多标签softmax + cross-entropy交叉熵损失函数详解及反向传播中的梯度求导

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摘要

本文求解 softmax + cross-entropy 在反向传播中的梯度.

相关

配套代码, 请参考文章 :

Python和PyTorch对比实现多标签softmax + cross-entropy交叉熵损失及反向传播

有关 softmax 的详细介绍, 请参考 :

softmax函数详解及反向传播中的梯度求导

有关 cross-entropy 的详细介绍, 请参考 :

通过案例详解cross-entropy交叉熵损失函数

系列文章索引 :

https://blog.csdn.net/oBrightLamp/article/details/85067981

正文

在大多数教程中, softmax 和 cross-entropy 总是一起出现, 求梯度的时候也是一起考虑.

softmax 和 cross-entropy 的梯度, 已经在上面的两篇文章中分别给出.

1. 题目

考虑一个输入向量 x, 经 softmax 函数归一化处理后得到向量 s 作为预测的概率分布, 已知向量 y 为真实的概率分布, 由 cross-entropy 函数计算得出误差值 error (标量 e ), 求 e 关于 x 的梯度.

x=(x1,x2,x3,⋯ ,xk)s=softmax(x)si=exi∑t=1kexte=crossEntropy(s,y)=−∑i=1kyilog(si) \quad\ x = (x_1, x_2, x_3, \cdots, x_k)\ \quad\ s = softmax(x)\ \quad\ s_{i} = \frac{e^{x_{i}}}{ \sum_{t = 1}^{k}e^{x_{t}}} \ \quad\ e = crossEntropy(s, y) = -\sum_{i = 1}^{k}y_{i}log(s_{i})\ x=(x1​,x2​,x3​,⋯,xk​)s=softmax(x)si​=∑t=1k​ext​exi​​e=crossEntropy(s,y)=−i=1∑k​yi​log(si​)

已知 :

∇e(s)=∂e∂s=(∂e∂s1,∂e∂s2,⋯ ,∂e∂sk)=(−y1s1,−y2s2,⋯ ,−yksk)  ∇s(x)=∂s∂x=(∂s1/∂x1∂s1/∂x2⋯∂s1/∂xk∂s2/∂x1∂s2/∂x2⋯∂s2/∂xk⋮⋮⋱⋮∂sk/∂x1∂sk/∂x2⋯∂sk/∂xk)=(−s1s1+s1−s1s2⋯−s1sk−s2s1−s2s2+s2⋯−s2sk⋮⋮⋱⋮−sks1−sks2⋯−sksk+sk) \nabla e_{(s)}=\frac{\partial e}{\partial s} =(\frac{\partial e}{\partial s_{1}},\frac{\partial e}{\partial s_{2}}, \cdots, \frac{\partial e}{\partial s_{k}}) =( -\frac{y_1}{s_1}, -\frac{y_2}{s_2},\cdots,-\frac{y_k}{s_k}) \ \;\ % ---------- \nabla s_{(x)}= \frac{\partial s}{\partial x}= \begin{pmatrix} \partial s_{1}/\partial x_{1}&\partial s_{1}/\partial x_{2}& \cdots&\partial s_{1}/\partial x_{k}\ \partial s_{2}/\partial x_{1}&\partial s_{2}/\partial x_{2}& \cdots&\partial s_{2}/\partial x_{k}\ \vdots & \vdots & \ddots & \vdots \ \partial s_{k}/\partial x_{1}&\partial s_{k}/\partial x_{2}& \cdots&\partial s_{k}/\partial x_{k}\ \end{pmatrix}= \begin{pmatrix} -s_{1}s_{1} + s_{1} & -s_{1}s_{2} & \cdots & -s_{1}s_{k} \ -s_{2}s_{1} & -s_{2}s_{2} + s_{2} & \cdots & -s_{2}s_{k} \ \vdots & \vdots & \ddots & \vdots \ -s_{k}s_{1} & -s_{k}s_{2} & \cdots & -s_{k}s_{k} + s_{k} \end{pmatrix} \ \quad\ ∇e(s)​=∂s∂e​=(∂s1​∂e​,∂s2​∂e​,⋯,∂sk​∂e​)=(−s1​y1​​,−s2​y2​​,⋯,−sk​yk​​)∇s(x)​=∂x∂s​=⎝⎜⎜⎜⎛​∂s1​/∂x1​∂s2​/∂x1​⋮∂sk​/∂x1​​∂s1​/∂x2​∂s2​/∂x2​⋮∂sk​/∂x2​​⋯⋯⋱⋯​∂s1​/∂xk​∂s2​/∂xk​⋮∂sk​/∂xk​​⎠⎟⎟⎟⎞​=⎝⎜⎜⎜⎛​−s1​s1​+s1​−s2​s1​⋮−sk​s1​​−s1​s2​−s2​s2​+s2​⋮−sk​s2​​⋯⋯⋱⋯​−s1​sk​−s2​sk​⋮−sk​sk​+sk​​⎠⎟⎟⎟⎞​

2. 求解过程 :

