用户需要迁移他们的代码以符合 CoW 规则。
pandas 的默认模式将对某些情况发出警告，这些情况将积极改变行为，从而改变用户预期的行为。
我们添加了另一种模式，例如
pd.options.mode.copy_on_write = "warn" 将会对每个会改变 CoW 行为的操作发出警告。我们预计这种模式会非常嘈杂，因为许多我们不认为会影响用户的情况也会发出警告。我们建议检查这种模式并分析警告，但不需要解决所有这些警告。以下列表的前两项是需要解决的唯一情况，以使现有代码与 CoW 兼容。
接下来的几个项目描述了用户可见的变化：
链接赋值永远不会起作用
应该使用loc作为替代。查看链接赋值部分获取更多细节。
访问 pandas 对象的底层数组将返回一个只读视图
In [10]: ser = pd.Series([1, 2, 3])
In [11]: ser.to_numpy()
Out[11]: array([1, 2, 3]) 这个示例返回一个 NumPy 数组，它是 Series 对象的一个视图。这个视图可以被修改，从而也会修改 pandas 对象。这不符合 CoW 规则。返回的数组被设置为不可写，以防止这种行为。创建这个数组的副本允许修改。如果你不再关心 pandas 对象，你也可以再次使数组可写。
有关只读 NumPy 数组的更多详细信息，请参阅相关部分。
一次只更新一个 pandas 对象
以下代码片段在没有 CoW 的情况下同时更新df和subset：
In [12]: df = pd.DataFrame({"foo": [1, 2, 3], "bar": [4, 5, 6]})
In [13]: subset = df["foo"]
In [14]: subset.iloc[0] = 100
In [15]: df
Out[15]:
foo bar
0 1 4
1 2 5
2 3 6 这在 CoW 中将不再可能，因为 CoW 规则明确禁止这样做。这包括将单个列更新为Series并依赖于更改传播回父DataFrame。如果需要此行为，可以使用loc或iloc将此语句重写为单个语句。DataFrame.where()是此情况的另一个合适的替代方案。
使用就地方法从DataFrame中选择的列更新也将不再起作用。
In [16]: df = pd.DataFrame({"foo": [1, 2, 3], "bar": [4, 5, 6]})
In [17]: df["foo"].replace(1, 5, inplace=True)
In [18]: df
Out[18]:
foo bar
0 1 4
1 2 5
2 3 6 这是另一种链式赋值的形式。通常可以以 2 种不同形式重写：
In [19]: df = pd.DataFrame({"foo": [1, 2, 3], "bar": [4, 5, 6]})
In [20]: df.replace({"foo": {1: 5}}, inplace=True)
In [21]: df
Out[21]:
foo bar
0 5 4
1 2 5
2 3 6 另一种选择是不使用inplace：
In [22]: df = pd.DataFrame({"foo": [1, 2, 3], "bar": [4, 5, 6]})
In [23]: df["foo"] = df["foo"].replace(1, 5)
In [24]: df
Out[24]:
foo bar
0 5 4
1 2 5
2 3 6 构造函数现在默认复制 NumPy 数组
Series 和 DataFrame 构造函数现在默认情况下将复制 NumPy 数组。这一变化是为了避免在 pandas 之外就地更改 NumPy 数组时改变 pandas 对象。您可以设置copy=False以避免此复制。
描述
CoW 意味着以任何方式从另一个 DataFrame 或 Series 派生的任何 DataFrame 或 Series 始终表现为副本。因此，我们只能通过修改对象本身来更改对象的值。CoW 不允许就地更新与另一个 DataFrame 或 Series 对象共享数据的 DataFrame 或 Series。
这样可以避免在修改值时产生副作用，因此大多数方法可以避免实际复制数据，只在必要时触发复制。
以下示例将在 CoW 下就地操作：
In [25]: df = pd.DataFrame({"foo": [1, 2, 3], "bar": [4, 5, 6]})
In [26]: df.iloc[0, 0] = 100
In [27]: df
Out[27]:
foo bar
0 100 4
1 2 5
2 3 6 对象df不与任何其他对象共享数据，因此在更新值时不会触发复制。相比之下，以下操作在 CoW 下触发数据的复制：
In [28]: df = pd.DataFrame({"foo": [1, 2, 3], "bar": [4, 5, 6]})
In [29]: df2 = df.reset_index(drop=True)
In [30]: df2.iloc[0, 0] = 100
In [31]: df
Out[31]:
foo bar
0 1 4
1 2 5
2 3 6
In [32]: df2
Out[32]:
foo bar
0 100 4
1 2 5
2 3 6 reset_index返回一个带有 CoW 的延迟复制，而不带 CoW 的复制数据。由于df和df2两个对象共享相同的数据，当修改df2时会触发复制。对象df仍然具有最初的值，而df2已被修改。
如果在执行reset_index操作后不再需要对象df，您可以通过将reset_index的输出分配给同一变量来模拟类似就地操作：
In [33]: df = pd.DataFrame({"foo": [1, 2, 3], "bar": [4, 5, 6]})
In [34]: df = df.reset_index(drop=True)
In [35]: df.iloc[0, 0] = 100
In [36]: df
Out[36]:
foo bar
0 100 4
1 2 5
2 3 6 当reset_index的结果被重新分配时，初始对象立即超出范围，因此df不与任何其他对象共享数据。在修改对象时不需要复制。这通常适用于写时复制优化中列出的所有方法。
以前，在操作视图时，会修改视图和父对象：
In [37]: with pd.option_context("mode.copy_on_write", False):
....: df = pd.DataFrame({"foo": [1, 2, 3], "bar": [4, 5, 6]})
....: view = df[:]
....: df.iloc[0, 0] = 100
....:
In [38]: df
Out[38]:
foo bar
0 100 4
1 2 5
2 3 6
In [39]: view
Out[39]:
foo bar
0 100 4
1 2 5
2 3 6 当df更改时触发拷贝，以避免突变view：
In [40]: df = pd.DataFrame({"foo": [1, 2, 3], "bar": [4, 5, 6]})
In [41]: view = df[:]
In [42]: df.iloc[0, 0] = 100
In [43]: df
Out[43]:
foo bar
0 100 4
1 2 5
2 3 6
In [44]: view
Out[44]:
foo bar
0 1 4
1 2 5
2 3 6 链式赋值
链式赋值引用一种技术，通过两个连续的索引操作来更新对象，例如。
In [45]: with pd.option_context("mode.copy_on_write", False):
....: df = pd.DataFrame({"foo": [1, 2, 3], "bar": [4, 5, 6]})
....: df["foo"][df["bar"] > 5] = 100
....: df
....: 当列bar大于 5 时，更新列foo。尽管如此，这违反了写时拷贝的原则，因为它必须在一步中修改视图df["foo"]和df。因此，链式赋值将始终无法工作，并在启用写时拷贝时引发ChainedAssignmentError警告：
In [46]: df = pd.DataFrame({"foo": [1, 2, 3], "bar": [4, 5, 6]})
In [47]: df["foo"][df["bar"] > 5] = 100 通过使用loc可以实现写时拷贝。
In [48]: df.loc[df["bar"] > 5, "foo"] = 100
```## 只读 NumPy 数组
如果数组与初始 DataFrame 共享数据，则访问 DataFrame 的底层 NumPy 数组将返回只读数组：
如果初始 DataFrame 由多个数组组成，则该数组是一个拷贝：
```py
In [49]: df = pd.DataFrame({"a": [1, 2], "b": [1.5, 2.5]})
In [50]: df.to_numpy()
Out[50]:
array([[1\. , 1.5],
[2\. , 2.5]]) 如果 DataFrame 仅由一个 NumPy 数组组成，则该数组与 DataFrame 共享数据：
In [51]: df = pd.DataFrame({"a": [1, 2], "b": [3, 4]})
In [52]: df.to_numpy()
Out[52]:
array([[1, 3],
[2, 4]]) 此数组是只读的，这意味着它不能就地修改：
In [53]: arr = df.to_numpy()
In [54]: arr[0, 0] = 100
---------------------------------------------------------------------------
ValueError Traceback (most recent call last)
Cell In[54], line 1
----> 1 arr[0, 0] = 100
ValueError: assignment destination is read-only 对于 Series 也是如此，因为 Series 始终由单个数组组成。
这有两种潜在的解决方案：
- 如果想避免更新与数组共享内存的 DataFrame，则手动触发拷贝。
- 使数组可写。这是一种性能更好的解决方案，但是绕过了写时拷贝规则，因此应谨慎使用。
In [55]: arr = df.to_numpy()
In [56]: arr.flags.writeable = True
In [57]: arr[0, 0] = 100
In [58]: arr
Out[58]:
array([[100, 3],
[ 2, 4]]) 避免模式
如果两个对象共享相同的数据，而您正在就地修改一个对象，则不会执行防御性拷贝。
In [59]: df = pd.DataFrame({"a": [1, 2, 3], "b": [4, 5, 6]})
In [60]: df2 = df.reset_index(drop=True)
In [61]: df2.iloc[0, 0] = 100 这将创建两个共享数据的对象，因此 setitem 操作将触发一个拷贝。如果不再需要初始对象df，则不需要这样做。简单地重新分配给相同的变量将使对象持有的引用无效。
In [62]: df = pd.DataFrame({"a": [1, 2, 3], "b": [4, 5, 6]})
In [63]: df = df.reset_index(drop=True)
In [64]: df.iloc[0, 0] = 100 在这个例子中不需要拷贝。创建多个引用会保持不必要的引用存在，因此会影响性能，因为写时拷贝。
写时拷贝优化
新的惰性拷贝机制，直到修改问题对象并且仅当该对象与另一个对象共享数据时才进行拷贝。此机制已添加到不需要底层数据拷贝的方法中。流行的例子有DataFrame.drop()用于axis=1和DataFrame.rename()。
当启用写时拷贝时，这些方法返回视图，与常规执行相比提供了显著的性能改进。 ## 如何启用写时拷贝
写时拷贝可以通过配置选项copy_on_write启用。该选项可以通过以下任一方式 __ 全局 __ 启用：
In [65]: pd.set_option("mode.copy_on_write", True)
In [66]: pd.options.mode.copy_on_write = True 先前的行为
pandas 的索引行为很难理解。一些操作返回视图，而另一些操作返回副本。根据操作的结果，改变一个对象可能会意外地改变另一个对象：
In [1]: df = pd.DataFrame({"foo": [1, 2, 3], "bar": [4, 5, 6]})
In [2]: subset = df["foo"]
In [3]: subset.iloc[0] = 100
In [4]: df
Out[4]:
foo bar
0 100 4
1 2 5
2 3 6 改变subset，例如更新其值，也会更新df。确切的行为很难预测。Copy-on-Write 解决了意外修改多个对象的问题，它明确禁止这种情况发生。启用 CoW 后，df保持不变：
In [5]: pd.options.mode.copy_on_write = True
In [6]: df = pd.DataFrame({"foo": [1, 2, 3], "bar": [4, 5, 6]})
In [7]: subset = df["foo"]
In [8]: subset.iloc[0] = 100
In [9]: df
Out[9]:
foo bar
0 1 4
1 2 5
2 3 6 接下来的部分将解释这意味着什么以及它如何影响现有应用程序。
迁移至 Copy-on-Write
在 pandas 3.0 中，Copy-on-Write 将成为默认且唯一模式。这意味着用户需要迁移其代码以符合 CoW 规则。
pandas 的默认模式将对某些情况发出警告，这些情况将积极改变行为，从而改变用户预期的行为。
我们添加了另一种模式，例如。
pd.options.mode.copy_on_write = "warn" 对于每个会改变行为的操作都会发出 CoW 警告。我们预计这种模式会非常嘈杂，因为许多我们不希望影响用户的情况也会发出警告。我们建议检查此模式并分析警告，但不需要解决所有这些警告。以下列表的前两项是需要解决的唯一情况，以使现有代码与 CoW 一起正常工作。
接下来的几个项目描述了用户可见的更改：
链式赋值永远不会起作用
应该使用loc作为替代方法。查看链式赋值部分以获取更多详细信息。
访问 pandas 对象的底层数组将返回一个只读视图
In [10]: ser = pd.Series([1, 2, 3])
In [11]: ser.to_numpy()
Out[11]: array([1, 2, 3]) 此示例返回一个 Series 对象的视图的 NumPy 数组。此视图可以被修改，从而也修改 pandas 对象。这不符合 CoW 规则。返回的数组设置为不可写，以防止这种行为。创建此数组的副本允许修改。如果不再关心 pandas 对象，也可以再次使数组可写。
查看关于只读 NumPy 数组的部分以获取更多详细信息。
一次只更新一个 pandas 对象
以下代码片段在没有 CoW 的情况下同时更新df和subset：
In [12]: df = pd.DataFrame({"foo": [1, 2, 3], "bar": [4, 5, 6]})
In [13]: subset = df["foo"]
In [14]: subset.iloc[0] = 100
In [15]: df
Out[15]:
foo bar
0 1 4
1 2 5
2 3 6 这在 CoW 下将不再可能，因为 CoW 规则明确禁止这样做。这包括更新单个列作为Series并依赖于更改传播回父DataFrame。如果需要此行为，可以将此语句重写为使用loc或iloc的单个语句。DataFrame.where()是此情况的另一个合适的替代方法。
使用就地方法从DataFrame中选择的列更新列也将不再起作用。
In [16]: df = pd.DataFrame({"foo": [1, 2, 3], "bar": [4, 5, 6]})
In [17]: df["foo"].replace(1, 5, inplace=True)
In [18]: df
Out[18]:
foo bar
0 1 4
1 2 5
2 3 6 这是另一种链式赋值的形式。这通常可以以 2 种不同的形式重写：
In [19]: df = pd.DataFrame({"foo": [1, 2, 3], "bar": [4, 5, 6]})
In [20]: df.replace({"foo": {1: 5}}, inplace=True)
In [21]: df
Out[21]:
foo bar
0 5 4
1 2 5
2 3 6 另一种选择是不使用inplace：
In [22]: df = pd.DataFrame({"foo": [1, 2, 3], "bar": [4, 5, 6]})
In [23]: df["foo"] = df["foo"].replace(1, 5)
In [24]: df
Out[24]:
foo bar
0 5 4
1 2 5
2 3 6 构造函数现在默认复制 NumPy 数组
当没有另行指定时，Series 和 DataFrame 构造函数现在默认复制 NumPy 数组。这一变更是为了避免在 pandas 之外原位更改 NumPy 数组时突变 pandas 对象。您可以设置copy=False来避免此复制。
描述
CoW 意味着以任何方式从另一个 DataFrame 或 Series 派生的任何 DataFrame 或 Series 都始终表现为副本。因此，我们只能通过修改对象本身来更改对象的值。CoW 不允许直接更新共享数据与另一个 DataFrame 或 Series 对象的 DataFrame 或 Series。
在修改值时避免副作用，因此，大多数方法可以避免实际复制数据，并且只在必要时触发复制。
以下示例将在 CoW 下进行就地操作：
In [25]: df = pd.DataFrame({"foo": [1, 2, 3], "bar": [4, 5, 6]})
In [26]: df.iloc[0, 0] = 100
In [27]: df
Out[27]:
foo bar
0 100 4
1 2 5
2 3 6 对象df不与任何其他对象共享数据，因此在更新值时不触发复制。相比之下，下面的操作在 CoW 下触发数据的复制：
In [28]: df = pd.DataFrame({"foo": [1, 2, 3], "bar": [4, 5, 6]})
In [29]: df2 = df.reset_index(drop=True)
In [30]: df2.iloc[0, 0] = 100
In [31]: df
Out[31]:
foo bar
0 1 4
1 2 5
2 3 6
In [32]: df2
Out[32]:
foo bar
0 100 4
1 2 5
2 3 6 reset_index返回一个带有 CoW 的延迟副本，而在没有 CoW 的情况下复制数据。由于df和df2这两个对象共享相同的数据，所以当修改df2时会触发复制。对象df仍然具有最初的相同值，而df2已经被修改。
如果在执行reset_index操作后不再需要对象df，则可以通过将reset_index的输出分配给同一变量来模拟类似于 inplace 的操作：
In [33]: df = pd.DataFrame({"foo": [1, 2, 3], "bar": [4, 5, 6]})
In [34]: df = df.reset_index(drop=True)
In [35]: df.iloc[0, 0] = 100
In [36]: df
Out[36]:
foo bar
0 100 4
1 2 5
2 3 6 当reset_index的结果重新分配时，初始对象就会超出范围，因此df与任何其他对象都不共享数据。在修改对象时，不需要复制。这通常对于列表中列出的所有方法都成立写时复制优化。
以前，在操作视图时，视图和父对象都会被修改：
In [37]: with pd.option_context("mode.copy_on_write", False):
....: df = pd.DataFrame({"foo": [1, 2, 3], "bar": [4, 5, 6]})
....: view = df[:]
....: df.iloc[0, 0] = 100
....:
In [38]: df
Out[38]:
foo bar
0 100 4
1 2 5
2 3 6
In [39]: view
Out[39]:
foo bar
0 100 4
1 2 5
2 3 6 当修改df时，CoW 会触发复制以避免同时更改view：
In [40]: df = pd.DataFrame({"foo": [1, 2, 3], "bar": [4, 5, 6]})
In [41]: view = df[:]
In [42]: df.iloc[0, 0] = 100
In [43]: df
Out[43]:
foo bar
0 100 4
1 2 5
2 3 6
In [44]: view
Out[44]:
foo bar
0 1 4
1 2 5
2 3 6 链式赋值
