cs231n assignment1 knn
实现compute_distances_two_loops
def compute_distances_two_loops(self, X):
"""
Compute the distance between each test point in X and each training point
in self.X_train using a nested loop over both the training data and the
test data.
Inputs:
- X: A numpy array of shape (num_test, D) containing test data.
Returns:
- dists: A numpy array of shape (num_test, num_train) where dists[i, j]
is the Euclidean distance between the ith test point and the jth training
point.
"""
num_test = X.shape[0]
num_train = self.X_train.shape[0]
dists = np.zeros((num_test, num_train))
for i in xrange(num_test):
for j in xrange(num_train):
#####################################################################
# TODO: #
# Compute the l2 distance between the ith test point and the jth #
# training point, and store the result in dists[i, j]. You should #
# not use a loop over dimension. #
#####################################################################
dists[i][j] = np.sqrt(np.sum(np.square(X[i,:] - self.X_train[j,:])))
#####################################################################
# END OF YOUR CODE #
#####################################################################
return dists实现predict_labels
def predict_labels(self, dists, k=1):
"""
Given a matrix of distances between test points and training points,
predict a label for each test point.
Inputs:
- dists: A numpy array of shape (num_test, num_train) where dists[i, j]
gives the distance betwen the ith test point and the jth training point.
Returns:
- y: A numpy array of shape (num_test,) containing predicted labels for the
test data, where y[i] is the predicted label for the test point X[i].
"""
num_test = dists.shape[0]
y_pred = np.zeros(num_test)
for i in xrange(num_test):
# A list of length k storing the labels of the k nearest neighbors to
# the ith test point.
closest_y = []
#########################################################################
# TODO: #
# Use the distance matrix to find the k nearest neighbors of the ith #
# testing point, and use self.y_train to find the labels of these #
# neighbors. Store these labels in closest_y. #
# Hint: Look up the function numpy.argsort. #
#########################################################################
sorted_vector = np.argsort(dists[i,:])
# print sorted_vector[:k]
closest_y = []
closest_y = self.y_train[sorted_vector[:k]]
# print closest_y
#########################################################################
# TODO: #
# Now that you have found the labels of the k nearest neighbors, you #
# need to find the most common label in the list closest_y of labels. #
# Store this label in y_pred[i]. Break ties by choosing the smaller #
# label. #
#########################################################################
# appear_times:
# key: the label
# value: the appear times of the label
appear_times = {}
for label in closest_y:
if label in appear_times:
appear_times[label] += 1
else:
appear_times[label] = 0
# find most commen label
y_pred[i] = max(appear_times, key=lambda x: appear_times[x])
#########################################################################
# END OF YOUR CODE #
#########################################################################
return y_pred预测准确率
k=1时
# Now implement the function predict_labels and run the code below:
# We use k = 1 (which is Nearest Neighbor).
y_test_pred = classifier.predict_labels(dists, k=1)
# Compute and print the fraction of correctly predicted examples
num_correct = np.sum(y_test_pred == y_test)
accuracy = float(num_correct) / num_test
print 'Got %d / %d correct => accuracy: %f' % (num_correct, num_test, accuracy)Got 137 / 500 correct => accuracy: 0.274000
k=5时
y_test_pred = classifier.predict_labels(dists, k=5)
num_correct = np.sum(y_test_pred == y_test)
accuracy = float(num_correct) / num_test
print 'Got %d / %d correct => accuracy: %f' % (num_correct, num_test, accuracy)Got 142 / 500 correct => accuracy: 0.284000
compute_distances_one_loop
def compute_distances_one_loop(self, X):
"""
Compute the distance between each test point in X and each training point
in self.X_train using a single loop over the test data.
Input / Output: Same as compute_distances_two_loops
"""
num_test = X.shape[0]
num_train = self.X_train.shape[0]
dists = np.zeros((num_test, num_train))
for i in xrange(num_test):
#######################################################################
# TODO: #
# Compute the l2 distance between the ith test point and all training #
# points, and store the result in dists[i, :]. #
#######################################################################
dists[i] = np.sqrt(np.sum(np.square(self.X_train - X[i]), axis=1))
#######################################################################
# END OF YOUR CODE #
#######################################################################
return distsDifference was: 0.000000 Good! The distance matrices are the same
计算结果的速度更快了，而且结果一致
此处用到了numpy的二维数组减一维数组
# 二维数组减一维数组
a = np.array([1,2,3])
b = np.ones([2,3])
a - barray([[ 0., 1., 2.], [ 0., 1., 2.]])
