这是 ARM Cortex M0 体系决定的，它只支持对齐访问 ( Aligned Access )。
如图1所示，与前人使用的randomized SMILES和 canonical SMILES不同，本研究提出的Root-aligned SMILES（R-SMILES），通过将输入和输出的根原子进行对齐的方式...参考资料 Root-aligned SMILES: A Tight Representation for Chemical Reaction Prediction. Chem. Sci. 2022.
乐固-签名APK-会出来一个aligned的apk包，在签名的时候你们还做了什么操作吗？
} \text {MOTA} = 1 - \frac{|\text {FN}| + |\text {FP}| + |\text {IDSW}|}{|\text {gtDet}|} \end{aligned...} \text {MOTP} = \frac{1}{|\text {TP}|}\sum _{\text {TP}}{ \mathcal {S}} \end{aligned}...} \begin{aligned} \text {MODA}&= 1 - \frac{|\text {FN}| + |\text {FP}|}{|\text {gtDet}|}&\\&= \frac{|...} \end{aligned} 可以发现，如果检测的Precision小于等于0.5的话，MODA就会为0，甚至出现负值，而检测的Recall小于等于0.5...}&\text {DetA}_\alpha = \frac{|\text {TP}|}{|\text {TP}| + |\text {FN}| + |\text {FP}|}&\end{aligned}
, int *cols_aligned, float *vals_aligned, int num_rows, int num_cols, const size_t num_non_zeros) {...= std_vec_2_aligned_array(i_idx, mem_align); int *cols_aligned = std_vec_2_aligned_array<int...(); sparse_matrix_t coo_mtx = create_coo_mtx(rows_aligned, cols_aligned, vals_aligned, num_rows,...); free(cols_aligned); free(vals_aligned); free(b_aligned); free(x); #else _aligned_free...(rows_aligned); _aligned_free(cols_aligned); _aligned_free(vals_aligned); _aligned_free(b_aligned
} 1.添项减项凑出需要的结构 例1.1 \begin{aligned}提示:&A \cdot B - 1 = B(A - 1) + B - 1\end{aligned} ---- \begin{aligned...\cos {\rm n}x}{x ^ 2}\end{aligned} 解答 解法 1 解法 2 \begin{aligned}解法1: \\&原式=\lim_{x \to 0} \dfrac{1 -...\sqrt[n]{\cos nx} }{x ^ 2}\end{aligned} 解答 \begin{aligned}&原式=\lim_{x \to 0} - \dfrac{\ln (\cos x...\sim -\frac{x^3}{3} \\ \end{aligned} 解答 \begin{aligned}&原式 = \lim_{x \to 0} \dfrac{\frac{1}{2}...\\end{aligned} 解答 \begin{aligned}原式&{=\Large \lim_{x \to 0} \frac{1}{x ^ 3}[{\rm e} ^ {x \cdot \ln
} \text {MOTA} = 1 - \frac{|\text {FN}| + |\text {FP}| + |\text {IDSW}|}{|\text {gtDet}|} \end{aligned...} \begin{aligned} \text {MODA}&= 1 - \frac{|\text {FN}| + |\text {FP}|}{|\text {gtDet}|}&\\&= \frac{|...} \end{aligned} MODA?=1?∣gtDet∣∣FN∣+∣FP∣?=∣TP∣+∣FN∣∣TP∣?∣FP∣?=DetRe?(2?DetPr1?)???...}&\text {DetA}_\alpha = \frac{|\text {TP}|}{|\text {TP}| + |\text {FN}| + |\text {FP}|}&\end{aligned}...}&\begin{aligned} \text {HOTA}_\alpha&\quad = \sqrt{\frac{\sum _{c \in \{\text {TP}\}} \mathcal {A}(c
\end{aligned} \tag{2} 这里面的an,bn系数怎么求呢？...+c \end{aligned} \tag{4} 有了上面这些基本公式，先来证明几个恒等式： \begin{aligned} \int_{0}^{2\pi}sin(mt)dt &=0 ,...(m \ne 0) \end{aligned} \tag{a} \begin{aligned} \int_{0}^{2\pi}cos(mt)dt &=0,(m \ne 0) \\ \end...{aligned} \tag{b} \begin{aligned} \int_{0}^{2\pi}sin(mt) \cdot cos(nt)dt &=0,(m \ne n , m \ne -n...) \\ \end{aligned} \tag{c} \begin{aligned} \int_{0}^{2\pi}cos(mt) \cdot cos(nt)dt &=0,(m \ne n
concordantly 0 times 13590 (0.10%) aligned concordantly exactly 1 time 7230 (0.05%) aligned...1695827 (27.69%) aligned exactly 1 time 4432 (0.07%) aligned >1 times 86.16% overall...1622752 (27.93%) aligned exactly 1 time 7228 (0.12%) aligned >1 times 85.29% overall...2110960 (27.05%) aligned exactly 1 time 20696 (0.27%) aligned >1 times 84.48% overall...1838519 (26.75%) aligned exactly 1 time 8174 (0.12%) aligned >1 times 85.32% overall
然后，我们将 c^{\frac{\log_a b}{\log_a c}} 进一步拆分，得到： \begin{aligned} c^{\frac{\log_a b}{\log_a c}} &= c^{\...log_c b} \ &= b \end{aligned} 最后，将上述结果代入原式中，有： \begin{aligned} a^{\log_b c} &= a^x \ &= c^{\frac{\log_a...{2^n} = \infty 考虑到： \begin{aligned} \frac{(n + 1)!}{n!} &= n + 1 \ &> 2 \end{aligned} 因此，n!...至少每乘以 2 次就要增长一倍，即： \begin{aligned} n!...展开，则有： \begin{aligned} n!
