【PyTorch】自动微分torch.auotograd()——.backword()分析

一、简介

torch.auotograd模块可实现任意标量值函数自动求导的类和函数。针对一个张量只需要设置参数requires_grad=True，通过相关计算即可输出其在传播过程中的梯度（导数）信息。

二、案例分析

分析1

在PyTorch中生成一个矩阵张量x，且$y=sum(x^2+2x+1)$，计算出y在x上的导数，程序如下：

import torch x = torch.tensor([[1.0, 2], [3, 4]], requires_grad=True) y = torch.sum(x ** 2 + 2 * x + 1) y.backward() print(x.grad) >>>tensor([[ 4., 6.], [ 8., 10.]])

首先，生成$[2\times 2]$维度的张量$x$，并计算$y=sum(x^2+2x+1)$，计算出的$y$是标量(scalar value)。 $$y=(x_{11}^2+2x_{11}+1)+(x_{13}^2+2x_{13}+1)+(x_{31}^2+2x_{31}+1)+(x_2^2+2x_{22}+1)$$

此时使用y.backward()即可自动计算出$y$在$x$每个元素上的导数。即： $$\left[\begin{array}{cc}\frac{\partial y}{\partial x_{11}}& \frac{\partial y}{\partial x_{12}}\\ \frac{\partial y}{\partial x_{21}}& \frac{\partial y}{\partial x_{22}}\end{array}\right]=\left[\begin{array}{cc}2x_{11}+2& 2x_{12}+2\\ 2x_{21}+2& 2x_{22}+2\end{array}\right]$$ 通过计算即可得出如下结果：

[[ 4., 6.], [ 8., 10.]]

分析2

为什么要使用sum()呢？可不可以去掉sum()呢？像这样： $$y=x^2+2x+1$$

尝试一下：

import torch x = torch.tensor([[1.0, 2], [3, 4]], requires_grad=True) y = x ** 2 + 2 * x + 1 y.backward() print(x.grad)

此时会报错，其含义大概就是.backward()只能对标量使用。 经过查阅相关内容，对代码进行改进：

import torch x = torch.tensor([[1.0, 2], [3, 4]], requires_grad=True) y = x ** 2 + 2 * x + 1 y.backward(gradient=torch.tensor([[1.0, 1], [1, 1]])) print(x.grad) >>>tensor([[ 4., 6.], [ 8., 10.]])

在y.backward()内添加一个$[2\times 2]$的单位向量，此时代码可成功运行。可以这么理解： $$\left[\begin{array}{cc}{y}_{11}& y_{12}\\ y_{21}& y_{22}\end{array}\right]=\left[\begin{array}{cc}(x_{11}^2+2x_{11}+1)& (x_{12}^2+2x_{12}+1)\\ (x_{21}^2+2x_{21}+1)& (x_{22}^2+2x_{22}+1)\end{array}\right]$$ y已经不再是一个标量，而是一个矩阵，此时的求导是对应元素分别求导。 $$\left[\begin{array}{cc}\frac{\partial y_{11}}{\partial x_{11}}& \frac{\partial y_{12}}{\partial x_{12}}\\ \frac{\partial y_{21}}{\partial x_{21}}& \frac{\partial y_{22}}{\partial x_{22}}\end{array}\right]=\left[\begin{array}{cc}2x_{11}+2& 2x_{12}+2\\ 2x_{21}+2& 2x_{22}+2\end{array}\right]$$ 如果将单位矩阵乘2：y.backward(gradient=torch.tensor([[2.0, 2], [2, 2]]))

import torch x = torch.tensor([[1.0, 2], [3, 4]], requires_grad=True) y = x ** 2 + 2 * x + 1 y.backward(gradient=torch.tensor([[2.0, 2], [2, 2]])) print(x.grad) >>>tensor([[ 8., 12.], [16., 20.]])

相当于： $$\left[\begin{array}{cc}{y}_{11}& y_{12}\\ y_{21}& y_{22}\end{array}\right]=\left[\begin{array}{cc}2(x_{11}^2+2x_{11}+1)& 2(x_{12}^2+2x_{12}+1)\\ 2(x_{21}^2+2x_{21}+1)& 2(x_{22}^2+2x_{22}+1)\end{array}\right]$$ 所以其导数也将是原值的二倍。 $$\left[\begin{array}{cc}\frac{\partial y_{11}}{\partial x_{11}}& \frac{\partial y_{12}}{\partial x_{12}}\\ \frac{\partial y_{21}}{\partial x_{21}}& \frac{\partial y_{22}}{\partial x_{22}}\end{array}\right]=\left[\begin{array}{cc}4x_{11}+4& 4x_{12}+4\\ 4x_{21}+4& 4x_{22}+4\end{array}\right]$$

分析3

想象力丰富一点，将前面两个案例相结合。

import torch x = torch.tensor([[1.0, 2], [3, 4]], requires_grad=True) y = x ** 2 + 2 * x + 1 z = torch.sum(x ** 2 + 2 * x + 1) y.backward(gradient=torch.tensor([[1.0, 1], [1, 1]])) z.backward() print(x.grad) >>>tensor([[ 8., 12.], [16., 20.]])

以此对y、z进行反向传播求导，其输出结果是两次求导的和。求导公式应为： $$\left[\begin{array}{cc}\frac{\partial y}{\partial x_{11}}+\frac{\partial y_{11}}{\partial x_{11}}& \frac{\partial y}{\partial x_{12}}+\frac{\partial y_{12}}{\partial x_{12}}\\ \frac{\partial y}{\partial x_{21}}+\frac{\partial y_{21}}{\partial x_{21}}& \frac{\partial y}{\partial x_{22}}+\frac{\partial y_{22}}{\partial x_{22}}\end{array}\right]=\left[\begin{array}{cc}2x_{11}+2& 2x_{12}+2\\ 2x_{21}+2& 2x_{22}+2\end{array}\right]+\left[\begin{array}{cc}2x_{11}+2& 2x_{12}+2\\ 2x_{21}+2& 2x_{22}+2\end{array}\right]$$