cross-validation
num_folds = 5
k_choices = [1, 3, 5, 8, 10, 12, 15, 20, 50, 100]
X_train_folds = []
y_train_folds = []
################################################################################
# TODO: #
# Split up the training data into folds. After splitting, X_train_folds and #
# y_train_folds should each be lists of length num_folds, where #
# y_train_folds[i] is the label vector for the points in X_train_folds[i]. #
# Hint: Look up the numpy array_split function. #
################################################################################
X_train_folds = np.array_split(X_train, num_folds)
Y_train_folds = np.array_split(y_train, num_folds)
################################################################################
# END OF YOUR CODE #
################################################################################
# A dictionary holding the accuracies for different values of k that we find
# when running cross-validation. After running cross-validation,
# k_to_accuracies[k] should be a list of length num_folds giving the different
# accuracy values that we found when using that value of k.
k_to_accuracies = {}
################################################################################
# TODO: #
# Perform k-fold cross validation to find the best value of k. For each #
# possible value of k, run the k-nearest-neighbor algorithm num_folds times, #
# where in each case you use all but one of the folds as training data and the #
# last fold as a validation set. Store the accuracies for all fold and all #
# values of k in the k_to_accuracies dictionary. #
################################################################################
for k in k_choices:
accuracy_sum = 0
k_to_accuracies[k] = []
for f in xrange(num_folds):
x_trai = np.array(X_train_folds[:f] + X_train_folds[f+1:])
y_trai = np.array(Y_train_folds[:f] + Y_train_folds[f+1:])
x_trai = x_trai.reshape(-1, x_trai.shape[2])
y_trai = y_trai.reshape(-1)
x_vali = np.array(X_train_folds[f])
y_vali = np.array(Y_train_folds[f])
classifier.train(x_trai, y_trai)
dists = classifier.compute_distances_no_loops(x_vali)
y_vali_pred = classifier.predict_labels(dists, k=k)
# Compute and print the fraction of correctly predicted examples
num_correct = np.sum(y_vali_pred == y_vali)
acc = float(num_correct) / y_vali.shape[0]
k_to_accuracies[k].append(acc)
################################################################################
# END OF YOUR CODE #
################################################################################
# Print out the computed accuracies
for k in sorted(k_to_accuracies):
for accuracy in k_to_accuracies[k]:
print 'k = %d, accuracy = %f' % (k, accuracy)k = 1, accuracy = 0.263000 k = 1, accuracy = 0.257000 k = 1, accuracy = 0.264000 k = 1, accuracy = 0.278000 k = 1, accuracy = 0.266000 k = 3, accuracy = 0.241000 k = 3, accuracy = 0.249000 k = 3, accuracy = 0.243000 k = 3, accuracy = 0.273000 k = 3, accuracy = 0.264000 k = 5, accuracy = 0.258000 k = 5, accuracy = 0.273000 k = 5, accuracy = 0.281000 k = 5, accuracy = 0.290000 k = 5, accuracy = 0.272000 k = 8, accuracy = 0.263000 k = 8, accuracy = 0.288000 k = 8, accuracy = 0.278000 k = 8, accuracy = 0.285000 k = 8, accuracy = 0.277000 k = 10, accuracy = 0.265000 k = 10, accuracy = 0.296000 k = 10, accuracy = 0.278000 k = 10, accuracy = 0.284000 k = 10, accuracy = 0.286000 k = 12, accuracy = 0.260000 k = 12, accuracy = 0.294000 k = 12, accuracy = 0.281000 k = 12, accuracy = 0.282000 k = 12, accuracy = 0.281000 k = 15, accuracy = 0.255000 k = 15, accuracy = 0.290000 k = 15, accuracy = 0.281000 k = 15, accuracy = 0.281000 k = 15, accuracy = 0.276000 k = 20, accuracy = 0.270000 k = 20, accuracy = 0.281000 k = 20, accuracy = 0.280000 k = 20, accuracy = 0.282000 k = 20, accuracy = 0.284000 k = 50, accuracy = 0.271000 k = 50, accuracy = 0.288000 k = 50, accuracy = 0.278000 k = 50, accuracy = 0.269000 k = 50, accuracy = 0.266000 k = 100, accuracy = 0.256000 k = 100, accuracy = 0.270000 k = 100, accuracy = 0.263000 k = 100, accuracy = 0.256000 k = 100, accuracy = 0.263000
选取最佳的k
# Based on the cross-validation results above, choose the best value for k,
# retrain the classifier using all the training data, and test it on the test
# data. You should be able to get above 28% accuracy on the test data.
best_k = 7
classifier = KNearestNeighbor()
classifier.train(X_train, y_train)
y_test_pred = classifier.predict(X_test, k=best_k)
# Compute and display the accuracy
num_correct = np.sum(y_test_pred == y_test)
accuracy = float(num_correct) / num_test
print 'Got %d / %d correct => accuracy: %f' % (num_correct, num_test, accuracy)Got 141 / 500 correct => accuracy: 0.282000
可见，k = 7比较好
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原始发表：2018.01.28 ，如有侵权请联系 cloudcommunity@tencent.com 删除
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