学着在shell下写循环 cd output_results #SAM装换为BAM samtools view -S -b -o sim_1_aligned.bam sim_1_aligned.sam...samtools view -S -b -o sim_2_aligned.bam sim_2_aligned.sam samtools view -S -b -o sim_3_aligned.bam...sim_3_aligned.sam #排序 samtools sort sim_1_aligned.bam -o sim_1_aligned.sorted.bam samtools sort sim_...2_aligned.bam -o sim_2_aligned.sorted.bam samtools sort sim_3_aligned.bam -o sim_3_aligned.sorted.bam.../Reference_genome/ecoli.fa sim_1_aligned.sorted.bam sim_2_aligned.sorted.bam sim_3_aligned.sorted.bam
concordantly 0 times 632294 (1.28%) aligned concordantly exactly 1 time 6389844 (12.97%) aligned...(0.00%) aligned discordantly 1 time ---- 42251014 pairs aligned 0 times concordantly or discordantly...33143 (0.04%) aligned exactly 1 time 396000 (0.47%) aligned >1 times 14.69% overall...concordantly 0 times 1965637 (4.65%) aligned concordantly exactly 1 time 2423616 (5.74%) aligned...400055 (0.53%) aligned exactly 1 time 225554 (0.30%) aligned >1 times 11.19% overall
然后，我们将 $c^{\frac{\log_a b}{\log_a c}}$ 进一步拆分，得到：$$\begin{aligned}c^{\frac{\log_a b}{\log_a c}} &= c^{...\log_c b} \ &= b\end{aligned}$$最后，将上述结果代入原式中，有：$$\begin{aligned}a^{\log_b c} &= a^x \ &= c^{\frac{\log_a...{2^n} = \infty$$考虑到：$$\begin{aligned}\frac{(n + 1)!}{n!} &= n + 1 \&> 2\end{aligned}$$因此，$n!...$ 至少每乘以 2 次就要增长一倍，即：$$\begin{aligned}n!...$ 展开，则有：$$\begin{aligned}n!
} A(x) & = \sum_{n=0}^{\infty} x^n = \frac{1}{1-x} \end{aligned} \{a_n\} , a_n = (-1)^n ; \begin{...aligned} A(x) & = \sum_{n=0}^{\infty} (-1)^n x^n = \frac{1}{1+x} \end{aligned} \{a_n\} , a_n = k^n..., k 为正整数 ; \begin{aligned} A(x) & = \sum_{n=0}^{\infty} k^n x^n = \frac{1}{1-kx} \end{aligned} 二项式系数相关...= ( 1 + x ) ^m \end{aligned} 组合数相关 : \{a_n\} , a_n = \dbinom{m+n-1}{n} , m,n 为正整数 ; \begin{aligned...) x^n \\ & = \frac{1}{{(1-x)}^2} \end{aligned}
部分语法可能与 LaTeX 略有不同，比如这里我想要的等号换行对其效果就不同 问题解决 ① CSDN 中首先要用 $$ 将想要输入的数学公式括起来 $$ 内容 $$ ② 之后用 \begin{aligned...} 与 \end{aligned} 将要多行输出的内容括起来 $$ \begin{aligned} 数学公式 \end{aligned} $$ ③ 输入数学公式，使用 \\ 进行换行，使用 & 进行等号位置对齐的控制...$$ \begin{aligned} d_{AB} &=\sqrt{(6-2)^2+(6-2)^2}\\ &=\sqrt{4^2+4^2}\\ &= 4\sqrt{2} \end{aligned} $...效果如下： \begin{aligned} d_{AB} &=\sqrt{(6-2)^2+(6-2)^2}\\ &=\sqrt{4^2+4^2}\\ &= 4\sqrt{2} \end{aligned}