链式赋值引用一种通过两个后续索引操作更新对象的技术，例如
In [45]: with pd.option_context("mode.copy_on_write", False):
....: df = pd.DataFrame({"foo": [1, 2, 3], "bar": [4, 5, 6]})
....: df["foo"][df["bar"] > 5] = 100
....: df
....: 当列bar大于 5 时，更新列foo。尽管如此，这违反了 CoW 原则，因为它需要一次性修改视图df["foo"]和df。因此，链式赋值始终不起作用，并在启用 CoW 时引发ChainedAssignmentError警告：
In [46]: df = pd.DataFrame({"foo": [1, 2, 3], "bar": [4, 5, 6]})
In [47]: df["foo"][df["bar"] > 5] = 100 使用loc可以使用写时复制来完成这个过程。
In [48]: df.loc[df["bar"] > 5, "foo"] = 100 只读 NumPy 数组
访问 DataFrame 的底层 NumPy 数组将返回一个只读数组，如果数组与初始 DataFrame 共享数据：
如果初始 DataFrame 包含多个数组，则数组是副本：
In [49]: df = pd.DataFrame({"a": [1, 2], "b": [1.5, 2.5]})
In [50]: df.to_numpy()
Out[50]:
array([[1\. , 1.5],
[2\. , 2.5]]) 如果 DataFrame 只包含一个 NumPy 数组，则该数组与 DataFrame 共享数据：
In [51]: df = pd.DataFrame({"a": [1, 2], "b": [3, 4]})
In [52]: df.to_numpy()
Out[52]:
array([[1, 3],
[2, 4]]) 此数组是只读的，这意味着它不能就地修改：
In [53]: arr = df.to_numpy()
In [54]: arr[0, 0] = 100
---------------------------------------------------------------------------
ValueError Traceback (most recent call last)
Cell In[54], line 1
----> 1 arr[0, 0] = 100
ValueError: assignment destination is read-only 对于 Series 也是如此，因为 Series 总是由单个数组组成。
有两种潜在的解决方案：
- 如果您想要避免更新与数组共享内存的 DataFrame，则手动触发复制。
- 使数组可写。这是一种更高效的解决方案，但是它绕过了写时复制规则，因此应谨慎使用。
In [55]: arr = df.to_numpy()
In [56]: arr.flags.writeable = True
In [57]: arr[0, 0] = 100
In [58]: arr
Out[58]:
array([[100, 3],
[ 2, 4]]) 避免的模式
如果两个对象在您就地修改一个对象时共享相同的数据，则不会执行防御性复制。
In [59]: df = pd.DataFrame({"a": [1, 2, 3], "b": [4, 5, 6]})
In [60]: df2 = df.reset_index(drop=True)
In [61]: df2.iloc[0, 0] = 100 这会创建两个共享数据的对象，因此 setitem 操作将触发复制。如果初始对象 df 不再需要，则不需要这样做。简单地重新分配给同一个变量将使对象持有的引用失效。
In [62]: df = pd.DataFrame({"a": [1, 2, 3], "b": [4, 5, 6]})
In [63]: df = df.reset_index(drop=True)
In [64]: df.iloc[0, 0] = 100 在这个例子中不需要复制。创建多个引用会保持不必要的引用活动，因此会通过写时复制对性能造成损害。
写时复制优化
新的惰性复制机制推迟了直到修改了问题对象并且仅在此对象与另一个对象共享数据时才复制该对象。此机制已添加到不需要复制底层数据的方法中。常见示例是DataFrame.drop()对于axis=1和DataFrame.rename()。
当启用写时复制（Copy-on-Write）时，这些方法返回视图，与常规执行相比，这提供了显著的性能改进。
如何启用写时复制
可以通过配置选项 copy_on_write 启用写时复制。该选项可以通过以下任一全局方式进行打开：
In [65]: pd.set_option("mode.copy_on_write", True)
In [66]: pd.options.mode.copy_on_write = True 合并，连接，串联和比较
pandas 提供了各种方法来合并和比较Series或DataFrame。
-
concat(): 将多个Series或DataFrame对象沿着共享的索引或列合并 -
DataFrame.join(): 沿着列合并多个DataFrame对象 -
DataFrame.combine_first(): 在相同位置使用非缺失值更新缺失值 -
merge(): 用类似 SQL 的方式合并两个Series或DataFrame对象 -
merge_ordered(): 沿着有序轴合并两个Series或DataFrame对象 -
merge_asof(): 通过近似匹配键而不是精确匹配键来合并两个Series或DataFrame对象 -
Series.compare()和DataFrame.compare(): 显示两个Series或DataFrame对象之间的值差异
concat()
concat()函数沿着一个轴连接任意数量的Series或DataFrame对象，同时在其他轴上执行可选的集合逻辑（并集或交集）索引。与numpy.concatenate类似，concat()接受一个同类型对象的列表或字典，并将它们连接起来。
In [1]: df1 = pd.DataFrame(
...: {
...: "A": ["A0", "A1", "A2", "A3"],
...: "B": ["B0", "B1", "B2", "B3"],
...: "C": ["C0", "C1", "C2", "C3"],
...: "D": ["D0", "D1", "D2", "D3"],
...: },
...: index=[0, 1, 2, 3],
...: )
...:
In [2]: df2 = pd.DataFrame(
...: {
...: "A": ["A4", "A5", "A6", "A7"],
...: "B": ["B4", "B5", "B6", "B7"],
...: "C": ["C4", "C5", "C6", "C7"],
...: "D": ["D4", "D5", "D6", "D7"],
...: },
...: index=[4, 5, 6, 7],
...: )
...:
In [3]: df3 = pd.DataFrame(
...: {
...: "A": ["A8", "A9", "A10", "A11"],
...: "B": ["B8", "B9", "B10", "B11"],
...: "C": ["C8", "C9", "C10", "C11"],
...: "D": ["D8", "D9", "D10", "D11"],
...: },
...: index=[8, 9, 10, 11],
...: )
...:
In [4]: frames = [df1, df2, df3]
In [5]: result = pd.concat(frames)
In [6]: result
Out[6]:
A B C D
0 A0 B0 C0 D0
1 A1 B1 C1 D1
2 A2 B2 C2 D2
3 A3 B3 C3 D3
4 A4 B4 C4 D4
5 A5 B5 C5 D5
6 A6 B6 C6 D6
7 A7 B7 C7 D7
8 A8 B8 C8 D8
9 A9 B9 C9 D9
10 A10 B10 C10 D10
11 A11 B11 C11 D11 注意
concat()会对数据进行完全复制，并且反复使用concat()可能会创建不必要的副本。在使用concat()之前，先将所有DataFrame或Series对象收集到一个列表中。
frames = [process_your_file(f) for f in files]
result = pd.concat(frames) 注意
当连接具有命名轴的DataFrame时，pandas 会尽可能保留这些索引/列名称。在所有输入共享一个公共名称的情况下，该名称将分配给结果。当输入名称不完全一致时，结果将没有名称。对于MultiIndex也是如此，但逻辑是逐级别分别应用的。
结果轴的连接逻辑
join关键字指定如何处理第一个DataFrame中不存在的轴值。
join='outer'取所有轴值的并集
In [7]: df4 = pd.DataFrame(
...: {
...: "B": ["B2", "B3", "B6", "B7"],
...: "D": ["D2", "D3", "D6", "D7"],
...: "F": ["F2", "F3", "F6", "F7"],
...: },
...: index=[2, 3, 6, 7],
...: )
...:
In [8]: result = pd.concat([df1, df4], axis=1)
In [9]: result
Out[9]:
A B C D B D F
0 A0 B0 C0 D0 NaN NaN NaN
1 A1 B1 C1 D1 NaN NaN NaN
2 A2 B2 C2 D2 B2 D2 F2
3 A3 B3 C3 D3 B3 D3 F3
6 NaN NaN NaN NaN B6 D6 F6
7 NaN NaN NaN NaN B7 D7 F7 join='inner'取轴值的交集
In [10]: result = pd.concat([df1, df4], axis=1, join="inner")
In [11]: result
Out[11]:
A B C D B D F
2 A2 B2 C2 D2 B2 D2 F2
3 A3 B3 C3 D3 B3 D3 F3 为了使用原始DataFrame的确切索引执行有效的“左”连接，结果可以重新索引。
In [12]: result = pd.concat([df1, df4], axis=1).reindex(df1.index)
In [13]: result
Out[13]:
A B C D B D F
0 A0 B0 C0 D0 NaN NaN NaN
1 A1 B1 C1 D1 NaN NaN NaN
2 A2 B2 C2 D2 B2 D2 F2
3 A3 B3 C3 D3 B3 D3 F3 ### 在连接轴上忽略索引
对于没有有意义索引的DataFrame对象，ignore_index会忽略重叠的索引。
In [14]: result = pd.concat([df1, df4], ignore_index=True, sort=False)
In [15]: result
Out[15]:
A B C D F
0 A0 B0 C0 D0 NaN
1 A1 B1 C1 D1 NaN
2 A2 B2 C2 D2 NaN
3 A3 B3 C3 D3 NaN
4 NaN B2 NaN D2 F2
5 NaN B3 NaN D3 F3
6 NaN B6 NaN D6 F6
7 NaN B7 NaN D7 F7 ### 将Series和DataFrame连接在一起
您可以连接一组Series和DataFrame对象。Series将转换为具??列名的DataFrame，列名为Series的名称。
In [16]: s1 = pd.Series(["X0", "X1", "X2", "X3"], name="X")
In [17]: result = pd.concat([df1, s1], axis=1)
In [18]: result
Out[18]:
A B C D X
0 A0 B0 C0 D0 X0
1 A1 B1 C1 D1 X1
2 A2 B2 C2 D2 X2
3 A3 B3 C3 D3 X3 未命名的Series将按顺序编号。
In [19]: s2 = pd.Series(["_0", "_1", "_2", "_3"])
In [20]: result = pd.concat([df1, s2, s2, s2], axis=1)
In [21]: result
Out[21]:
A B C D 0 1 2
0 A0 B0 C0 D0 _0 _0 _0
1 A1 B1 C1 D1 _1 _1 _1
2 A2 B2 C2 D2 _2 _2 _2
3 A3 B3 C3 D3 _3 _3 _3 ignore_index=True 将删除所有名称引用。
In [22]: result = pd.concat([df1, s1], axis=1, ignore_index=True)
In [23]: result
Out[23]:
0 1 2 3 4
0 A0 B0 C0 D0 X0
1 A1 B1 C1 D1 X1
2 A2 B2 C2 D2 X2
3 A3 B3 C3 D3 X3 结果的keys
keys 参数将向结果索引或列添加另一个轴级别（创建一个MultiIndex），将特定键与每个原始DataFrame关联。
In [24]: result = pd.concat(frames, keys=["x", "y", "z"])
In [25]: result
Out[25]:
A B C D
x 0 A0 B0 C0 D0
1 A1 B1 C1 D1
2 A2 B2 C2 D2
3 A3 B3 C3 D3
y 4 A4 B4 C4 D4
5 A5 B5 C5 D5
6 A6 B6 C6 D6
7 A7 B7 C7 D7
z 8 A8 B8 C8 D8
9 A9 B9 C9 D9
10 A10 B10 C10 D10
11 A11 B11 C11 D11
In [26]: result.loc["y"]
Out[26]:
A B C D
4 A4 B4 C4 D4
5 A5 B5 C5 D5
6 A6 B6 C6 D6
7 A7 B7 C7 D7 keys 参数可以在基于现有Series创建新DataFrame时覆盖列名。
In [27]: s3 = pd.Series([0, 1, 2, 3], name="foo")
In [28]: s4 = pd.Series([0, 1, 2, 3])
In [29]: s5 = pd.Series([0, 1, 4, 5])
In [30]: pd.concat([s3, s4, s5], axis=1)
Out[30]:
foo 0 1
0 0 0 0
1 1 1 1
2 2 2 4
3 3 3 5
In [31]: pd.concat([s3, s4, s5], axis=1, keys=["red", "blue", "yellow"])
Out[31]:
red blue yellow
0 0 0 0
1 1 1 1
2 2 2 4
3 3 3 5 您还可以将字典传递给concat()，在这种情况下，除非指定了其他keys参数，否则将使用字典键作为keys参数：
In [32]: pieces = {"x": df1, "y": df2, "z": df3}
In [33]: result = pd.concat(pieces)
In [34]: result
Out[34]:
A B C D
x 0 A0 B0 C0 D0
1 A1 B1 C1 D1
2 A2 B2 C2 D2
3 A3 B3 C3 D3
y 4 A4 B4 C4 D4
5 A5 B5 C5 D5
6 A6 B6 C6 D6
7 A7 B7 C7 D7
z 8 A8 B8 C8 D8
9 A9 B9 C9 D9
10 A10 B10 C10 D10
11 A11 B11 C11 D11 In [35]: result = pd.concat(pieces, keys=["z", "y"])
In [36]: result
Out[36]:
A B C D
z 8 A8 B8 C8 D8
9 A9 B9 C9 D9
10 A10 B10 C10 D10
11 A11 B11 C11 D11
y 4 A4 B4 C4 D4
5 A5 B5 C5 D5
6 A6 B6 C6 D6
7 A7 B7 C7 D7 创建的MultiIndex具有从传递的键和DataFrame片段的索引构建的级别：
In [37]: result.index.levels
Out[37]: FrozenList([['z', 'y'], [4, 5, 6, 7, 8, 9, 10, 11]]) levels 参数允许指定与keys相关联的结果级别
In [38]: result = pd.concat(
....: pieces, keys=["x", "y", "z"], levels=[["z", "y", "x", "w"]], names=["group_key"]
....: )
....:
In [39]: result
Out[39]:
A B C D
group_key
x 0 A0 B0 C0 D0
1 A1 B1 C1 D1
2 A2 B2 C2 D2
3 A3 B3 C3 D3
y 4 A4 B4 C4 D4
5 A5 B5 C5 D5
6 A6 B6 C6 D6
7 A7 B7 C7 D7
z 8 A8 B8 C8 D8
9 A9 B9 C9 D9
10 A10 B10 C10 D10
11 A11 B11 C11 D11 In [40]: result.index.levels
Out[40]: FrozenList([['z', 'y', 'x', 'w'], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11]]) 将行附加到DataFrame
如果您有一个要附加为单行到 DataFrame 的 Series，您可以将该行转换为 DataFrame 并使用 concat()
In [41]: s2 = pd.Series(["X0", "X1", "X2", "X3"], index=["A", "B", "C", "D"])
In [42]: result = pd.concat([df1, s2.to_frame().T], ignore_index=True)
In [43]: result
Out[43]:
A B C D
0 A0 B0 C0 D0
1 A1 B1 C1 D1
2 A2 B2 C2 D2
3 A3 B3 C3 D3
4 X0 X1 X2 X3 ## merge()
merge() 执行类似于关系数据库（如 SQL）的连接操作。熟悉 SQL 但是对 pandas 新手的用户可以参考与 SQL 的比较。
连接类型
merge() 实现了常见的 SQL 风格的连接操作。
- 一对一：在它们的索引上连接两个
DataFrame对象，这些索引必须包含唯一值。 - 一对多：将唯一索引与不同
DataFrame中的一个或多个列进行连接。 - 多对多：在列上连接列。
注意
当在列上连接列时，可能是多对多的连接，传递的 DataFrame 对象上的任何索引将被丢弃。
对于多对多的连接，如果一个键组合在两个表中出现多次，DataFrame 将具有相关数据的笛卡尔积。
In [44]: left = pd.DataFrame(
....: {
....: "key": ["K0", "K1", "K2", "K3"],
....: "A": ["A0", "A1", "A2", "A3"],
....: "B": ["B0", "B1", "B2", "B3"],
....: }
....: )
....:
In [45]: right = pd.DataFrame(
....: {
....: "key": ["K0", "K1", "K2", "K3"],
....: "C": ["C0", "C1", "C2", "C3"],
....: "D": ["D0", "D1", "D2", "D3"],
....: }
....: )
....:
In [46]: result = pd.merge(left, right, on="key")
In [47]: result
Out[47]:
key A B C D
0 K0 A0 B0 C0 D0
1 K1 A1 B1 C1 D1
2 K2 A2 B2 C2 D2
3 K3 A3 B3 C3 D3 对于 merge() 的 how 参数指定了哪些键包含在结果表中。如果一个键组合不存在于左表或右表中，连接表中的值将为 NA。以下是 how 选项及其 SQL 等效名称的摘要：
合并方法 | SQL 连接名称 | 描述 |
|---|---|---|
left | LEFT OUTER JOIN | 仅使用左框架中的键 |
right | RIGHT OUTER JOIN | 仅使用右框架中的键 |
outer | FULL OUTER JOIN | 使用两个框架的键的并集 |
