ARTS-S pytorch中backward函数的gradient参数做用

导数偏导数的数学定义

参考资料1和2中对导数偏导数的定义都很是明确.导数和偏导数都是函数对自变量而言.从数学定义上讲,求导或者求偏导只有函数对自变量,其他任何状况都是错的.可是不少机器学习的资料和开源库都涉及到标量对向量求导.好比下面这个pytorch的例子.

import torch
x = torch.tensor([1.0, 2.0, 3.0], requires_grad=True)
y = x ** 2 + 2
z = torch.sum(y)
z.backward()
print(x.grad)

简单解释下,设\(x=[x_1,x_2,x_3]\),则
\[ \begin{equation*} z=x_1^2+x_2^2+x_3^2+6 \end{equation*} \]
则
\[ \begin{equation*} \frac{\partial z}{\partial x_1}=2x_1 \end{equation*} \]
\[ \begin{equation*} \frac{\partial z}{\partial x_2}=2x_2 \end{equation*} \]
\[ \begin{equation*} \frac{\partial z}{\partial x_3}=2x_3 \end{equation*} \]
将\(x_1=1.0\),\(x_2=2.0\),\(x_3=3.0\)代入就能够获得
\[ \begin{equation*} (\frac{\partial z}{\partial x_1},\frac{\partial z}{\partial x_2},\frac{\partial z}{\partial x_3})=(2x_1,2x_2,2x_3)=(2.0,4.0,6.0) \end{equation*} \]
结果是和pytorch的输出是同样的.反过来想一想,其实所谓的"标量对向量求导"本质上是函数对各个自变量求导,这里只是把各个自变量当作一个向量.和数学上的定义并不矛盾.

backward的gradient参数做用

如今有以下问题,已知
\[ \begin{equation*} y_1=x_1x_2x_3 \end{equation*} \]
\[ \begin{equation*} y_2=x_1+x_2+x_3 \end{equation*} \]
\[ \begin{equation*} y_3=x_1+x_2x_3 \end{equation*} \]
\[ \begin{equation*} A=f(y_1,y_2,y_3) \end{equation*} \]
其中函数\(f(y_1,y_2,y_3)\)的具体定义未知,如今求
\[ \begin{equation*} \frac{\partial A}{\partial x_1}=? \end{equation*} \]
\[ \begin{equation*} \frac{\partial A}{\partial x_2}=? \end{equation*} \]
\[ \begin{equation*} \frac{\partial A}{\partial x_3}=? \end{equation*} \]
根据参考资料2中讲的多元复合函数的求导法则.
\[ \begin{equation*} \frac{\partial A}{\partial x_1}=\frac{\partial A}{\partial y_1}\frac{\partial y_1}{\partial x_1}+\frac{\partial A}{\partial y_2}\frac{\partial y_2}{\partial x_1}+\frac{\partial A}{\partial y_3}\frac{\partial y_3}{\partial x_1} \end{equation*} \]
\[ \begin{equation*} \frac{\partial A}{\partial x_2}=\frac{\partial A}{\partial y_1}\frac{\partial y_1}{\partial x_2}+\frac{\partial A}{\partial y_2}\frac{\partial y_2}{\partial x_2}+\frac{\partial A}{\partial y_3}\frac{\partial y_3}{\partial x_2} \end{equation*} \]
\[ \begin{equation*} \frac{\partial A}{\partial x_3}=\frac{\partial A}{\partial y_1}\frac{\partial y_1}{\partial x_3}+\frac{\partial A}{\partial y_2}\frac{\partial y_2}{\partial x_3}+\frac{\partial A}{\partial y_3}\frac{\partial y_3}{\partial x_3} \end{equation*} \]
上面3个等式能够写成矩阵相乘的形式.以下
\[ \begin{equation}\label{simple} [\frac{\partial A}{\partial x_1},\frac{\partial A}{\partial x_2},\frac{\partial A}{\partial x_3}]= [\frac{\partial A}{\partial y_1},\frac{\partial A}{\partial y_2},\frac{\partial A}{\partial y_3}] \left[ \begin{matrix} \frac{\partial y_1}{\partial x_1} & \frac{\partial y_1}{\partial x_2} & \frac{\partial A}{\partial x_3} \\ \frac{\partial y_2}{\partial x_1} & \frac{\partial y_2}{\partial x_2} & \frac{\partial A}{\partial x_3} \\ \frac{\partial y_3}{\partial x_1} & \frac{\partial y_3}{\partial x_2} & \frac{\partial A}{\partial x_3} \end{matrix} \right] \end{equation} \]
其中
\[ \begin{equation*} \left[ \begin{matrix} \frac{\partial y_1}{\partial x_1} & \frac{\partial y_1}{\partial x_2} & \frac{\partial y_1}{\partial x_3} \\ \frac{\partial y_2}{\partial x_1} & \frac{\partial y_2}{\partial x_2} & \frac{\partial y_2}{\partial x_3} \\ \frac{\partial y_3}{\partial x_1} & \frac{\partial y_3}{\partial x_2} & \frac{\partial y_3}{\partial x_3} \end{matrix} \right] \end{equation*} \]
叫做雅可比(Jacobian)式.雅可比式能够根据已知条件求出.如今只要知道\([\frac{\partial A}{\partial y_1},\frac{\partial A}{\partial y_2},\frac{\partial A}{\partial y_3}]\)的值,哪怕不知道\(f(y_1,y_2,y_3)\)的具体形式也能求出来\([\frac{\partial A}{\partial x_1},\frac{\partial A}{\partial x_2},\frac{\partial A}{\partial x_3}]\). 那如今的如今的问题是:
怎么样才能求出
\[ \begin{equation*} [\frac{\partial A}{\partial y_1},\frac{\partial A}{\partial y_2},\frac{\partial A}{\partial y_3}] \end{equation*} \]
答案是由pytorch的backward函数的gradient参数提供.这就是gradient参数的做用. 参数gradient能解决什么问题,有什么实际的做用呢?说实话,由于我才接触到pytorch,还真没有见过现实中怎么用gradient参数.可是目前能够经过数学意义来理解,就是能够忽略复合函数某个位置以前的全部函数 的具体形式,直接给定一个梯度来求得对各个自变量的偏导.
上面各个方程用代码表示以下所示:

# coding utf-8
import torch

x1 = torch.tensor(1, requires_grad=True, dtype=torch.float)
x2 = torch.tensor(2, requires_grad=True, dtype=torch.float)
x3 = torch.tensor(3, requires_grad=True, dtype=torch.float)
y = torch.randn(3)
y[0] = x1 * x2 * x3
y[1] = x1 + x2 + x3
y[2] = x1 + x2 * x3
x = torch.tensor([x1, x2, x3])
y.backward(torch.tensor([0.1, 0.2, 0.3], dtype=torch.float))
print(x1.grad)
print(x2.grad)
print(x3.grad)

按照上用的推导方法
\[ \begin{equation*} \begin{split} [\frac{\partial A}{\partial x_1},\frac{\partial A}{\partial x_2},\frac{\partial A}{\partial x_3}] &=[\frac{\partial A}{\partial y_1},\frac{\partial A}{\partial y_2},\frac{\partial A}{\partial y_3}] \left[ \begin{matrix} x_2x_3 & x_1x_3 & x_1x_2 \\ 1 & 1 & 1 \\ 1 & x_3 & x_2 \end{matrix} \right] &=[0.1,0.2,0.3] \left[ \begin{matrix} 6 & 3 & 2 \\ 1 & 1 & 1 \\ 1 & 3 & 2 \end{matrix} \right] &=[1.1,1.4,1.0] \end{split} \end{equation*} \]
和代码的运行结果是同样的.

参考资料

1. 同济大学数学系,高等数学第七版上册,高等教育出版社,p75-76, 2015.
2. 同济大学数学系,高等数学第七版下册,高等教育出版社,p78-80,p88-91, 2015.
3. Calculus,Thirteenth Edition,p822, 2013.
4. 详解Pytorch 自动微分里的（vector-Jacobian product）
5. PyTorch 的 backward 为何有一个 grad_variables 参数？）