((aligned (8))); unsigned long InEchoReps __attribute__ ((aligned (8))); unsigned long OutEchos __...attribute__ ((aligned (8))); unsigned long OutEchoReps __attribute__ ((aligned (8))); unsigned long...InTimestamps __attribute__ ((aligned (8))); unsigned long InTimestampReps __attribute__ ((aligned (...__attribute__ ((aligned (8))); unsigned long OutAddrMasks __attribute__ ((aligned (8))); unsigned...__attribute__ ((aligned (8))); unsigned long OutTimeExcds __attribute__ ((aligned (8))); unsigned
比如GPL21827这个平台： 60898 reads; of these: 60898 (100.00%) were unpaired; of these: 59099 (97.05%) aligned...0 times 1753 (2.88%) aligned exactly 1 time 46 (0.08%) aligned >1 times 2.95% overall alignment...0 times 811 (0.48%) aligned exactly 1 time 149 (0.09%) aligned >1 times 0.56% overall alignment...acc=GPL8180 380 reads; of these: 380 (100.00%) were unpaired; of these: 215 (56.58%) aligned 0...times 138 (36.32%) aligned exactly 1 time 27 (7.11%) aligned >1 times 43.42% overall alignment
修改次数：1 历史修改内容： 1.0.2 修改离散型条件概率密度公式 1.0.1 修改联合分布函数的性质公式 联合分布函数 F(x,y)=P{X\le x, Y\le y} 性质： $$ \begin{aligned...} &F(+\infty,+\infty)=1 \ F(-\infty, -\infty)=&F(x, -\infty)=F(-\infty, y)=0 \end{aligned} $$ 二维随机变量边缘分布律...} P_1(x)=\int_{-\infty}^{+\infty}p(x,y)dy \ P_2(y)=\int_{-\infty}^{+\infty}p(x,y)dx \end{aligned} $$...二维随机变量函数的分布 和分布（$Z=X+Y$） $$ \begin{aligned} F_Z(z)&=\int_{-\infty}^{z}g(u)du \ g(u)&=\int_{-\infty}^{...+\infty}f(x,u-x)dx \ f_Z(z)&=g(z)= \int_{-\infty}^{+\infty}f(x,z-x)dx \end{aligned} $$ 相互独立，则： f_Z(
rk = 0 = \binom {r-1}{k-1} 1.1.3+ 相伴恒等式 Companion Identity (r-k)\binom rk=r\binom {r-1}k 证明： \begin{aligned...}(r-k)\binom rk & =(r-k)\binom r {r-k}\\& = r\binom {r-1}{r-k-1}\\& = r\binom {r-1}k\end{aligned} 1.1.4...n=0 时，左边 =\binom 0m=[m=0]=\binom 1{m+1}= 右边 时， \begin{aligned}\sum_{0\leq k\leq n+1}\binom km & =\sum...\\& = \binom rk \binom {r-k}{m-k}\end{aligned} 若 m<kk<00。...首先有： \begin{aligned} [z^n](1+z)^{2n} & =\sum_{k=0}^{2n}\binom {2n}kz^k\\ & =\binom {2n}n \end{aligned
接着在做 self-attention 之前，会用词嵌入向量计算 q, k, v 向量同时加入位置信息，函数公式表达如下： \begin{aligned} q_m=f_q(x_m,m) \\ k_n=f_k...(x_n,n) \\ v_n=f_v(x_n,n) \end{aligned} 其中 qm 表示第 m 个 token 对应的词向量 xm 集成位置信息 m 之后的 query 向量。...假定现在词嵌入向量的维度是两维 d=2，这样就可以利用上2维度平面上的向量的几何性质，然后论文中提出了一个满足上述关系的 f 和 g 的形式如下： \begin{aligned} f_q(x_m,m)=...则上述 f 和 g 公式中的 \begin{aligned} e^{im\theta}=\cos (m\theta) + i\sin (m\theta) \\ e^{in\theta}=\cos (n...} z=a+ib\\ z^*=a-ib \end{aligned} 所以可得： \begin{aligned} W_qx_m = q_m = q_m^{(1)} + iq_m^{(2)} \\ W_kx_n
领取专属 10元无门槛券
手把手带您无忧上云
云服务器
即时通信 IM
ICP备案
对象存储
实时音视频