inner | INNER JOIN | 使用两个框架中键的交集 |
cross | CROSS JOIN | 创建两个框架行的笛卡尔积 |
In [48]: left = pd.DataFrame(
....: {
....: "key1": ["K0", "K0", "K1", "K2"],
....: "key2": ["K0", "K1", "K0", "K1"],
....: "A": ["A0", "A1", "A2", "A3"],
....: "B": ["B0", "B1", "B2", "B3"],
....: }
....: )
....:
In [49]: right = pd.DataFrame(
....: {
....: "key1": ["K0", "K1", "K1", "K2"],
....: "key2": ["K0", "K0", "K0", "K0"],
....: "C": ["C0", "C1", "C2", "C3"],
....: "D": ["D0", "D1", "D2", "D3"],
....: }
....: )
....:
In [50]: result = pd.merge(left, right, how="left", on=["key1", "key2"])
In [51]: result
Out[51]:
key1 key2 A B C D
0 K0 K0 A0 B0 C0 D0
1 K0 K1 A1 B1 NaN NaN
2 K1 K0 A2 B2 C1 D1
3 K1 K0 A2 B2 C2 D2
4 K2 K1 A3 B3 NaN NaN In [52]: result = pd.merge(left, right, how="right", on=["key1", "key2"])
In [53]: result
Out[53]:
key1 key2 A B C D
0 K0 K0 A0 B0 C0 D0
1 K1 K0 A2 B2 C1 D1
2 K1 K0 A2 B2 C2 D2
3 K2 K0 NaN NaN C3 D3 In [54]: result = pd.merge(left, right, how="outer", on=["key1", "key2"])
In [55]: result
Out[55]:
key1 key2 A B C D
0 K0 K0 A0 B0 C0 D0
1 K0 K1 A1 B1 NaN NaN
2 K1 K0 A2 B2 C1 D1
3 K1 K0 A2 B2 C2 D2
4 K2 K0 NaN NaN C3 D3
5 K2 K1 A3 B3 NaN NaN In [56]: result = pd.merge(left, right, how="inner", on=["key1", "key2"])
In [57]: result
Out[57]:
key1 key2 A B C D
0 K0 K0 A0 B0 C0 D0
1 K1 K0 A2 B2 C1 D1
2 K1 K0 A2 B2 C2 D2 In [58]: result = pd.merge(left, right, how="cross")
In [59]: result
Out[59]:
key1_x key2_x A B key1_y key2_y C D
0 K0 K0 A0 B0 K0 K0 C0 D0
1 K0 K0 A0 B0 K1 K0 C1 D1
2 K0 K0 A0 B0 K1 K0 C2 D2
3 K0 K0 A0 B0 K2 K0 C3 D3
4 K0 K1 A1 B1 K0 K0 C0 D0
.. ... ... .. .. ... ... .. ..
11 K1 K0 A2 B2 K2 K0 C3 D3
12 K2 K1 A3 B3 K0 K0 C0 D0
13 K2 K1 A3 B3 K1 K0 C1 D1
14 K2 K1 A3 B3 K1 K0 C2 D2
15 K2 K1 A3 B3 K2 K0 C3 D3
[16 rows x 8 columns] 如果MultiIndex的名称与DataFrame中的列名对应，则可以使用Series和具有MultiIndex的DataFrame。在合并之前，使用Series.reset_index()将Series转换为DataFrame
In [60]: df = pd.DataFrame({"Let": ["A", "B", "C"], "Num": [1, 2, 3]})
In [61]: df
Out[61]:
Let Num
0 A 1
1 B 2
2 C 3
In [62]: ser = pd.Series(
....: ["a", "b", "c", "d", "e", "f"],
....: index=pd.MultiIndex.from_arrays(
....: [["A", "B", "C"] * 2, [1, 2, 3, 4, 5, 6]], names=["Let", "Num"]
....: ),
....: )
....:
In [63]: ser
Out[63]:
Let Num
A 1 a
B 2 b
C 3 c
A 4 d
B 5 e
C 6 f
dtype: object
In [64]: pd.merge(df, ser.reset_index(), on=["Let", "Num"])
Out[64]:
Let Num 0
0 A 1 a
1 B 2 b
2 C 3 c 在具有重复连接键的DataFrame中执行外连接
In [65]: left = pd.DataFrame({"A": [1, 2], "B": [2, 2]})
In [66]: right = pd.DataFrame({"A": [4, 5, 6], "B": [2, 2, 2]})
In [67]: result = pd.merge(left, right, on="B", how="outer")
In [68]: result
Out[68]:
A_x B A_y
0 1 2 4
1 1 2 5
2 1 2 6
3 2 2 4
4 2 2 5
5 2 2 6 警告
在重复键上进行合并会显著增加结果的维度，并可能导致内存溢出。
合并键的唯一性
validate参数检查合并键的唯一性。在执行合并操作之前检查键的唯一性可以防止内存溢出和意外键重复。
In [69]: left = pd.DataFrame({"A": [1, 2], "B": [1, 2]})
In [70]: right = pd.DataFrame({"A": [4, 5, 6], "B": [2, 2, 2]})
In [71]: result = pd.merge(left, right, on="B", how="outer", validate="one_to_one")
---------------------------------------------------------------------------
MergeError Traceback (most recent call last)
Cell In[71], line 1
----> 1 result = pd.merge(left, right, on="B", how="outer", validate="one_to_one")
File ~/work/pandas/pandas/pandas/core/reshape/merge.py:170, in merge(left, right, how, on, left_on, right_on, left_index, right_index, sort, suffixes, copy, indicator, validate)
155 return _cross_merge(
156 left_df,
157 right_df,
(...)
167 copy=copy,
168 )
169 else:
--> 170 op = _MergeOperation(
171 left_df,
172 right_df,
173 how=how,
174 on=on,
175 left_on=left_on,
176 right_on=right_on,
177 left_index=left_index,
178 right_index=right_index,
179 sort=sort,
180 suffixes=suffixes,
181 indicator=indicator,
182 validate=validate,
183 )
184 return op.get_result(copy=copy)
File ~/work/pandas/pandas/pandas/core/reshape/merge.py:813, in _MergeOperation.__init__(self, left, right, how, on, left_on, right_on, left_index, right_index, sort, suffixes, indicator, validate)
809 # If argument passed to validate,
810 # check if columns specified as unique
811 # are in fact unique.
812 if validate is not None:
--> 813 self._validate_validate_kwd(validate)
File ~/work/pandas/pandas/pandas/core/reshape/merge.py:1657, in _MergeOperation._validate_validate_kwd(self, validate)
1653 raise MergeError(
1654 "Merge keys are not unique in left dataset; not a one-to-one merge"
1655 )
1656 if not right_unique:
-> 1657 raise MergeError(
1658 "Merge keys are not unique in right dataset; not a one-to-one merge"
1659 )
1661 elif validate in ["one_to_many", "1:m"]:
1662 if not left_unique:
MergeError: Merge keys are not unique in right dataset; not a one-to-one merge 如果用户意识到右侧DataFrame中存在重复项，但希望确保左侧DataFrame中没有重复项，则可以使用validate='one_to_many'参数，这样不会引发异常。
In [72]: pd.merge(left, right, on="B", how="outer", validate="one_to_many")
Out[72]:
A_x B A_y
0 1 1 NaN
1 2 2 4.0
2 2 2 5.0
3 2 2 6.0
```### 合并结果指示器
`merge()`接受参数`indicator`。如果为`True`，则将向输出对象添加一个名为`_merge`的分类列，其取值为：
> | 观察来源 | `_merge`值 |
> | --- | --- |
> | 仅在`'left'`数据框中的合并键 | `left_only` |
> | 仅在`'right'`数据框中的合并键 | `right_only` |
> | 两个数据框中的合并键 | `both` |
```py
In [73]: df1 = pd.DataFrame({"col1": [0, 1], "col_left": ["a", "b"]})
In [74]: df2 = pd.DataFrame({"col1": [1, 2, 2], "col_right": [2, 2, 2]})
In [75]: pd.merge(df1, df2, on="col1", how="outer", indicator=True)
Out[75]:
col1 col_left col_right _merge
0 0 a NaN left_only
1 1 b 2.0 both
2 2 NaN 2.0 right_only
3 2 NaN 2.0 right_only 字符串参数indicator将使用该值作为指示器列的名称。
In [76]: pd.merge(df1, df2, on="col1", how="outer", indicator="indicator_column")
Out[76]:
col1 col_left col_right indicator_column
0 0 a NaN left_only
1 1 b 2.0 both
2 2 NaN 2.0 right_only
3 2 NaN 2.0 right_only 重叠值列
合并suffixes参数接受一个字符串列表的元组，用于附加到输入DataFrame中重叠列名以消除结果列的歧义：
In [77]: left = pd.DataFrame({"k": ["K0", "K1", "K2"], "v": [1, 2, 3]})
In [78]: right = pd.DataFrame({"k": ["K0", "K0", "K3"], "v": [4, 5, 6]})
In [79]: result = pd.merge(left, right, on="k")
In [80]: result
Out[80]:
k v_x v_y
0 K0 1 4
1 K0 1 5 In [81]: result = pd.merge(left, right, on="k", suffixes=("_l", "_r"))
In [82]: result
Out[82]:
k v_l v_r
0 K0 1 4
1 K0 1 5 DataFrame.join()
DataFrame.join()将多个可能具有不同索引的列的DataFrame合并为单个结果DataFrame。
In [83]: left = pd.DataFrame(
....: {"A": ["A0", "A1", "A2"], "B": ["B0", "B1", "B2"]}, index=["K0", "K1", "K2"]
....: )
....:
In [84]: right = pd.DataFrame(
....: {"C": ["C0", "C2", "C3"], "D": ["D0", "D2", "D3"]}, index=["K0", "K2", "K3"]
....: )
....:
In [85]: result = left.join(right)
In [86]: result
Out[86]:
A B C D
K0 A0 B0 C0 D0
K1 A1 B1 NaN NaN
K2 A2 B2 C2 D2 In [87]: result = left.join(right, how="outer")
In [88]: result
Out[88]:
A B C D
K0 A0 B0 C0 D0
K1 A1 B1 NaN NaN
K2 A2 B2 C2 D2
K3 NaN NaN C3 D3 In [89]: result = left.join(right, how="inner")
In [90]: result
Out[90]:
A B C D
K0 A0 B0 C0 D0
K2 A2 B2 C2 D2 DataFrame.join()接受一个可选的on参数，可以是要对齐的列或多个列名。
In [91]: left = pd.DataFrame(
....: {
....: "A": ["A0", "A1", "A2", "A3"],
....: "B": ["B0", "B1", "B2", "B3"],
....: "key": ["K0", "K1", "K0", "K1"],
....: }
....: )
....:
In [92]: right = pd.DataFrame({"C": ["C0", "C1"], "D": ["D0", "D1"]}, index=["K0", "K1"])
In [93]: result = left.join(right, on="key")
In [94]: result
Out[94]:
A B key C D
0 A0 B0 K0 C0 D0
1 A1 B1 K1 C1 D1
2 A2 B2 K0 C0 D0
3 A3 B3 K1 C1 D1 In [95]: result = pd.merge(
....: left, right, left_on="key", right_index=True, how="left", sort=False
....: )
....:
In [96]: result
Out[96]:
A B key C D
0 A0 B0 K0 C0 D0
1 A1 B1 K1 C1 D1
2 A2 B2 K0 C0 D0
3 A3 B3 K1 C1 D1 要在多个键上连接，传递的DataFrame必须具有MultiIndex:
In [97]: left = pd.DataFrame(
....: {
....: "A": ["A0", "A1", "A2", "A3"],
....: "B": ["B0", "B1", "B2", "B3"],
....: "key1": ["K0", "K0", "K1", "K2"],
....: "key2": ["K0", "K1", "K0", "K1"],
....: }
....: )
....:
In [98]: index = pd.MultiIndex.from_tuples(
....: [("K0", "K0"), ("K1", "K0"), ("K2", "K0"), ("K2", "K1")]
....: )
....:
In [99]: right = pd.DataFrame(
....: {"C": ["C0", "C1", "C2", "C3"], "D": ["D0", "D1", "D2", "D3"]}, index=index
....: )
....:
In [100]: result = left.join(right, on=["key1", "key2"])
In [101]: result
Out[101]:
A B key1 key2 C D
0 A0 B0 K0 K0 C0 D0
1 A1 B1 K0 K1 NaN NaN
2 A2 B2 K1 K0 C1 D1
3 A3 B3 K2 K1 C3 D3 DataFrame.join的默认行为是执行左连接，仅使用调用DataFrame中找到的键。其他连接类型可以通过how指定。
In [102]: result = left.join(right, on=["key1", "key2"], how="inner")
In [103]: result
Out[103]:
A B key1 key2 C D
0 A0 B0 K0 K0 C0 D0
2 A2 B2 K1 K0 C1 D1
3 A3 B3 K2 K1 C3 D3 ### 将单个索引连接到多重索引
你可以将一个具有Index的DataFrame与具有MultiIndex的DataFrame在一个级别上连接。Index的name将与MultiIndex的级别名称匹配。
In [104]: left = pd.DataFrame(
.....: {"A": ["A0", "A1", "A2"], "B": ["B0", "B1", "B2"]},
.....: index=pd.Index(["K0", "K1", "K2"], name="key"),
.....: )
.....:
In [105]: index = pd.MultiIndex.from_tuples(
.....: [("K0", "Y0"), ("K1", "Y1"), ("K2", "Y2"), ("K2", "Y3")],
.....: names=["key", "Y"],
.....: )
.....:
In [106]: right = pd.DataFrame(
.....: {"C": ["C0", "C1", "C2", "C3"], "D": ["D0", "D1", "D2", "D3"]},
.....: index=index,
.....: )
.....:
In [107]: result = left.join(right, how="inner")
In [108]: result
Out[108]:
A B C D
key Y
K0 Y0 A0 B0 C0 D0
K1 Y1 A1 B1 C1 D1
K2 Y2 A2 B2 C2 D2
Y3 A2 B2 C3 D3 ### 与两个MultiIndex连接
输入参数的MultiIndex必须完全用于连接，并且是左参数中索引的子集。
In [109]: leftindex = pd.MultiIndex.from_product(
.....: [list("abc"), list("xy"), [1, 2]], names=["abc", "xy", "num"]
.....: )
.....:
In [110]: left = pd.DataFrame({"v1": range(12)}, index=leftindex)
In [111]: left