∂e∂xi=∂e∂s1∂s1∂xi+∂e∂s2∂s2∂xi+∂e∂s3∂s3∂xi+⋯+∂e∂sk∂sk∂xi \frac{\partial e}{\partial x_i} = \frac{\partial e}{\partial s_1}\frac{\partial s_1}{\partial x_i} +\frac{\partial e}{\partial s_2}\frac{\partial s_2}{\partial x_i} +\frac{\partial e}{\partial s_3}\frac{\partial s_3}{\partial x_i} + \cdots +\frac{\partial e}{\partial s_k}\frac{\partial s_k}{\partial x_i}\ ∂xi​∂e​=∂s1​∂e​∂xi​∂s1​​+∂s2​∂e​∂xi​∂s2​​+∂s3​∂e​∂xi​∂s3​​+⋯+∂sk​∂e​∂xi​∂sk​​

展开 ∂e/∂xi\partial e/\partial x_i∂e/∂xi​ 可得 e 关于 X 的梯度向量 :

∇e(x)=(∂e∂s1,∂e∂s2,∂e∂s3,⋯ ,∂e∂sk)(∂s1/∂x1∂s1/∂x2⋯∂s1/∂xk∂s2/∂x1∂s2/∂x2⋯∂s2/∂xk⋮⋮⋱⋮∂sk/∂x1∂sk/∂x2⋯∂sk/∂xk)  ∇e(x)=∇e(s)∇s(x) \nabla e_{(x)} = (\frac{\partial e}{\partial s_1},\frac{\partial e}{\partial s_2},\frac{\partial e}{\partial s_3}, \cdots ,\frac{\partial e}{\partial s_k}) \begin{pmatrix} \partial s_{1}/\partial x_{1}&\partial s_{1}/\partial x_{2}& \cdots&\partial s_{1}/\partial x_{k}\ \partial s_{2}/\partial x_{1}&\partial s_{2}/\partial x_{2}& \cdots&\partial s_{2}/\partial x_{k}\ \vdots & \vdots & \ddots & \vdots \ \partial s_{k}/\partial x_{1}&\partial s_{k}/\partial x_{2}& \cdots&\partial s_{k}/\partial x_{k}\ \end{pmatrix}\ \;\ \nabla e_{(x)} =\nabla e_{(s)} \nabla s_{(x)}\ ∇e(x)​=(∂s1​∂e​,∂s2​∂e​,∂s3​∂e​,⋯,∂sk​∂e​)⎝⎜⎜⎜⎛​∂s1​/∂x1​∂s2​/∂x1​⋮∂sk​/∂x1​​∂s1​/∂x2​∂s2​/∂x2​⋮∂sk​/∂x2​​⋯⋯⋱⋯​∂s1​/∂xk​∂s2​/∂xk​⋮∂sk​/∂xk​​⎠⎟⎟⎟⎞​∇e(x)​=∇e(s)​∇s(x)​

由于 :

∇e(s)=(−y1s1,−y2s2,⋯ ,−yksk)  ∇s(x)=(−s1s1+s1−s1s2⋯−s1sk−s2s1−s2s2+s2⋯−s2sk⋮⋮⋱⋮−sks1−sks2⋯−sksk+sk) \nabla e_{(s)}=( -\frac{y_1}{s_1}, -\frac{y_2}{s_2},\cdots,-\frac{y_k}{s_k})\ \;\ \nabla s_{(x)} =\begin{pmatrix} -s_{1}s_{1} + s_{1} & -s_{1}s_{2} & \cdots & -s_{1}s_{k} \ -s_{2}s_{1} & -s_{2}s_{2} + s_{2} & \cdots & -s_{2}s_{k} \ \vdots & \vdots & \ddots & \vdots \ -s_{k}s_{1} & -s_{k}s_{2} & \cdots & -s_{k}s_{k} + s_{k} \end{pmatrix} ∇e(s)​=(−s1​y1​​,−s2​y2​​,⋯,−sk​yk​​)∇s(x)​=⎝⎜⎜⎜⎛​−s1​s1​+s1​−s2​s1​⋮−sk​s1​​−s1​s2​−s2​s2​+s2​⋮−sk​s2​​⋯⋯⋱⋯​−s1​sk​−s2​sk​⋮−sk​sk​+sk​​⎠⎟⎟⎟⎞​

得 :

∇e(x)=(s1∑t=1kyt−y1,  s2∑t=1kyt−y2,⋯ ,si∑t=1kyt−yi)  ∂e∂xi=si∑t=1kyt−yi \nabla e_{(x)}= (s_1\sum_{t = 1}^{k}y_t- y_1, \;s_2\sum_{t = 1}^{k}y_t- y_2,\cdots,s_i\sum_{t = 1}^{k}y_t- y_i)\ \;\ \frac{\partial e}{\partial x_i} =s_i\sum_{t = 1}^{k}y_t- y_i ∇e(x)​=(s1​t=1∑k​yt​−y1​,s2​t=1∑k​yt​−y2​,⋯,si​t=1∑k​yt​−yi​)∂xi​∂e​=si​t=1∑k​yt​−yi​

结论:

将 softmax 和 cross-entropy 放在一起使用, 可以大大减少梯度求解的计算量.