Out[111]:
v1
abc xy num
a x 1 0
2 1
y 1 2
2 3
b x 1 4
2 5
y 1 6
2 7
c x 1 8
2 9
y 1 10
2 11
In [112]: rightindex = pd.MultiIndex.from_product(
.....: [list("abc"), list("xy")], names=["abc", "xy"]
.....: )
.....:
In [113]: right = pd.DataFrame({"v2": [100 * i for i in range(1, 7)]}, index=rightindex)
In [114]: right
Out[114]:
v2
abc xy
a x 100
y 200
b x 300
y 400
c x 500
y 600
In [115]: left.join(right, on=["abc", "xy"], how="inner")
Out[115]:
v1 v2
abc xy num
a x 1 0 100
2 1 100
y 1 2 200
2 3 200
b x 1 4 300
2 5 300
y 1 6 400
2 7 400
c x 1 8 500
2 9 500
y 1 10 600
2 11 600 In [116]: leftindex = pd.MultiIndex.from_tuples(
.....: [("K0", "X0"), ("K0", "X1"), ("K1", "X2")], names=["key", "X"]
.....: )
.....:
In [117]: left = pd.DataFrame(
.....: {"A": ["A0", "A1", "A2"], "B": ["B0", "B1", "B2"]}, index=leftindex
.....: )
.....:
In [118]: rightindex = pd.MultiIndex.from_tuples(
.....: [("K0", "Y0"), ("K1", "Y1"), ("K2", "Y2"), ("K2", "Y3")], names=["key", "Y"]
.....: )
.....:
In [119]: right = pd.DataFrame(
.....: {"C": ["C0", "C1", "C2", "C3"], "D": ["D0", "D1", "D2", "D3"]}, index=rightindex
.....: )
.....:
In [120]: result = pd.merge(
.....: left.reset_index(), right.reset_index(), on=["key"], how="inner"
.....: ).set_index(["key", "X", "Y"])
.....:
In [121]: result
Out[121]:
A B C D
key X Y
K0 X0 Y0 A0 B0 C0 D0
X1 Y0 A1 B1 C0 D0
K1 X2 Y1 A2 B2 C1 D1 ### 在列和索引级别的组合上合并
作为on、left_on和right_on参数传递的字符串可以引用列名或索引级别名。这使得在不重置索引的情况下，可以在索引级别和列的组合上合并DataFrame实例。
In [122]: left_index = pd.Index(["K0", "K0", "K1", "K2"], name="key1")
In [123]: left = pd.DataFrame(
.....: {
.....: "A": ["A0", "A1", "A2", "A3"],
.....: "B": ["B0", "B1", "B2", "B3"],
.....: "key2": ["K0", "K1", "K0", "K1"],
.....: },
.....: index=left_index,
.....: )
.....:
In [124]: right_index = pd.Index(["K0", "K1", "K2", "K2"], name="key1")
In [125]: right = pd.DataFrame(
.....: {
.....: "C": ["C0", "C1", "C2", "C3"],
.....: "D": ["D0", "D1", "D2", "D3"],
.....: "key2": ["K0", "K0", "K0", "K1"],
.....: },
.....: index=right_index,
.....: )
.....:
In [126]: result = left.merge(right, on=["key1", "key2"])
In [127]: result
Out[127]:
A B key2 C D
key1
K0 A0 B0 K0 C0 D0
K1 A2 B2 K0 C1 D1
K2 A3 B3 K1 C3 D3 注意
当DataFrame在匹配两个参数中的索引级别的字符串上进行连接时，索引级别将保留为结果DataFrame中的索引级别。
注意
当只使用MultiIndex的一些级别连接DataFrame时，多余的级别将从结果连接中删除。要保留这些级别，请在连接之前对这些级别名称使用DataFrame.reset_index()将这些级别移动到列中。 ### 连接多个DataFrame
也可以将DataFrame的列表或元组传递给join()，以在它们的索引上将它们连接在一起。
In [128]: right2 = pd.DataFrame({"v": [7, 8, 9]}, index=["K1", "K1", "K2"])
In [129]: result = left.join([right, right2]) ### DataFrame.combine_first()
DataFrame.combine_first() 用另一个DataFrame中的非缺失值更新一个DataFrame中的缺失值，位置对应。
In [130]: df1 = pd.DataFrame(
.....: [[np.nan, 3.0, 5.0], [-4.6, np.nan, np.nan], [np.nan, 7.0, np.nan]]
.....: )
.....:
In [131]: df2 = pd.DataFrame([[-42.6, np.nan, -8.2], [-5.0, 1.6, 4]], index=[1, 2])
In [132]: result = df1.combine_first(df2)
In [133]: result
Out[133]:
0 1 2
0 NaN 3.0 5.0
1 -4.6 NaN -8.2
2 -5.0 7.0 4.0 ## merge_ordered()
merge_ordered() 将有序数据（如数值或时间序列数据）与可选的使用fill_method填充缺失数据合并。
In [134]: left = pd.DataFrame(
.....: {"k": ["K0", "K1", "K1", "K2"], "lv": [1, 2, 3, 4], "s": ["a", "b", "c", "d"]}
.....: )
.....:
In [135]: right = pd.DataFrame({"k": ["K1", "K2", "K4"], "rv": [1, 2, 3]})
In [136]: pd.merge_ordered(left, right, fill_method="ffill", left_by="s")
Out[136]:
k lv s rv
0 K0 1.0 a NaN
1 K1 1.0 a 1.0
2 K2 1.0 a 2.0
3 K4 1.0 a 3.0
4 K1 2.0 b 1.0
5 K2 2.0 b 2.0
6 K4 2.0 b 3.0
7 K1 3.0 c 1.0
8 K2 3.0 c 2.0
9 K4 3.0 c 3.0
10 K1 NaN d 1.0
11 K2 4.0 d 2.0
12 K4 4.0 d 3.0
```## `merge_asof()`
`merge_asof()` 类似于有序的左连接，不同之处在于匹配的是最近的键而不是相等的键。对于`left` `DataFrame`中的每一行，选择`right` `DataFrame`中最后一行，其中`on`键小于左侧的键。两个`DataFrame`必须按键排序。
可选地，`merge_asof()`可以通过在`by`键上匹配来执行分组合并，同时在`on`键上找到最近的匹配。
```py
In [137]: trades = pd.DataFrame(
.....: {
.....: "time": pd.to_datetime(
.....: [
.....: "20160525 13:30:00.023",
.....: "20160525 13:30:00.038",
.....: "20160525 13:30:00.048",
.....: "20160525 13:30:00.048",
.....: "20160525 13:30:00.048",
.....: ]
.....: ),
.....: "ticker": ["MSFT", "MSFT", "GOOG", "GOOG", "AAPL"],
.....: "price": [51.95, 51.95, 720.77, 720.92, 98.00],
.....: "quantity": [75, 155, 100, 100, 100],
.....: },
.....: columns=["time", "ticker", "price", "quantity"],
.....: )
.....:
In [138]: quotes = pd.DataFrame(
.....: {
.....: "time": pd.to_datetime(
.....: [
.....: "20160525 13:30:00.023",
.....: "20160525 13:30:00.023",
.....: "20160525 13:30:00.030",
.....: "20160525 13:30:00.041",
.....: "20160525 13:30:00.048",
.....: "20160525 13:30:00.049",
.....: "20160525 13:30:00.072",
.....: "20160525 13:30:00.075",
.....: ]
.....: ),
.....: "ticker": ["GOOG", "MSFT", "MSFT", "MSFT", "GOOG", "AAPL", "GOOG", "MSFT"],
.....: "bid": [720.50, 51.95, 51.97, 51.99, 720.50, 97.99, 720.50, 52.01],
.....: "ask": [720.93, 51.96, 51.98, 52.00, 720.93, 98.01, 720.88, 52.03],
.....: },
.....: columns=["time", "ticker", "bid", "ask"],
.....: )
.....:
In [139]: trades
Out[139]:
time ticker price quantity
0 2016-05-25 13:30:00.023 MSFT 51.95 75
1 2016-05-25 13:30:00.038 MSFT 51.95 155
2 2016-05-25 13:30:00.048 GOOG 720.77 100
3 2016-05-25 13:30:00.048 GOOG 720.92 100
4 2016-05-25 13:30:00.048 AAPL 98.00 100
In [140]: quotes
Out[140]:
time ticker bid ask
0 2016-05-25 13:30:00.023 GOOG 720.50 720.93
1 2016-05-25 13:30:00.023 MSFT 51.95 51.96
2 2016-05-25 13:30:00.030 MSFT 51.97 51.98
3 2016-05-25 13:30:00.041 MSFT 51.99 52.00
4 2016-05-25 13:30:00.048 GOOG 720.50 720.93
5 2016-05-25 13:30:00.049 AAPL 97.99 98.01
6 2016-05-25 13:30:00.072 GOOG 720.50 720.88
7 2016-05-25 13:30:00.075 MSFT 52.01 52.03
In [141]: pd.merge_asof(trades, quotes, on="time", by="ticker")
Out[141]:
time ticker price quantity bid ask
0 2016-05-25 13:30:00.023 MSFT 51.95 75 51.95 51.96
1 2016-05-25 13:30:00.038 MSFT 51.95 155 51.97 51.98
2 2016-05-25 13:30:00.048 GOOG 720.77 100 720.50 720.93
3 2016-05-25 13:30:00.048 GOOG 720.92 100 720.50 720.93
4 2016-05-25 13:30:00.048 AAPL 98.00 100 NaN NaN merge_asof() 在报价时间和交易时间之间的2ms内合并。
In [142]: pd.merge_asof(trades, quotes, on="time", by="ticker", tolerance=pd.Timedelta("2ms"))
Out[142]:
time ticker price quantity bid ask
0 2016-05-25 13:30:00.023 MSFT 51.95 75 51.95 51.96
1 2016-05-25 13:30:00.038 MSFT 51.95 155 NaN NaN
2 2016-05-25 13:30:00.048 GOOG 720.77 100 720.50 720.93
3 2016-05-25 13:30:00.048 GOOG 720.92 100 720.50 720.93
4 2016-05-25 13:30:00.048 AAPL 98.00 100 NaN NaN merge_asof() 在报价时间和交易时间之间的10ms内合并，并排除时间上的精确匹配。请注意，尽管我们排除了精确匹配（报价），但之前的报价确实传播到那个时间点。
In [143]: pd.merge_asof(
.....: trades,
.....: quotes,
.....: on="time",
.....: by="ticker",
.....: tolerance=pd.Timedelta("10ms"),
.....: allow_exact_matches=False,
.....: )
.....:
Out[143]:
time ticker price quantity bid ask
0 2016-05-25 13:30:00.023 MSFT 51.95 75 NaN NaN
1 2016-05-25 13:30:00.038 MSFT 51.95 155 51.97 51.98
2 2016-05-25 13:30:00.048 GOOG 720.77 100 NaN NaN
3 2016-05-25 13:30:00.048 GOOG 720.92 100 NaN NaN
4 2016-05-25 13:30:00.048 AAPL 98.00 100 NaN NaN
```## `compare()`
`Series.compare()` 和 `DataFrame.compare()` 方法允许您比较两个分别是`DataFrame`或`Series`的对象，并总结它们的差异。
```py
In [144]: df = pd.DataFrame(
.....: {
.....: "col1": ["a", "a", "b", "b", "a"],
.....: "col2": [1.0, 2.0, 3.0, np.nan, 5.0],
.....: "col3": [1.0, 2.0, 3.0, 4.0, 5.0],
.....: },
.....: columns=["col1", "col2", "col3"],
.....: )
.....:
In [145]: df
Out[145]:
col1 col2 col3
0 a 1.0 1.0
1 a 2.0 2.0
2 b 3.0 3.0
3 b NaN 4.0
4 a 5.0 5.0
In [146]: df2 = df.copy()
In [147]: df2.loc[0, "col1"] = "c"
In [148]: df2.loc[2, "col3"] = 4.0
In [149]: df2
Out[149]:
col1 col2 col3
0 c 1.0 1.0
1 a 2.0 2.0
2 b 3.0 4.0
3 b NaN 4.0
4 a 5.0 5.0
In [150]: df.compare(df2)
Out[150]:
col1 col3
self other self other
0 a c NaN NaN
2 NaN NaN 3.0 4.0 默认情况下，如果两个对应的值相等，它们将显示为NaN。此外，如果整行/列中的所有值都相等，则该行/列将从结果中省略。剩余的差异将对齐在列上。
在行上堆叠差异。
In [151]: df.compare(df2, align_axis=0)
Out[151]:
col1 col3
0 self a NaN
other c NaN
2 self NaN 3.0
other NaN 4.0 保持所有原始行和列，使用keep_shape=True。
In [152]: df.compare(df2, keep_shape=True)
Out[152]:
col1 col2 col3
self other self other self other
0 a c NaN NaN NaN NaN
1 NaN NaN NaN NaN NaN NaN
2 NaN NaN NaN NaN 3.0 4.0
3 NaN NaN NaN NaN NaN NaN
4 NaN NaN NaN NaN NaN NaN 保留所有原始值，即使它们相等。
In [153]: df.compare(df2, keep_shape=True, keep_equal=True)
Out[153]:
col1 col2 col3
self other self other self other
0 a c 1.0 1.0 1.0 1.0
1 a a 2.0 2.0 2.0 2.0
2 b b 3.0 3.0 3.0 4.0
3 b b NaN NaN 4.0 4.0
4 a a 5.0 5.0 5.0 5.0
```## `concat()`
`concat()`函数沿着一个轴连接任意数量的`Series`或`DataFrame`对象，同时在其他轴上执行可选的集合逻辑（并集或交集）索引。像`numpy.concatenate`一样，`concat()`接受同类型对象的列表或字典，并将它们连接起来。
```py
In [1]: df1 = pd.DataFrame(
...: {
...: "A": ["A0", "A1", "A2", "A3"],
...: "B": ["B0", "B1", "B2", "B3"],
...: "C": ["C0", "C1", "C2", "C3"],
...: "D": ["D0", "D1", "D2", "D3"],
...: },
...: index=[0, 1, 2, 3],
...: )
...:
In [2]: df2 = pd.DataFrame(
...: {
...: "A": ["A4", "A5", "A6", "A7"],
...: "B": ["B4", "B5", "B6", "B7"],
...: "C": ["C4", "C5", "C6", "C7"],
...: "D": ["D4", "D5", "D6", "D7"],
...: },
...: index=[4, 5, 6, 7],
...: )
...:
In [3]: df3 = pd.DataFrame(
...: {
...: "A": ["A8", "A9", "A10", "A11"],
...: "B": ["B8", "B9", "B10", "B11"],
...: "C": ["C8", "C9", "C10", "C11"],
...: "D": ["D8", "D9", "D10", "D11"],
...: },
...: index=[8, 9, 10, 11],
...: )
...:
In [4]: frames = [df1, df2, df3]
In [5]: result = pd.concat(frames)
In [6]: result
Out[6]:
A B C D
0 A0 B0 C0 D0
1 A1 B1 C1 D1
2 A2 B2 C2 D2
3 A3 B3 C3 D3
4 A4 B4 C4 D4
5 A5 B5 C5 D5
6 A6 B6 C6 D6
7 A7 B7 C7 D7
8 A8 B8 C8 D8
9 A9 B9 C9 D9
10 A10 B10 C10 D10
11 A11 B11 C11 D11 注意
concat()会对数据进行完全复制，并且反复使用concat()可能会创建不必要的副本。在使用concat()之前，先将所有DataFrame或Series对象收集到一个列表中。
frames = [process_your_file(f) for f in files]
result = pd.concat(frames) 注意
当连接具有命名轴的DataFrame时，pandas 会尽可能保留这些索引/列名称。在所有输入共享一个公共名称的情况下，该名称将分配给结果。当输入名称不完全一致时，结果将不具有名称。对于MultiIndex也是如此，但逻辑是逐级别分别应用的。
结果轴的连接逻辑
join关键字指定如何处理第一个DataFrame中不存在的轴值。
join='outer'取所有轴值的并集
In [7]: df4 = pd.DataFrame(
...: {
...: "B": ["B2", "B3", "B6", "B7"],
...: "D": ["D2", "D3", "D6", "D7"],
...: "F": ["F2", "F3", "F6", "F7"],
...: },
...: index=[2, 3, 6, 7],
...: )
...:
In [8]: result = pd.concat([df1, df4], axis=1)
In [9]: result
Out[9]:
A B C D B D F
0 A0 B0 C0 D0 NaN NaN NaN
1 A1 B1 C1 D1 NaN NaN NaN
2 A2 B2 C2 D2 B2 D2 F2
3 A3 B3 C3 D3 B3 D3 F3
6 NaN NaN NaN NaN B6 D6 F6
7 NaN NaN NaN NaN B7 D7 F7 join='inner'取轴值的交集
In [10]: result = pd.concat([df1, df4], axis=1, join="inner")
In [11]: result
Out[11]:
A B C D B D F
2 A2 B2 C2 D2 B2 D2 F2
3 A3 B3 C3 D3 B3 D3 F3 要使用原始DataFrame的精确索引执行有效的“左连接”，结果可以重新索引。
In [12]: result = pd.concat([df1, df4], axis=1).reindex(df1.index)
In [13]: result
Out[13]:
A B C D B D F
0 A0 B0 C0 D0 NaN NaN NaN
1 A1 B1 C1 D1 NaN NaN NaN
2 A2 B2 C2 D2 B2 D2 F2
3 A3 B3 C3 D3 B3 D3 F3 ### 在连接轴上忽略索引
对于没有有意义索引的DataFrame对象，ignore_index会忽略重叠的索引。
In [14]: result = pd.concat([df1, df4], ignore_index=True, sort=False)
In [15]: result
Out[15]:
A B C D F
0 A0 B0 C0 D0 NaN
1 A1 B1 C1 D1 NaN
2 A2 B2 C2 D2 NaN
3 A3 B3 C3 D3 NaN
4 NaN B2 NaN D2 F2
5 NaN B3 NaN D3 F3
6 NaN B6 NaN D6 F6
7 NaN B7 NaN D7 F7 ### 将Series和DataFrame连接在一起
你可以连接一组Series和DataFrame对象。Series将被转换为DataFrame，列名为Series的名称。
In [16]: s1 = pd.Series(["X0", "X1", "X2", "X3"], name="X")
In [17]: result = pd.concat([df1, s1], axis=1)
In [18]: result
Out[18]:
A B C D X
0 A0 B0 C0 D0 X0
1 A1 B1 C1 D1 X1
2 A2 B2 C2 D2 X2
3 A3 B3 C3 D3 X3 未命名的Series将按顺序编号。
In [19]: s2 = pd.Series(["_0", "_1", "_2", "_3"])
In [20]: result = pd.concat([df1, s2, s2, s2], axis=1)
In [21]: result
Out[21]:
A B C D 0 1 2
0 A0 B0 C0 D0 _0 _0 _0
1 A1 B1 C1 D1 _1 _1 _1
2 A2 B2 C2 D2 _2 _2 _2
3 A3 B3 C3 D3 _3 _3 _3 ignore_index=True将删除所有名称引用。
In [22]: result = pd.concat([df1, s1], axis=1, ignore_index=True)
In [23]: result
Out[23]:
0 1 2 3 4
0 A0 B0 C0 D0 X0
1 A1 B1 C1 D1 X1
2 A2 B2 C2 D2 X2
3 A3 B3 C3 D3 X3 结果的keys
keys参数会为结果的索引或列添加另一个轴级别（创建一个MultiIndex），将特定键与每个原始DataFrame关联起来。
In [24]: result = pd.concat(frames, keys=["x", "y", "z"])
In [25]: result
Out[25]:
A B C D
x 0 A0 B0 C0 D0
1 A1 B1 C1 D1
2 A2 B2 C2 D2
3 A3 B3 C3 D3
y 4 A4 B4 C4 D4
5 A5 B5 C5 D5
6 A6 B6 C6 D6
7 A7 B7 C7 D7
z 8 A8 B8 C8 D8
9 A9 B9 C9 D9
10 A10 B10 C10 D10
11 A11 B11 C11 D11
In [26]: result.loc["y"]
Out[26]:
A B C D
4 A4 B4 C4 D4
5 A5 B5 C5 D5
6 A6 B6 C6 D6
7 A7 B7 C7 D7 keys参数可以在基于现有Series创建新DataFrame时覆盖列名。
In [27]: s3 = pd.Series([0, 1, 2, 3], name="foo")
In [28]: s4 = pd.Series([0, 1, 2, 3])
In [29]: s5 = pd.Series([0, 1, 4, 5])
In [30]: pd.concat([s3, s4, s5], axis=1)
Out[30]:
foo 0 1
0 0 0 0
1 1 1 1
2 2 2 4
3 3 3 5
In [31]: pd.concat([s3, s4, s5], axis=1, keys=["red", "blue", "yellow"])
Out[31]:
red blue yellow
0 0 0 0
1 1 1 1
2 2 2 4
3 3 3 5 你也可以向concat()传递一个字典，此时字典键将用于keys参数，除非指定了其他keys参数：
In [32]: pieces = {"x": df1, "y": df2, "z": df3}
In [33]: result = pd.concat(pieces)
In [34]: result
Out[34]:
A B C D
x 0 A0 B0 C0 D0
1 A1 B1 C1 D1
2 A2 B2 C2 D2
3 A3 B3 C3 D3
y 4 A4 B4 C4 D4
5 A5 B5 C5 D5
6 A6 B6 C6 D6
7 A7 B7 C7 D7
z 8 A8 B8 C8 D8
9 A9 B9 C9 D9
10 A10 B10 C10 D10
11 A11 B11 C11 D11 In [35]: result = pd.concat(pieces, keys=["z", "y"])
In [36]: result
Out[36]:
A B C D
z 8 A8 B8 C8 D8
9 A9 B9 C9 D9
10 A10 B10 C10 D10
11 A11 B11 C11 D11
y 4 A4 B4 C4 D4
5 A5 B5 C5 D5
6 A6 B6 C6 D6
7 A7 B7 C7 D7 创建的MultiIndex具有从传递的键和DataFrame片段的索引构建的级别：
In [37]: result.index.levels
Out[37]: FrozenList([['z', 'y'], [4, 5, 6, 7, 8, 9, 10, 11]]) levels参数允许指定与keys关联的结果级别
In [38]: result = pd.concat(
....: pieces, keys=["x", "y", "z"], levels=[["z", "y", "x", "w"]], names=["group_key"]
....: )
....:
In [39]: result
Out[39]:
A B C D
group_key
x 0 A0 B0 C0 D0
1 A1 B1 C1 D1
2 A2 B2 C2 D2
3 A3 B3 C3 D3
y 4 A4 B4 C4 D4
5 A5 B5 C5 D5
6 A6 B6 C6 D6
7 A7 B7 C7 D7
z 8 A8 B8 C8 D8
9 A9 B9 C9 D9
10 A10 B10 C10 D10
11 A11 B11 C11 D11 In [40]: result.index.levels
Out[40]: FrozenList([['z', 'y', 'x', 'w'], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11]]) 将行附加到DataFrame
如果您有一个想要附加为单行到DataFrame的Series，您可以将行转换为DataFrame并使用concat()
In [41]: s2 = pd.Series(["X0", "X1", "X2", "X3"], index=["A", "B", "C", "D"])
In [42]: result = pd.concat([df1, s2.to_frame().T], ignore_index=True)
In [43]: result
Out[43]:
A B C D
0 A0 B0 C0 D0
1 A1 B1 C1 D1
2 A2 B2 C2 D2
3 A3 B3 C3 D3
4 X0 X1 X2 X3 结果轴的连接逻辑
join 关键字指定如何处理第一个DataFrame中不存在的轴值。
join='outer'取所有轴值的并集
In [7]: df4 = pd.DataFrame(
...: {
...: "B": ["B2", "B3", "B6", "B7"],
...: "D": ["D2", "D3", "D6", "D7"],
...: "F": ["F2", "F3", "F6", "F7"],
...: },
...: index=[2, 3, 6, 7],
...: )
...:
In [8]: result = pd.concat([df1, df4], axis=1)
In [9]: result
Out[9]:
A B C D B D F
0 A0 B0 C0 D0 NaN NaN NaN
1 A1 B1 C1 D1 NaN NaN NaN
2 A2 B2 C2 D2 B2 D2 F2
3 A3 B3 C3 D3 B3 D3 F3
6 NaN NaN NaN NaN B6 D6 F6
7 NaN NaN NaN NaN B7 D7 F7 join='inner'取轴值的交集
In [10]: result = pd.concat([df1, df4], axis=1, join="inner")
In [11]: result
Out[11]:
A B C D B D F
2 A2 B2 C2 D2 B2 D2 F2
3 A3 B3 C3 D3 B3 D3 F3 要使用原始DataFrame的确切索引执行有效的“左”连接，结果可以重新索引。
In [12]: result = pd.concat([df1, df4], axis=1).reindex(df1.index)
In [13]: result
Out[13]:
A B C D B D F
0 A0 B0 C0 D0 NaN NaN NaN
1 A1 B1 C1 D1 NaN NaN NaN
2 A2 B2 C2 D2 B2 D2 F2
3 A3 B3 C3 D3 B3 D3 F3 ### 在连接轴上忽略索引
对于没有有意义索引的DataFrame对象，ignore_index会忽略重叠的索引。
In [14]: result = pd.concat([df1, df4], ignore_index=True, sort=False)
In [15]: result
Out[15]:
A B C D F
0 A0 B0 C0 D0 NaN
1 A1 B1 C1 D1 NaN
2 A2 B2 C2 D2 NaN
3 A3 B3 C3 D3 NaN
4 NaN B2 NaN D2 F2
5 NaN B3 NaN D3 F3
6 NaN B6 NaN D6 F6
7 NaN B7 NaN D7 F7 ### 将Series和DataFrame连接在一起
您可以连接一组Series和DataFrame对象。Series将转换为具有列名的DataFrame，列名为Series的名称。
In [16]: s1 = pd.Series(["X0", "X1", "X2", "X3"], name="X")
In [17]: result = pd.concat([df1, s1], axis=1)
In [18]: result
Out[18]:
A B C D X
0 A0 B0 C0 D0 X0
1 A1 B1 C1 D1 X1
2 A2 B2 C2 D2 X2
3 A3 B3 C3 D3 X3 未命名的Series将按顺序编号。
In [19]: s2 = pd.Series(["_0", "_1", "_2", "_3"])
In [20]: result = pd.concat([df1, s2, s2, s2], axis=1)
In [21]: result
Out[21]:
A B C D 0 1 2
0 A0 B0 C0 D0 _0 _0 _0
1 A1 B1 C1 D1 _1 _1 _1
2 A2 B2 C2 D2 _2 _2 _2
3 A3 B3 C3 D3 _3 _3 _3 ignore_index=True将删除所有名称引用。
In [22]: result = pd.concat([df1, s1], axis=1, ignore_index=True)
In [23]: result
Out[23]:
0 1 2 3 4
0 A0 B0 C0 D0 X0
1 A1 B1 C1 D1 X1
2 A2 B2 C2 D2 X2
3 A3 B3 C3 D3 X3 结果keys
keys参数将另一个轴级别添加到结果索引或列（创建一个MultiIndex），将特定键与每个原始DataFrame关联起来。
In [24]: result = pd.concat(frames, keys=["x", "y", "z"])
In [25]: result
Out[25]:
A B C D
x 0 A0 B0 C0 D0
1 A1 B1 C1 D1
2 A2 B2 C2 D2
3 A3 B3 C3 D3
y 4 A4 B4 C4 D4
5 A5 B5 C5 D5
6 A6 B6 C6 D6
7 A7 B7 C7 D7
z 8 A8 B8 C8 D8
9 A9 B9 C9 D9
10 A10 B10 C10 D10
11 A11 B11 C11 D11
In [26]: result.loc["y"]
Out[26]:
A B C D
4 A4 B4 C4 D4
5 A5 B5 C5 D5
6 A6 B6 C6 D6
7 A7 B7 C7 D7 keys参数可以在基于现有Series创建新DataFrame时覆盖列名。
In [27]: s3 = pd.Series([0, 1, 2, 3], name="foo")
In [28]: s4 = pd.Series([0, 1, 2, 3])
In [29]: s5 = pd.Series([0, 1, 4, 5])
In [30]: pd.concat([s3, s4, s5], axis=1)
Out[30]:
foo 0 1
0 0 0 0
1 1 1 1
2 2 2 4
3 3 3 5
In [31]: pd.concat([s3, s4, s5], axis=1, keys=["red", "blue", "yellow"])
Out[31]:
red blue yellow
0 0 0 0
1 1 1 1
2 2 2 4
3 3 3 5 你也可以将字典传递给concat()，在这种情况下，除非指定了其他keys参数，否则字典键将用于keys参数：
In [32]: pieces = {"x": df1, "y": df2, "z": df3}
In [33]: result = pd.concat(pieces)
In [34]: result
Out[34]:
A B C D
x 0 A0 B0 C0 D0
1 A1 B1 C1 D1
2 A2 B2 C2 D2
3 A3 B3 C3 D3
y 4 A4 B4 C4 D4
5 A5 B5 C5 D5
6 A6 B6 C6 D6
7 A7 B7 C7 D7
z 8 A8 B8 C8 D8
9 A9 B9 C9 D9
10 A10 B10 C10 D10
11 A11 B11 C11 D11 In [35]: result = pd.concat(pieces, keys=["z", "y"])
In [36]: result
Out[36]:
A B C D
z 8 A8 B8 C8 D8
9 A9 B9 C9 D9
10 A10 B10 C10 D10
11 A11 B11 C11 D11
y 4 A4 B4 C4 D4
5 A5 B5 C5 D5
6 A6 B6 C6 D6
7 A7 B7 C7 D7 创建的MultiIndex具有从传递的键和DataFrame片段的索引构造的级别：
In [37]: result.index.levels
Out[37]: FrozenList([['z', 'y'], [4, 5, 6, 7, 8, 9, 10, 11]]) levels参数允许指定与keys关联的结果级别
In [38]: result = pd.concat(
....: pieces, keys=["x", "y", "z"], levels=[["z", "y", "x", "w"]], names=["group_key"]
....: )
....:
In [39]: result
Out[39]:
A B C D
group_key
x 0 A0 B0 C0 D0
1 A1 B1 C1 D1
2 A2 B2 C2 D2
3 A3 B3 C3 D3
y 4 A4 B4 C4 D4
5 A5 B5 C5 D5
6 A6 B6 C6 D6
7 A7 B7 C7 D7
z 8 A8 B8 C8 D8
9 A9 B9 C9 D9
10 A10 B10 C10 D10
11 A11 B11 C11 D11 In [40]: result.index.levels
Out[40]: FrozenList([['z', 'y', 'x', 'w'], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11]]) 将行附加到DataFrame
如果你有一个要附加为单行到DataFrame的Series，你可以将该行转换为DataFrame并使用concat()
In [41]: s2 = pd.Series(["X0", "X1", "X2", "X3"], index=["A", "B", "C", "D"])
In [42]: result = pd.concat([df1, s2.to_frame().T], ignore_index=True)
In [43]: result
Out[43]:
A B C D
0 A0 B0 C0 D0
1 A1 B1 C1 D1
2 A2 B2 C2 D2
3 A3 B3 C3 D3
4 X0 X1 X2 X3 ## merge()
merge()执行类似于关系数据库（如 SQL）的连接操作。熟悉 SQL 但是对 pandas 新手的用户可以参考与 SQL 的比较。
合并类型
merge()实现常见的 SQL 风格连接操作。
- 一对一：在它们的索引上连接两个
DataFrame对象，这些索引必须包含唯一值。 - 多对一：将唯一索引与不同
DataFrame中的一个或多个列连接。 - 多对多：在列上进行列连接。
注意
当在列上进行列连接时，可能是多对多的连接，传递的DataFrame对象上的任何索引将被丢弃。
对于多对多连接，如果一个键组合在两个表中出现多次，则DataFrame将具有相关数据的笛卡尔积。
In [44]: left = pd.DataFrame(
....: {
....: "key": ["K0", "K1", "K2", "K3"],
....: "A": ["A0", "A1", "A2", "A3"],
....: "B": ["B0", "B1", "B2", "B3"],
....: }
....: )
....:
In [45]: right = pd.DataFrame(
....: {
....: "key": ["K0", "K1", "K2", "K3"],
....: "C": ["C0", "C1", "C2", "C3"],
....: "D": ["D0", "D1", "D2", "D3"],
....: }
....: )
....:
In [46]: result = pd.merge(left, right, on="key")
In [47]: result
Out[47]:
key A B C D
0 K0 A0 B0 C0 D0
1 K1 A1 B1 C1 D1
2 K2 A2 B2 C2 D2
3 K3 A3 B3 C3 D3 merge()的how参数指定了包含在结果表中的键。如果一个键组合不存在于左表或右表中，那么连接表中的值将为NA。以下是how选项及其 SQL 等效名称的摘要：
合并方法 | SQL 连接名称 | 描述 |
|---|---|---|
left | LEFT OUTER JOIN | 仅使用左侧框架的键 |
right | RIGHT OUTER JOIN | 仅使用右侧框架的键 |
outer | FULL OUTER JOIN | 使用两个框架的键的并集 |
inner | INNER JOIN | 使用两个框架键的交集 |
cross | CROSS JOIN | 创建两个框架行的笛卡尔积 |
In [48]: left = pd.DataFrame(
....: {
....: "key1": ["K0", "K0", "K1", "K2"],
....: "key2": ["K0", "K1", "K0", "K1"],
....: "A": ["A0", "A1", "A2", "A3"],
....: "B": ["B0", "B1", "B2", "B3"],
....: }
....: )
....:
In [49]: right = pd.DataFrame(
....: {
....: "key1": ["K0", "K1", "K1", "K2"],
....: "key2": ["K0", "K0", "K0", "K0"],
....: "C": ["C0", "C1", "C2", "C3"],
....: "D": ["D0", "D1", "D2", "D3"],
....: }
....: )
....:
In [50]: result = pd.merge(left, right, how="left", on=["key1", "key2"])
In [51]: result
Out[51]:
key1 key2 A B C D
0 K0 K0 A0 B0 C0 D0
1 K0 K1 A1 B1 NaN NaN
2 K1 K0 A2 B2 C1 D1
3 K1 K0 A2 B2 C2 D2
4 K2 K1 A3 B3 NaN NaN In [52]: result = pd.merge(left, right, how="right", on=["key1", "key2"])
In [53]: result
Out[53]:
key1 key2 A B C D
0 K0 K0 A0 B0 C0 D0
1 K1 K0 A2 B2 C1 D1
2 K1 K0 A2 B2 C2 D2
3 K2 K0 NaN NaN C3 D3 In [54]: result = pd.merge(left, right, how="outer", on=["key1", "key2"])
In [55]: result
Out[55]:
key1 key2 A B C D
0 K0 K0 A0 B0 C0 D0
1 K0 K1 A1 B1 NaN NaN
2 K1 K0 A2 B2 C1 D1
3 K1 K0 A2 B2 C2 D2
4 K2 K0 NaN NaN C3 D3
5 K2 K1 A3 B3 NaN NaN In [56]: result = pd.merge(left, right, how="inner", on=["key1", "key2"])
In [57]: result
Out[57]:
key1 key2 A B C D
0 K0 K0 A0 B0 C0 D0
1 K1 K0 A2 B2 C1 D1
2 K1 K0 A2 B2 C2 D2 In [58]: result = pd.merge(left, right, how="cross")
In [59]: result
Out[59]:
key1_x key2_x A B key1_y key2_y C D
0 K0 K0 A0 B0 K0 K0 C0 D0
1 K0 K0 A0 B0 K1 K0 C1 D1
2 K0 K0 A0 B0 K1 K0 C2 D2
3 K0 K0 A0 B0 K2 K0 C3 D3
4 K0 K1 A1 B1 K0 K0 C0 D0
.. ... ... .. .. ... ... .. ..
11 K1 K0 A2 B2 K2 K0 C3 D3
12 K2 K1 A3 B3 K0 K0 C0 D0
13 K2 K1 A3 B3 K1 K0 C1 D1
14 K2 K1 A3 B3 K1 K0 C2 D2
15 K2 K1 A3 B3 K2 K0 C3 D3
[16 rows x 8 columns] 如果MultiIndex的名称与DataFrame的列对应，则可以使用Series.reset_index()将Series转换为DataFrame，然后再进行合并。
In [60]: df = pd.DataFrame({"Let": ["A", "B", "C"], "Num": [1, 2, 3]})
In [61]: df
Out[61]:
Let Num
0 A 1
1 B 2
2 C 3
In [62]: ser = pd.Series(
....: ["a", "b", "c", "d", "e", "f"],
....: index=pd.MultiIndex.from_arrays(
....: [["A", "B", "C"] * 2, [1, 2, 3, 4, 5, 6]], names=["Let", "Num"]
....: ),
....: )
....:
In [63]: ser
Out[63]:
Let Num
A 1 a
B 2 b
C 3 c
A 4 d
B 5 e
C 6 f
dtype: object
In [64]: pd.merge(df, ser.reset_index(), on=["Let", "Num"])
Out[64]:
Let Num 0
0 A 1 a
1 B 2 b
2 C 3 c 在DataFrame中具有重复连接键执行外连接
In [65]: left = pd.DataFrame({"A": [1, 2], "B": [2, 2]})
In [66]: right = pd.DataFrame({"A": [4, 5, 6], "B": [2, 2, 2]})
In [67]: result = pd.merge(left, right, on="B", how="outer")
In [68]: result
Out[68]:
A_x B A_y
0 1 2 4
1 1 2 5
2 1 2 6
3 2 2 4
4 2 2 5
5 2 2 6 警告
在重复键上合并会显著增加结果的维度，并可能导致内存溢出。
合并键的唯一性
validate 参数检查合并键的唯一性。在合并操作之前检查键的唯一性，可以防止内存溢出和意外键重复。
In [69]: left = pd.DataFrame({"A": [1, 2], "B": [1, 2]})
In [70]: right = pd.DataFrame({"A": [4, 5, 6], "B": [2, 2, 2]})
In [71]: result = pd.merge(left, right, on="B", how="outer", validate="one_to_one")
---------------------------------------------------------------------------
MergeError Traceback (most recent call last)
Cell In[71], line 1
----> 1 result = pd.merge(left, right, on="B", how="outer", validate="one_to_one")
File ~/work/pandas/pandas/pandas/core/reshape/merge.py:170, in merge(left, right, how, on, left_on, right_on, left_index, right_index, sort, suffixes, copy, indicator, validate)
155 return _cross_merge(
156 left_df,
157 right_df,
(...)
167 copy=copy,
168 )
169 else:
--> 170 op = _MergeOperation(
171 left_df,
172 right_df,
173 how=how,
174 on=on,
175 left_on=left_on,
176 right_on=right_on,
177 left_index=left_index,
178 right_index=right_index,
179 sort=sort,
180 suffixes=suffixes,
181 indicator=indicator,
182 validate=validate,
183 )
184 return op.get_result(copy=copy)
File ~/work/pandas/pandas/pandas/core/reshape/merge.py:813, in _MergeOperation.__init__(self, left, right, how, on, left_on, right_on, left_index, right_index, sort, suffixes, indicator, validate)
809 # If argument passed to validate,
810 # check if columns specified as unique
811 # are in fact unique.
812 if validate is not None:
--> 813 self._validate_validate_kwd(validate)
File ~/work/pandas/pandas/pandas/core/reshape/merge.py:1657, in _MergeOperation._validate_validate_kwd(self, validate)
1653 raise MergeError(
1654 "Merge keys are not unique in left dataset; not a one-to-one merge"
1655 )
1656 if not right_unique:
-> 1657 raise MergeError(
1658 "Merge keys are not unique in right dataset; not a one-to-one merge"
1659 )
1661 elif validate in ["one_to_many", "1:m"]:
1662 if not left_unique:
MergeError: Merge keys are not unique in right dataset; not a one-to-one merge 如果用户知道右侧 DataFrame 中存在重复项，但希望确保左侧 DataFrame 中没有重复项，可以使用 validate='one_to_many' 参数，而不会引发异常。
In [72]: pd.merge(left, right, on="B", how="outer", validate="one_to_many")
Out[72]:
A_x B A_y
0 1 1 NaN
1 2 2 4.0
2 2 2 5.0
3 2 2 6.0
```### 合并结果指示器
`merge()` 接受参数 `indicator`。如果为 `True`，则将添加一个名为 `_merge` 的分类列到输出对象中，其值为：
> | 观察来源 | `_merge` 值 |
> | --- | --- |
> | 仅在 `'left'` 框架中的合并键 | `left_only` |
> | 仅在 `'right'` 框架中的合并键 | `right_only` |
> | 两个框架中的合并键 | `both` |
```py
In [73]: df1 = pd.DataFrame({"col1": [0, 1], "col_left": ["a", "b"]})
In [74]: df2 = pd.DataFrame({"col1": [1, 2, 2], "col_right": [2, 2, 2]})
In [75]: pd.merge(df1, df2, on="col1", how="outer", indicator=True)
Out[75]:
col1 col_left col_right _merge
0 0 a NaN left_only
1 1 b 2.0 both
2 2 NaN 2.0 right_only
3 2 NaN 2.0 right_only indicator 的字符串参数将用作指示器列的名称。
In [76]: pd.merge(df1, df2, on="col1", how="outer", indicator="indicator_column")
Out[76]:
col1 col_left col_right indicator_column
0 0 a NaN left_only
1 1 b 2.0 both
2 2 NaN 2.0 right_only
3 2 NaN 2.0 right_only 重叠值列
合并 suffixes 参数接受一个字符串列表元组，以附加到输入 DataFrame 中重叠列名称以消除结果列的歧义：
In [77]: left = pd.DataFrame({"k": ["K0", "K1", "K2"], "v": [1, 2, 3]})
In [78]: right = pd.DataFrame({"k": ["K0", "K0", "K3"], "v": [4, 5, 6]})
In [79]: result = pd.merge(left, right, on="k")
In [80]: result
Out[80]:
k v_x v_y
0 K0 1 4
1 K0 1 5 In [81]: result = pd.merge(left, right, on="k", suffixes=("_l", "_r"))
In [82]: result
Out[82]:
k v_l v_r
0 K0 1 4
1 K0 1 5 合并类型
merge() 实现常见的 SQL 风格连接操作。
- 一对一：在它们的索引上连接两个
DataFrame对象，这些对象必须包含唯一值。 - 多对一：将唯一索引与不同
DataFrame中的一个或多个列连接。 - 多对多：在列上连接列。
注意
在列上连接列时，可能是多对多连接，传递的 DataFrame 对象上的任何索引将被丢弃。
对于多对多连接，如果在两个表中一个键组合出现多次，DataFrame 将具有相关数据的笛卡尔积。
In [44]: left = pd.DataFrame(
....: {
....: "key": ["K0", "K1", "K2", "K3"],
....: "A": ["A0", "A1", "A2", "A3"],
....: "B": ["B0", "B1", "B2", "B3"],
....: }
....: )
....:
In [45]: right = pd.DataFrame(
....: {
....: "key": ["K0", "K1", "K2", "K3"],
....: "C": ["C0", "C1", "C2", "C3"],
....: "D": ["D0", "D1", "D2", "D3"],
....: }
....: )
....:
In [46]: result = pd.merge(left, right, on="key")
In [47]: result
Out[47]:
key A B C D
0 K0 A0 B0 C0 D0
1 K1 A1 B1 C1 D1
2 K2 A2 B2 C2 D2
3 K3 A3 B3 C3 D3 merge()的how参数指定了哪些键包含在结果表中。如果一个键组合在左表或右表中都不存在，则连接表中的值将为NA。以下是how选项及其 SQL 等效名称的摘要：
合并方法 | SQL 连接名称 | 描述 |
|---|---|---|
left | LEFT OUTER JOIN | 仅使用左侧框架的键 |
right | RIGHT OUTER JOIN | 仅使用右侧框架的键 |
outer | FULL OUTER JOIN | 使用两个框架的键的并集 |
inner | INNER JOIN | 使用两个框架的键的交集 |
cross | CROSS JOIN | 创建两个框架行的笛卡尔积 |
In [48]: left = pd.DataFrame(
....: {
....: "key1": ["K0", "K0", "K1", "K2"],
....: "key2": ["K0", "K1", "K0", "K1"],
....: "A": ["A0", "A1", "A2", "A3"],
....: "B": ["B0", "B1", "B2", "B3"],
....: }
....: )
....:
In [49]: right = pd.DataFrame(
....: {
....: "key1": ["K0", "K1", "K1", "K2"],
....: "key2": ["K0", "K0", "K0", "K0"],
....: "C": ["C0", "C1", "C2", "C3"],
....: "D": ["D0", "D1", "D2", "D3"],
....: }
....: )
....:
In [50]: result = pd.merge(left, right, how="left", on=["key1", "key2"])
In [51]: result
Out[51]:
key1 key2 A B C D
0 K0 K0 A0 B0 C0 D0
1 K0 K1 A1 B1 NaN NaN
2 K1 K0 A2 B2 C1 D1
3 K1 K0 A2 B2 C2 D2
4 K2 K1 A3 B3 NaN NaN In [52]: result = pd.merge(left, right, how="right", on=["key1", "key2"])
In [53]: result
Out[53]:
key1 key2 A B C D
0 K0 K0 A0 B0 C0 D0
1 K1 K0 A2 B2 C1 D1
2 K1 K0 A2 B2 C2 D2
3 K2 K0 NaN NaN C3 D3 In [54]: result = pd.merge(left, right, how="outer", on=["key1", "key2"])
In [55]: result
Out[55]:
key1 key2 A B C D
0 K0 K0 A0 B0 C0 D0
1 K0 K1 A1 B1 NaN NaN
2 K1 K0 A2 B2 C1 D1
3 K1 K0 A2 B2 C2 D2
4 K2 K0 NaN NaN C3 D3
5 K2 K1 A3 B3 NaN NaN In [56]: result = pd.merge(left, right, how="inner", on=["key1", "key2"])
In [57]: result
Out[57]:
key1 key2 A B C D
0 K0 K0 A0 B0 C0 D0
1 K1 K0 A2 B2 C1 D1
2 K1 K0 A2 B2 C2 D2 In [58]: result = pd.merge(left, right, how="cross")
In [59]: result
Out[59]:
key1_x key2_x A B key1_y key2_y C D
0 K0 K0 A0 B0 K0 K0 C0 D0
1 K0 K0 A0 B0 K1 K0 C1 D1
2 K0 K0 A0 B0 K1 K0 C2 D2
3 K0 K0 A0 B0 K2 K0 C3 D3
4 K0 K1 A1 B1 K0 K0 C0 D0
.. ... ... .. .. ... ... .. ..
11 K1 K0 A2 B2 K2 K0 C3 D3
12 K2 K1 A3 B3 K0 K0 C0 D0
13 K2 K1 A3 B3 K1 K0 C1 D1
14 K2 K1 A3 B3 K1 K0 C2 D2
15 K2 K1 A3 B3 K2 K0 C3 D3
[16 rows x 8 columns] 如果MultiIndex的名称与DataFrame中的列对应，则可以使用Series和DataFrame。在合并之前，使用Series.reset_index()将Series转换为DataFrame
In [60]: df = pd.DataFrame({"Let": ["A", "B", "C"], "Num": [1, 2, 3]})
In [61]: df
Out[61]:
Let Num
0 A 1
1 B 2
2 C 3
In [62]: ser = pd.Series(
....: ["a", "b", "c", "d", "e", "f"],
....: index=pd.MultiIndex.from_arrays(
....: [["A", "B", "C"] * 2, [1, 2, 3, 4, 5, 6]], names=["Let", "Num"]
....: ),
....: )
....:
In [63]: ser
Out[63]:
Let Num
A 1 a
B 2 b
C 3 c
A 4 d
B 5 e
C 6 f
dtype: object
In [64]: pd.merge(df, ser.reset_index(), on=["Let", "Num"])
Out[64]:
Let Num 0
0 A 1 a
1 B 2 b
2 C 3 c 在DataFrame中执行具有重复连接键的外部连接
In [65]: left = pd.DataFrame({"A": [1, 2], "B": [2, 2]})
In [66]: right = pd.DataFrame({"A": [4, 5, 6], "B": [2, 2, 2]})
In [67]: result = pd.merge(left, right, on="B", how="outer")
In [68]: result
Out[68]:
A_x B A_y
0 1 2 4
1 1 2 5
2 1 2 6
3 2 2 4
4 2 2 5
5 2 2 6 警告
在重复键上合并会显著增加结果的维度，并可能导致内存溢出。
合并键唯一性
validate参数检查合并键的唯一性。在合并操作之前检查键的唯一性，可以防止内存溢出和意外键重复。
In [69]: left = pd.DataFrame({"A": [1, 2], "B": [1, 2]})
In [70]: right = pd.DataFrame({"A": [4, 5, 6], "B": [2, 2, 2]})
In [71]: result = pd.merge(left, right, on="B", how="outer", validate="one_to_one")
---------------------------------------------------------------------------
MergeError Traceback (most recent call last)
Cell In[71], line 1
----> 1 result = pd.merge(left, right, on="B", how="outer", validate="one_to_one")
File ~/work/pandas/pandas/pandas/core/reshape/merge.py:170, in merge(left, right, how, on, left_on, right_on, left_index, right_index, sort, suffixes, copy, indicator, validate)
155 return _cross_merge(
156 left_df,
157 right_df,
(...)
167 copy=copy,
168 )
169 else:
--> 170 op = _MergeOperation(
171 left_df,
172 right_df,
173 how=how,
174 on=on,
175 left_on=left_on,
176 right_on=right_on,
177 left_index=left_index,
178 right_index=right_index,
179 sort=sort,
180 suffixes=suffixes,
181 indicator=indicator,
182 validate=validate,
183 )
184 return op.get_result(copy=copy)
File ~/work/pandas/pandas/pandas/core/reshape/merge.py:813, in _MergeOperation.__init__(self, left, right, how, on, left_on, right_on, left_index, right_index, sort, suffixes, indicator, validate)
809 # If argument passed to validate,
810 # check if columns specified as unique
811 # are in fact unique.
812 if validate is not None:
--> 813 self._validate_validate_kwd(validate)
File ~/work/pandas/pandas/pandas/core/reshape/merge.py:1657, in _MergeOperation._validate_validate_kwd(self, validate)
1653 raise MergeError(
1654 "Merge keys are not unique in left dataset; not a one-to-one merge"
1655 )
1656 if not right_unique:
-> 1657 raise MergeError(
1658 "Merge keys are not unique in right dataset; not a one-to-one merge"
1659 )
1661 elif validate in ["one_to_many", "1:m"]:
1662 if not left_unique:
MergeError: Merge keys are not unique in right dataset; not a one-to-one merge 如果用户意识到右侧 DataFrame 中存在重复项，但希望确保左侧 DataFrame 中没有重复项，则可以使用 validate='one_to_many' 参数，而不会引发异常。
In [72]: pd.merge(left, right, on="B", how="outer", validate="one_to_many")
Out[72]:
A_x B A_y
0 1 1 NaN
1 2 2 4.0
2 2 2 5.0
3 2 2 6.0 合并结果指示器
merge() 接受参数 indicator。如果为 True，则会向输出对象添加一个名为 _merge 的分类列，其取值为：
观察来源
_merge值仅在'left'框架中的合并键left_only仅在'right'框架中的合并键right_only两个框架中的合并键both
In [73]: df1 = pd.DataFrame({"col1": [0, 1], "col_left": ["a", "b"]})
In [74]: df2 = pd.DataFrame({"col1": [1, 2, 2], "col_right": [2, 2, 2]})
In [75]: pd.merge(df1, df2, on="col1", how="outer", indicator=True)
Out[75]:
col1 col_left col_right _merge
0 0 a NaN left_only
1 1 b 2.0 both
2 2 NaN 2.0 right_only
3 2 NaN 2.0 right_only indicator 的字符串参数将使用该值作为指示器列的名称。
In [76]: pd.merge(df1, df2, on="col1", how="outer", indicator="indicator_column")
Out[76]:
col1 col_left col_right indicator_column
0 0 a NaN left_only
1 1 b 2.0 both
2 2 NaN 2.0 right_only
3 2 NaN 2.0 right_only 重叠值列
合并 suffixes 参数接受一个字符串列表的元组，以附加到输入 DataFrame 中重叠列名称以消除结果列的歧义：
In [77]: left = pd.DataFrame({"k": ["K0", "K1", "K2"], "v": [1, 2, 3]})
In [78]: right = pd.DataFrame({"k": ["K0", "K0", "K3"], "v": [4, 5, 6]})
In [79]: result = pd.merge(left, right, on="k")
In [80]: result
Out[80]:
k v_x v_y
0 K0 1 4
1 K0 1 5 In [81]: result = pd.merge(left, right, on="k", suffixes=("_l", "_r"))
In [82]: result
Out[82]:
k v_l v_r
0 K0 1 4
1 K0 1 5 DataFrame.join()
DataFrame.join() 将多个、可能具有不同索引的 DataFrame 的列合并为单个结果 DataFrame。
In [83]: left = pd.DataFrame(
....: {"A": ["A0", "A1", "A2"], "B": ["B0", "B1", "B2"]}, index=["K0", "K1", "K2"]
....: )
....:
In [84]: right = pd.DataFrame(
....: {"C": ["C0", "C2", "C3"], "D": ["D0", "D2", "D3"]}, index=["K0", "K2", "K3"]
....: )
....:
In [85]: result = left.join(right)
In [86]: result
Out[86]:
A B C D
K0 A0 B0 C0 D0
K1 A1 B1 NaN NaN
K2 A2 B2 C2 D2 In [87]: result = left.join(right, how="outer")
In [88]: result
Out[88]:
A B C D
K0 A0 B0 C0 D0
K1 A1 B1 NaN NaN
K2 A2 B2 C2 D2
K3 NaN NaN C3 D3 In [89]: result = left.join(right, how="inner")
In [90]: result
Out[90]:
A B C D
K0 A0 B0 C0 D0
K2 A2 B2 C2 D2 DataFrame.join() 接受一个可选的 on 参数，该参数可以是要对齐的列或多个列名，传递的 DataFrame 将对齐。
In [91]: left = pd.DataFrame(
....: {
....: "A": ["A0", "A1", "A2", "A3"],
....: "B": ["B0", "B1", "B2", "B3"],
....: "key": ["K0", "K1", "K0", "K1"],
....: }
....: )
....:
In [92]: right = pd.DataFrame({"C": ["C0", "C1"], "D": ["D0", "D1"]}, index=["K0", "K1"])
In [93]: result = left.join(right, on="key")
In [94]: result
Out[94]:
A B key C D
0 A0 B0 K0 C0 D0
1 A1 B1 K1 C1 D1
2 A2 B2 K0 C0 D0
3 A3 B3 K1 C1 D1 In [95]: result = pd.merge(
....: left, right, left_on="key", right_index=True, how="left", sort=False
....: )
....:
In [96]: result
Out[96]:
A B key C D
0 A0 B0 K0 C0 D0
1 A1 B1 K1 C1 D1
2 A2 B2 K0 C0 D0
3 A3 B3 K1 C1 D1 要根据多个键进行连接，传递的 DataFrame 必须具有 MultiIndex：
In [97]: left = pd.DataFrame(
....: {
....: "A": ["A0", "A1", "A2", "A3"],
....: "B": ["B0", "B1", "B2", "B3"],
....: "key1": ["K0", "K0", "K1", "K2"],
....: "key2": ["K0", "K1", "K0", "K1"],
....: }
....: )
....:
In [98]: index = pd.MultiIndex.from_tuples(
....: [("K0", "K0"), ("K1", "K0"), ("K2", "K0"), ("K2", "K1")]
....: )
....:
In [99]: right = pd.DataFrame(
....: {"C": ["C0", "C1", "C2", "C3"], "D": ["D0", "D1", "D2", "D3"]}, index=index
....: )
....:
In [100]: result = left.join(right, on=["key1", "key2"])
In [101]: result
Out[101]:
A B key1 key2 C D
0 A0 B0 K0 K0 C0 D0
1 A1 B1 K0 K1 NaN NaN
2 A2 B2 K1 K0 C1 D1
3 A3 B3 K2 K1 C3 D3 DataFrame.join的默认行为是执行左连接，仅使用调用DataFrame中找到的键。可以使用how指定其他连接类型。
In [102]: result = left.join(right, on=["key1", "key2"], how="inner")
In [103]: result
Out[103]:
A B key1 key2 C D
0 A0 B0 K0 K0 C0 D0
2 A2 B2 K1 K0 C1 D1
3 A3 B3 K2 K1 C3 D3 ### 将单个索引连接到多重索引
您可以将具有MultiIndex的Index与具有级别的DataFrame连接。Index的name将与MultiIndex的级别名称匹配。
In [104]: left = pd.DataFrame(
.....: {"A": ["A0", "A1", "A2"], "B": ["B0", "B1", "B2"]},
.....: index=pd.Index(["K0", "K1", "K2"], name="key"),
.....: )
.....:
In [105]: index = pd.MultiIndex.from_tuples(
.....: [("K0", "Y0"), ("K1", "Y1"), ("K2", "Y2"), ("K2", "Y3")],
.....: names=["key", "Y"],
.....: )
.....:
In [106]: right = pd.DataFrame(
.....: {"C": ["C0", "C1", "C2", "C3"], "D": ["D0", "D1", "D2", "D3"]},
.....: index=index,
.....: )
.....:
In [107]: result = left.join(right, how="inner")
In [108]: result
Out[108]:
A B C D
key Y
K0 Y0 A0 B0 C0 D0
K1 Y1 A1 B1 C1 D1
K2 Y2 A2 B2 C2 D2
Y3 A2 B2 C3 D3 ### 与两个MultiIndex连接
输入参数的MultiIndex必须完全在连接中使用，并且是左参数中索引的子集。
In [109]: leftindex = pd.MultiIndex.from_product(
.....: [list("abc"), list("xy"), [1, 2]], names=["abc", "xy", "num"]
.....: )
.....:
In [110]: left = pd.DataFrame({"v1": range(12)}, index=leftindex)
In [111]: left
Out[111]:
v1
abc xy num
a x 1 0
2 1
y 1 2
2 3
b x 1 4
2 5
y 1 6
2 7
c x 1 8
2 9
y 1 10
2 11
In [112]: rightindex = pd.MultiIndex.from_product(
.....: [list("abc"), list("xy")], names=["abc", "xy"]
.....: )
.....:
In [113]: right = pd.DataFrame({"v2": [100 * i for i in range(1, 7)]}, index=rightindex)
In [114]: right
Out[114]:
v2
abc xy
a x 100
y 200
b x 300
y 400
c x 500
y 600
In [115]: left.join(right, on=["abc", "xy"], how="inner")
Out[115]:
v1 v2
abc xy num
a x 1 0 100
2 1 100
y 1 2 200
2 3 200
b x 1 4 300
2 5 300
y 1 6 400
2 7 400
c x 1 8 500
2 9 500
y 1 10 600
2 11 600 In [116]: leftindex = pd.MultiIndex.from_tuples(
.....: [("K0", "X0"), ("K0", "X1"), ("K1", "X2")], names=["key", "X"]
.....: )
.....:
In [117]: left = pd.DataFrame(
.....: {"A": ["A0", "A1", "A2"], "B": ["B0", "B1", "B2"]}, index=leftindex
.....: )
.....:
In [118]: rightindex = pd.MultiIndex.from_tuples(
.....: [("K0", "Y0"), ("K1", "Y1"), ("K2", "Y2"), ("K2", "Y3")], names=["key", "Y"]
.....: )
.....:
In [119]: right = pd.DataFrame(
.....: {"C": ["C0", "C1", "C2", "C3"], "D": ["D0", "D1", "D2", "D3"]}, index=rightindex
.....: )
.....:
In [120]: result = pd.merge(
.....: left.reset_index(), right.reset_index(), on=["key"], how="inner"
.....: ).set_index(["key", "X", "Y"])
.....:
In [121]: result
Out[121]:
A B C D
key X Y
K0 X0 Y0 A0 B0 C0 D0
X1 Y0 A1 B1 C0 D0
K1 X2 Y1 A2 B2 C1 D1 ### 在列和索引级别的组合上合并
作为on、left_on和right_on参数传递的字符串可以引用列名或索引级别名称。这使得在不重置索引的情况下，可以在索引级别和列的组合上合并DataFrame实例。
In [122]: left_index = pd.Index(["K0", "K0", "K1", "K2"], name="key1")
In [123]: left = pd.DataFrame(
.....: {
.....: "A": ["A0", "A1", "A2", "A3"],
.....: "B": ["B0", "B1", "B2", "B3"],
.....: "key2": ["K0", "K1", "K0", "K1"],
.....: },
.....: index=left_index,
.....: )
.....:
In [124]: right_index = pd.Index(["K0", "K1", "K2", "K2"], name="key1")
In [125]: right = pd.DataFrame(
.....: {
.....: "C": ["C0", "C1", "C2", "C3"],
.....: "D": ["D0", "D1", "D2", "D3"],
.....: "key2": ["K0", "K0", "K0", "K1"],
.....: },
.....: index=right_index,
.....: )
.....:
In [126]: result = left.merge(right, on=["key1", "key2"])
In [127]: result
Out[127]:
A B key2 C D
key1
K0 A0 B0 K0 C0 D0
K1 A2 B2 K0 C1 D1
K2 A3 B3 K1 C3 D3 注意
当DataFrame在两个参数中匹配索引级别的字符串上进行连接时，索引级别将保留为结果DataFrame中的索引级别。
注意
当仅使用MultiIndex的一些级别来连接DataFrame时，结果连接中将删除额外的级别。要保留这些级别，请在连接之前对这些级别名称使用DataFrame.reset_index()将这些级别移动到列中。 ### 连接多个DataFrame
一个：class：DataFrame``的列表或元组也可以传递给join()，以便根据它们的索引将它们连接在一起。
In [128]: right2 = pd.DataFrame({"v": [7, 8, 9]}, index=["K1", "K1", "K2"])
In [129]: result = left.join([right, right2]) ### DataFrame.combine_first()
DataFrame.combine_first()将一个DataFrame中的??失值更新为另一个DataFrame中相应位置的非缺失值。
In [130]: df1 = pd.DataFrame(
.....: [[np.nan, 3.0, 5.0], [-4.6, np.nan, np.nan], [np.nan, 7.0, np.nan]]
.....: )
.....:
In [131]: df2 = pd.DataFrame([[-42.6, np.nan, -8.2], [-5.0, 1.6, 4]], index=[1, 2])
In [132]: result = df1.combine_first(df2)
In [133]: result
Out[133]:
0 1 2
0 NaN 3.0 5.0
1 -4.6 NaN -8.2
2 -5.0 7.0 4.0 ### 将单个索引连接到多重索引
您可以将一个带有Index的DataFrame与具有MultiIndex的DataFrame在一个级别上连接。Index的name将与MultiIndex的级别名称匹配。
In [104]: left = pd.DataFrame(
.....: {"A": ["A0", "A1", "A2"], "B": ["B0", "B1", "B2"]},
.....: index=pd.Index(["K0", "K1", "K2"], name="key"),
.....: )
.....:
In [105]: index = pd.MultiIndex.from_tuples(
.....: [("K0", "Y0"), ("K1", "Y1"), ("K2", "Y2"), ("K2", "Y3")],
.....: names=["key", "Y"],
.....: )
.....:
In [106]: right = pd.DataFrame(
.....: {"C": ["C0", "C1", "C2", "C3"], "D": ["D0", "D1", "D2", "D3"]},
.....: index=index,
.....: )
.....:
In [107]: result = left.join(right, how="inner")
In [108]: result
Out[108]:
A B C D
key Y
K0 Y0 A0 B0 C0 D0
K1 Y1 A1 B1 C1 D1
K2 Y2 A2 B2 C2 D2
Y3 A2 B2 C3 D3 ### 与两个MultiIndex连接
输入参数的MultiIndex必须完全在连接中使用，并且是左侧参数中索引的子集。
In [109]: leftindex = pd.MultiIndex.from_product(
.....: [list("abc"), list("xy"), [1, 2]], names=["abc", "xy", "num"]
.....: )
.....:
In [110]: left = pd.DataFrame({"v1": range(12)}, index=leftindex)
In [111]: left
Out[111]:
v1
abc xy num
a x 1 0
2 1
y 1 2
2 3
b x 1 4
2 5
y 1 6
2 7
c x 1 8
2 9
y 1 10
2 11
In [112]: rightindex = pd.MultiIndex.from_product(
.....: [list("abc"), list("xy")], names=["abc", "xy"]
.....: )
.....:
In [113]: right = pd.DataFrame({"v2": [100 * i for i in range(1, 7)]}, index=rightindex)
In [114]: right
Out[114]:
v2
abc xy
a x 100
y 200
b x 300
y 400
c x 500
y 600
In [115]: left.join(right, on=["abc", "xy"], how="inner")
Out[115]:
v1 v2
abc xy num
a x 1 0 100
2 1 100
y 1 2 200
2 3 200
b x 1 4 300
2 5 300
y 1 6 400
2 7 400
c x 1 8 500
2 9 500
y 1 10 600
2 11 600 In [116]: leftindex = pd.MultiIndex.from_tuples(
.....: [("K0", "X0"), ("K0", "X1"), ("K1", "X2")], names=["key", "X"]
.....: )
.....:
In [117]: left = pd.DataFrame(
.....: {"A": ["A0", "A1", "A2"], "B": ["B0", "B1", "B2"]}, index=leftindex
.....: )
.....:
In [118]: rightindex = pd.MultiIndex.from_tuples(
.....: [("K0", "Y0"), ("K1", "Y1"), ("K2", "Y2"), ("K2", "Y3")], names=["key", "Y"]
.....: )
.....:
In [119]: right = pd.DataFrame(
.....: {"C": ["C0", "C1", "C2", "C3"], "D": ["D0", "D1", "D2", "D3"]}, index=rightindex
.....: )
.....:
In [120]: result = pd.merge(
.....: left.reset_index(), right.reset_index(), on=["key"], how="inner"
.....: ).set_index(["key", "X", "Y"])
.....:
In [121]: result
Out[121]:
A B C D
key X Y
K0 X0 Y0 A0 B0 C0 D0
X1 Y0 A1 B1 C0 D0
K1 X2 Y1 A2 B2 C1 D1 ### 在列和索引级别的组合上合并
作为on、left_on和right_on参数传递的字符串可以引用列名或索引级别名称。这使得可以在不重置索引的情况下，根据索引级别和列的组合来合并DataFrame实例。
In [122]: left_index = pd.Index(["K0", "K0", "K1", "K2"], name="key1")
In [123]: left = pd.DataFrame(
.....: {
.....: "A": ["A0", "A1", "A2", "A3"],
.....: "B": ["B0", "B1", "B2", "B3"],
.....: "key2": ["K0", "K1", "K0", "K1"],
.....: },
.....: index=left_index,
.....: )
.....:
In [124]: right_index = pd.Index(["K0", "K1", "K2", "K2"], name="key1")
In [125]: right = pd.DataFrame(
.....: {
.....: "C": ["C0", "C1", "C2", "C3"],
.....: "D": ["D0", "D1", "D2", "D3"],
.....: "key2": ["K0", "K0", "K0", "K1"],
.....: },
.....: index=right_index,
.....: )
.....:
In [126]: result = left.merge(right, on=["key1", "key2"])
In [127]: result
Out[127]:
A B key2 C D
key1
K0 A0 B0 K0 C0 D0
K1 A2 B2 K0 C1 D1
K2 A3 B3 K1 C3 D3 注意
当在两个参数中都匹配索引级别的字符串上连接DataFrame时，索引级别将作为结果DataFrame中的索引级别保留。
注意
当仅使用MultiIndex的部分级别连接DataFrame时，结果连接中的额外级别将被丢弃。要保留这些级别，请在连接之前对这些级别名称使用DataFrame.reset_index()将这些级别移动到列中。
连接多个DataFrame
也可以将DataFrame的列表或元组传递给join()，以便根据它们的索引将它们连接在一起。
In [128]: right2 = pd.DataFrame({"v": [7, 8, 9]}, index=["K1", "K1", "K2"])
In [129]: result = left.join([right, right2]) ### DataFrame.combine_first()
DataFrame.combine_first()将一个DataFrame中的缺失值更新为另一个DataFrame中相应位置的非缺失值。
In [130]: df1 = pd.DataFrame(
.....: [[np.nan, 3.0, 5.0], [-4.6, np.nan, np.nan], [np.nan, 7.0, np.nan]]
.....: )
.....:
In [131]: df2 = pd.DataFrame([[-42.6, np.nan, -8.2], [-5.0, 1.6, 4]], index=[1, 2])
In [132]: result = df1.combine_first(df2)
In [133]: result
Out[133]:
0 1 2
0 NaN 3.0 5.0
1 -4.6 NaN -8.2
2 -5.0 7.0 4.0 ## merge_ordered()
merge_ordered()将顺序数据（如数字或时间序列数据）与可选的使用fill_method填充缺失数据的数据合并。
In [134]: left = pd.DataFrame(
.....: {"k": ["K0", "K1", "K1", "K2"], "lv": [1, 2, 3, 4], "s": ["a", "b", "c", "d"]}
.....: )
.....:
In [135]: right = pd.DataFrame({"k": ["K1", "K2", "K4"], "rv": [1, 2, 3]})
In [136]: pd.merge_ordered(left, right, fill_method="ffill", left_by="s")
Out[136]:
k lv s rv
0 K0 1.0 a NaN
1 K1 1.0 a 1.0
2 K2 1.0 a 2.0
3 K4 1.0 a 3.0
4 K1 2.0 b 1.0
5 K2 2.0 b 2.0
6 K4 2.0 b 3.0
7 K1 3.0 c 1.0
8 K2 3.0 c 2.0
9 K4 3.0 c 3.0
10 K1 NaN d 1.0
11 K2 4.0 d 2.0
12 K4 4.0 d 3.0 merge_asof()
merge_asof() 类似于有序的左连接，只是匹配的是最近的键而不是相等的键。对于left DataFrame中的每一行，选择right DataFrame中的最后一行，其中 on 键小于左侧的键。两个DataFrame必须按键排序。
可选地，merge_asof() 可以通过在 on 键上最接近的匹配的同时匹配 by 键来执行分组合并。
In [137]: trades = pd.DataFrame(
.....: {
.....: "time": pd.to_datetime(
.....: [
.....: "20160525 13:30:00.023",
.....: "20160525 13:30:00.038",
.....: "20160525 13:30:00.048",
.....: "20160525 13:30:00.048",
.....: "20160525 13:30:00.048",
.....: ]
.....: ),
.....: "ticker": ["MSFT", "MSFT", "GOOG", "GOOG", "AAPL"],
.....: "price": [51.95, 51.95, 720.77, 720.92, 98.00],
.....: "quantity": [75, 155, 100, 100, 100],
.....: },
.....: columns=["time", "ticker", "price", "quantity"],
.....: )
.....:
In [138]: quotes = pd.DataFrame(
.....: {
.....: "time": pd.to_datetime(
.....: [
.....: "20160525 13:30:00.023",
.....: "20160525 13:30:00.023",
.....: "20160525 13:30:00.030",
.....: "20160525 13:30:00.041",
.....: "20160525 13:30:00.048",
.....: "20160525 13:30:00.049",
.....: "20160525 13:30:00.072",
.....: "20160525 13:30:00.075",
.....: ]
.....: ),
.....: "ticker": ["GOOG", "MSFT", "MSFT", "MSFT", "GOOG", "AAPL", "GOOG", "MSFT"],
.....: "bid": [720.50, 51.95, 51.97, 51.99, 720.50, 97.99, 720.50, 52.01],
.....: "ask": [720.93, 51.96, 51.98, 52.00, 720.93, 98.01, 720.88, 52.03],
.....: },
.....: columns=["time", "ticker", "bid", "ask"],
.....: )
.....:
In [139]: trades
Out[139]:
time ticker price quantity
0 2016-05-25 13:30:00.023 MSFT 51.95 75
1 2016-05-25 13:30:00.038 MSFT 51.95 155
2 2016-05-25 13:30:00.048 GOOG 720.77 100
3 2016-05-25 13:30:00.048 GOOG 720.92 100
4 2016-05-25 13:30:00.048 AAPL 98.00 100
In [140]: quotes
Out[140]:
time ticker bid ask
0 2016-05-25 13:30:00.023 GOOG 720.50 720.93
1 2016-05-25 13:30:00.023 MSFT 51.95 51.96
2 2016-05-25 13:30:00.030 MSFT 51.97 51.98
3 2016-05-25 13:30:00.041 MSFT 51.99 52.00
4 2016-05-25 13:30:00.048 GOOG 720.50 720.93
5 2016-05-25 13:30:00.049 AAPL 97.99 98.01
6 2016-05-25 13:30:00.072 GOOG 720.50 720.88
7 2016-05-25 13:30:00.075 MSFT 52.01 52.03
In [141]: pd.merge_asof(trades, quotes, on="time", by="ticker")
Out[141]:
time ticker price quantity bid ask
0 2016-05-25 13:30:00.023 MSFT 51.95 75 51.95 51.96
1 2016-05-25 13:30:00.038 MSFT 51.95 155 51.97 51.98
2 2016-05-25 13:30:00.048 GOOG 720.77 100 720.50 720.93
3 2016-05-25 13:30:00.048 GOOG 720.92 100 720.50 720.93
4 2016-05-25 13:30:00.048 AAPL 98.00 100 NaN NaN merge_asof() 在报价时间和交易时间之间的2ms内。
In [142]: pd.merge_asof(trades, quotes, on="time", by="ticker", tolerance=pd.Timedelta("2ms"))
Out[142]:
time ticker price quantity bid ask
0 2016-05-25 13:30:00.023 MSFT 51.95 75 51.95 51.96
1 2016-05-25 13:30:00.038 MSFT 51.95 155 NaN NaN
2 2016-05-25 13:30:00.048 GOOG 720.77 100 720.50 720.93
3 2016-05-25 13:30:00.048 GOOG 720.92 100 720.50 720.93
4 2016-05-25 13:30:00.048 AAPL 98.00 100 NaN NaN merge_asof() 在报价时间和交易时间之间的10ms内，并排除时间上的精确匹配。请注意，尽管我们排除了精确匹配（报价），但之前的报价确实传播到那个时间点。
In [143]: pd.merge_asof(
.....: trades,
.....: quotes,
.....: on="time",
.....: by="ticker",
.....: tolerance=pd.Timedelta("10ms"),
.....: allow_exact_matches=False,
.....: )
.....:
Out[143]:
time ticker price quantity bid ask
0 2016-05-25 13:30:00.023 MSFT 51.95 75 NaN NaN
1 2016-05-25 13:30:00.038 MSFT 51.95 155 51.97 51.98
2 2016-05-25 13:30:00.048 GOOG 720.77 100 NaN NaN
3 2016-05-25 13:30:00.048 GOOG 720.92 100 NaN NaN
4 2016-05-25 13:30:00.048 AAPL 98.00 100 NaN NaN compare()
Series.compare() 和 DataFrame.compare() 方法允许您比较两个分别是DataFrame或Series的对象，并总结它们的差异。
In [144]: df = pd.DataFrame(
.....: {
.....: "col1": ["a", "a", "b", "b", "a"],
.....: "col2": [1.0, 2.0, 3.0, np.nan, 5.0],
.....: "col3": [1.0, 2.0, 3.0, 4.0, 5.0],
.....: },
.....: columns=["col1", "col2", "col3"],
.....: )
.....:
In [145]: df
Out[145]:
col1 col2 col3
0 a 1.0 1.0
1 a 2.0 2.0
2 b 3.0 3.0
3 b NaN 4.0
4 a 5.0 5.0
In [146]: df2 = df.copy()
In [147]: df2.loc[0, "col1"] = "c"
In [148]: df2.loc[2, "col3"] = 4.0
In [149]: df2
Out[149]:
col1 col2 col3
0 c 1.0 1.0
1 a 2.0 2.0
2 b 3.0 4.0
3 b NaN 4.0
4 a 5.0 5.0
In [150]: df.compare(df2)
Out[150]:
col1 col3
self other self other
0 a c NaN NaN
2 NaN NaN 3.0 4.0 默认情况下，如果两个对应的值相等，它们将显示为 NaN。此外，如果整行/列中的所有值都相等，则该行/列将从结果中省略。剩余的差异将对齐在列上。
将差异堆叠在行上。
In [151]: df.compare(df2, align_axis=0)
Out[151]:
col1 col3
0 self a NaN
other c NaN
2 self NaN 3.0
other NaN 4.0 保留所有原始行和列，使用 keep_shape=True
In [152]: df.compare(df2, keep_shape=True)
Out[152]:
col1 col2 col3
self other self other self other
0 a c NaN NaN NaN NaN
1 NaN NaN NaN NaN NaN NaN
2 NaN NaN NaN NaN 3.0 4.0
3 NaN NaN NaN NaN NaN NaN
4 NaN NaN NaN NaN NaN NaN 保留所有原始值，即使它们相等。
In [153]: df.compare(df2, keep_shape=True, keep_equal=True)
Out[153]:
col1 col2 col3
self other self other self other
0 a c 1.0 1.0 1.0 1.0
1 a a 2.0 2.0 2.0 2.0
2 b b 3.0 3.0 3.0 4.0
3 b b NaN NaN 4.0 4.0
4 a a 5.0 5.0 5.0 5.0 本文参与?腾讯云自媒体分享计划，分享自作者个人站点/博客。
原始发表：2024-04-25，如有侵权请联系 cloudcommunity@tencent.com 删除
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