pytorch backward使用解析

目录

  • 前言
  • backward函数官方文档
  • backward理解
    • Jacobian矩阵
    • vector-Jacobian product的计算
    • vector-Jacobian product的例子理解
  • 输入和输出为标量或向量时的计算
    • 输入为标量,输出为标量
    • 输入为标量,输出为向量
    • 输入为向量,输出为标量
    • 输入为标量,输出为向量
  • 额外例子:输出为标量,gradient为向量
    • 输入为标量,输出为标量,gradient为向量
    • 输入为向量,输出为标量,gradient为向量
  • 总结

前言

torch版本为v1.13。

backward函数官方文档

torch.Tensor.backward:计算当前tensor相对于图的叶子的梯度。

叶子可以理解为自己创建的变量。使用链式法则,图是可微的。如果张量不是标量,并且需要梯度,该函数还需要指定gradient。它应该是匹配类型和位置的tensor,包含微分函数相对于self的梯度(???)。

这个函数在叶子中累计梯度,在调用它之前,可能需要将.grad属性归零或将它们设置为 None 。

参数:

  • gradient (Tensor or None) – 关于tensor的梯度。如果它是一个tensor,会自动转为不需要grad的Tensor,除非create_graph为True。None值可以指定为标量Tensor或不需要grad的Tensor。如果None值是可接受的,那么该参数是可选参数。
  • retain_graph (bool, optional) – 如果为False,用于计算grads的graph将被释放。注意,在几乎所有情况下,都不需要将此选项设置为 True,而且通常可以以更有效的方式解决。 默认为 create_graph 的值。
  • create_graph (bool, optional) – 默认为False。如果为True,导数的图将会被构建,允许计算更高阶的导数衍生品(derivative products)。
  • inputs (sequence of Tensor) – 将梯度累积到inputs 的 .grad 中。其他的Tensors将会被忽略。如果未提供,则梯度将累积到用于计算 attr::tensors 的所有叶张量中。

backward理解

pytorch backward使用解析_第1张图片

Jacobian矩阵

参考:wolfram-Jacobian
x = ( x 1 , x 2 , ⋯   , x n ) T \bold{x}=(x_1,x_2,\cdots,x_n)^T x=(x1,x2,,xn)T y = f ( x ) \bold{y}=f(\bold{x}) y=f(x),则有:
y = [ y 1 y 2 ⋮ y m ] = [ f 1 ( x ) f 2 ( x ) ⋮ f m ( x ) ] = [ f 1 ( x 1 , x 2 , ⋯   , x n ) f 2 ( x 1 , x 2 , ⋯   , x n ) ⋮ f m ( x 1 , x 2 , ⋯   , x n ) ] \bold{y}= \begin{bmatrix} y_1 \\ y_2 \\ \vdots \\ y_m \\ \end{bmatrix}= \begin{bmatrix} f_1(\bold{x}) \\ f_2(\bold{x}) \\ \vdots \\ f_m(\bold{x}) \\ \end{bmatrix}= \begin{bmatrix} f_1(x_1,x_2,\cdots,x_n) \\ f_2(x_1,x_2,\cdots,x_n) \\ \vdots \\ f_m(x_1,x_2,\cdots,x_n) \\ \end{bmatrix} y= y1y2ym = f1(x)f2(x)fm(x) = f1(x1,x2,,xn)f2(x1,x2,,xn)fm(x1,x2,,xn) ,则 y \bold{y} y 关于 x \bold{x} x 的梯度是一个雅可比矩阵:
J ( x 1 , x 2 , ⋯   , x n ) = ∂ ( y 1 , ⋯   , y m ) ∂ ( x 1 , ⋯   , x n ) = [ ∂ y 1 ∂ x 1 ∂ y 1 ∂ x 2 ⋯ ∂ y 1 ∂ x n ∂ y 2 ∂ x 1 ∂ y 2 ∂ x 2 ⋯ ∂ y 2 ∂ x n ⋮ ⋮ ⋱ ⋮ ∂ y m ∂ x 1 ∂ y m ∂ x 2 ⋯ ∂ y m ∂ x n ] J(x_1,x_2,\cdots,x_n) = {\frac {\partial(y_1,\cdots,y_m)} {\partial(x_1,\cdots,x_n)}}= \begin{bmatrix} \frac{\partial y_{1}}{\partial x_{1}} & \frac{\partial y_{1}}{\partial x_{2}} & \cdots &\frac{\partial y_{1}}{\partial x_{n}} \\ \frac{\partial y_{2}}{\partial x_{1}}& \frac{\partial y_{2}}{\partial x_{2}} & \cdots & \frac{\partial y_{2}}{\partial x_{n}}\\ \vdots &\vdots & \ddots & \vdots \\ \frac{\partial y_{m}}{\partial x_{1}}& \frac{\partial y_{m}}{\partial x_{2}} &\cdots & \frac{\partial y_{m}}{\partial x_{n}} \end{bmatrix} J(x1,x2,,xn)=(x1,,xn)(y1,,ym)= x1y1x1y2x1ymx2y1x2y2x2ymxny1xny2xnym

vector-Jacobian product的计算

参考:
The “gradient” argument in Pytorch’s “backward” function
详解Pytorch 自动微分里的(vector-Jacobian product)-知乎

backward(torch.autograd.backward或Tensor.backward)实现的是vector-Jacobian product,即矢量-雅可比积,Jacobian容易理解,这里的vector(设为 v v v)就是backward的gradient参数,有很多种理解:

  • y i y_i yi x i x_i xi 的偏导数沿 v v v上的投影, v v v的默认方向为 v = ( 1 , 1 , ⋯   , 1 ) v=(1,1,\cdots,1) v=(1,1,,1)
  • 各个分量函数关于 x i x_i xi偏好的权重

值得注意的是, v v v 的维度与输出保持一致,可正可负

所以,vector-Jacobian product的形式为:
v ⋅ J = [ v 1 , v 2 , ⋯   , v m ] ⋅ [ ∂ y 1 ∂ x 1 ∂ y 1 ∂ x 2 ⋯ ∂ y 1 ∂ x n ∂ y 2 ∂ x 1 ∂ y 2 ∂ x 2 ⋯ ∂ y 2 ∂ x n ⋮ ⋮ ⋱ ⋮ ∂ y m ∂ x 1 ∂ y m ∂ x 2 ⋯ ∂ y m ∂ x n ] = [ ∑ i = 0 m ∂ y i ∂ x 1 v i ,     ∑ i = 0 m ∂ y i ∂ x 2 v i ,     ⋯   ,     ∑ i = 0 m ∂ y i ∂ x n v i ] \begin{aligned} \bold{v} \cdot J &= [v_1,v_2,\cdots,v_m]\cdot \begin{bmatrix} \frac{\partial y_{1}}{\partial x_{1}} & \frac{\partial y_{1}}{\partial x_{2}} & \cdots &\frac{\partial y_{1}}{\partial x_{n}} \\ \frac{\partial y_{2}}{\partial x_{1}}& \frac{\partial y_{2}}{\partial x_{2}} & \cdots & \frac{\partial y_{2}}{\partial x_{n}}\\ \vdots &\vdots & \ddots & \vdots \\ \frac{\partial y_{m}}{\partial x_{1}}& \frac{\partial y_{m}}{\partial x_{2}} &\cdots & \frac{\partial y_{m}}{\partial x_{n}} \end{bmatrix}\\ &= \begin{bmatrix} \sum_{i=0}^{m}{\frac {\partial y_i} {\partial x_1}v_i},\,\,\, \sum_{i=0}^{m}{\frac {\partial y_i} {\partial x_2}v_i},\,\,\, \cdots,\,\,\, \sum_{i=0}^{m}{\frac {\partial y_i} {\partial x_n}v_i} \end{bmatrix} \end{aligned} vJ=[v1,v2,,vm] x1y1x1y2x1ymx2y1x2y2x2ymxny1xny2xnym =[i=0mx1yivi,i=0mx2yivi,,i=0mxnyivi]

这就是输出进行backward之后,叶子张量的.grad值。

vector-Jacobian product的例子理解

y = x 2 y=x^2 y=x2为例进行解释:
代码1:

x = torch.tensor([1,2,3.], requires_grad=True)
y = x**2
y.backward(gradient=torch.tensor([1,1,1.]))
print(x.grad)
"""
输出:
tensor([2., 4., 6.])
"""

代码2:

x = torch.tensor([1,2,3.], requires_grad=True)
y = x**2
y.backward(gradient=torch.tensor([10,-10,20.]))
print(x.grad)
"""
输出:
tensor([ 20., -40., 120.])
"""

可以发现,代码2相比于代码1,结果放大了gradient倍。设上述代码中的 x = ( x 1 , x 2 , x 3 ) \bold{x}=(x_1,x_2,x_3) x=(x1,x2,x3),则 y = ( y 1 , y 2 , y 3 ) = ( x 1 2 , x 2 2 , x 3 2 ) \bold{y}=(y_1,y_2,y_3)=(x_1^2,x_2^2,x_3^2) y=(y1,y2,y3)=(x12,x22,x32),gradient为 v = ( v 1 , v 2 , v 3 ) \bold{v}=(v_1,v_2,v_3) v=(v1,v2,v3)。使用vector-Jacobian product公式可得:

v ⋅ J = [ v 1 , v 2 , v 3 ] [ ∂ y 1 ∂ x 1 ∂ y 1 ∂ x 2 ∂ y 1 ∂ x n ∂ y 2 ∂ x 1 ∂ y 2 ∂ x 2 ∂ y 2 ∂ x n ∂ y 3 ∂ x 1 ∂ y 3 ∂ x 2 ∂ y 3 ∂ x n ] = [ ∑ i = 0 3 ∂ y i ∂ x 1 v i ,     ∑ i = 0 3 ∂ y i ∂ x 2 v i ,     ∑ i = 0 3 ∂ y i ∂ x 3 v i ] = [ 2 x 1 v 1 , 2 x 2 v 2 , 2 x 3 v 3 ] = [ 2 v 1 , 4 v 2 , 6 v 3 ] \begin{aligned} \bold{v} \cdot J &= [v_1,v_2,v_3] \begin{bmatrix} \frac{\partial y_{1}}{\partial x_{1}} & \frac{\partial y_{1}}{\partial x_{2}} &\frac{\partial y_{1}}{\partial x_{n}} \\ \frac{\partial y_{2}}{\partial x_{1}}& \frac{\partial y_{2}}{\partial x_{2}} & \frac{\partial y_{2}}{\partial x_{n}}\\ \frac{\partial y_{3}}{\partial x_{1}}& \frac{\partial y_{3}}{\partial x_{2}} & \frac{\partial y_{3}}{\partial x_{n}} \end{bmatrix}\\ &= \begin{bmatrix} \sum_{i=0}^{3}{\frac {\partial y_i} {\partial x_1}v_i},\,\,\, \sum_{i=0}^{3}{\frac {\partial y_i} {\partial x_2}v_i},\,\,\, \sum_{i=0}^{3}{\frac {\partial y_i} {\partial x_3}v_i} \end{bmatrix} \\ &=\begin{bmatrix}2x_1v_1,2x_2v_2,2x_3v_3\end{bmatrix} \\ &=\begin{bmatrix}2v_1,4v_2,6v_3\end{bmatrix} \end{aligned} vJ=[v1,v2,v3] x1y1x1y2x1y3x2y1x2y2x2y3xny1xny2xny3 =[i=03x1yivi,i=03x2yivi,i=03x3yivi]=[2x1v1,2x2v2,2x3v3]=[2v1,4v2,6v3]

分别将 v = ( 1 , 1 , 1. ) \bold{v}=(1,1,1.) v=(1,1,1.) v = ( 10 , − 10 , 20. ) \bold{v}=(10,-10,20.) v=(10,10,20.) 带入可得代码结果。

输入和输出为标量或向量时的计算

输入为标量,输出为标量

代码:

x = torch.tensor(2., requires_grad=True)
y = x**2+x
y.backward(gradient=torch.tensor(1.))
print(x.grad)
"""
输出:
tensor([5.])
"""

解释:
x = x 1 \bold{x}=x_1 x=x1,则 y = y 1 = x 1 2 + x 1 \bold{y}=y_1=x_1^2+x_1 y=y1=x12+x1,gradient为 v = v 1 \bold{v}=v_1 v=v1,则:
v ⋅ J = [ v 1 ] ⋅ [ ∂ y 1 ∂ x 1 ] = [ v 1 ] ⋅ [ 2 x 1 + 1 ] = [ 1 ] ⋅ [ 2 × 2 + 1 ] = 5 \begin{aligned} \bold{v} \cdot J &= [v_1]\cdot[\frac {\partial y_1} {\partial x_1}]\\ &=[v_1]\cdot[2x_1+1]\\ &=[1]\cdot[2\times2+1]\\ &=5 \end{aligned} vJ=[v1][x1y1]=[v1][2x1+1]=[1][2×2+1]=5

输入为标量,输出为向量

代码:

x = torch.tensor(1., requires_grad=True)
y = torch.empty(2)
y[0] = x**2
y[1] = x**3
y.backward(gradient=torch.tensor([1,2.]))
print(x.grad)
"""
输出:
tensor(8.)
"""

解释:
x = x 1 \bold{x}=x_1 x=x1,则 y = [ y 1 , y 2 ] = [ x 1 2 , x 1 3 ] \bold{y}=[y_1,y_2]=[x_1^2,x_1^3] y=[y1,y2]=[x12,x13],gradient为 v = [ v 1 , v 2 ] \bold{v}=[v_1,v_2] v=[v1,v2],则:
v ⋅ J = [ v 1 , v 2 ] ⋅ [ ∂ y 1 ∂ x 1 ∂ y 2 ∂ x 1 ] = [ v 1 , v 2 ] ⋅ [ 2 x 1 3 x 1 2 ] = [ 1 , 2 ] ⋅ [ 2 3 ] = 8 \begin{aligned} \bold{v} \cdot J &= [v_1,v_2]\cdot \begin{bmatrix} \frac {\partial y_1} {\partial x_1} \\ \frac {\partial y_2} {\partial x_1} \end{bmatrix}\\ &= [v_1,v_2]\cdot \begin{bmatrix} 2x_1 \\ 3x_1^2 \end{bmatrix}\\ &= [1,2]\cdot \begin{bmatrix} 2 \\ 3 \end{bmatrix}\\ &=8 \end{aligned} vJ=[v1,v2][x1y1x1y2]=[v1,v2][2x13x12]=[1,2][23]=8


输入为向量,输出为标量

代码:

x = torch.tensor([1.,2,3], requires_grad=True)
y = torch.sum(x**2)
y.backward()
print(x.grad)
"""
输出:
tensor([2., 4., 6.])
"""

解释:
x = [ x 1 , x 2 , x 3 ] \bold{x}=[x_1,x_2,x_3] x=[x1,x2,x3],则 y = y 1 = x 1 2 + x 2 2 + x 3 2 \bold{y}=y_1=x_1^2+x_2^2+x_3^2 y=y1=x12+x22+x32,gradient为 v = [ v 1 ] \bold{v}=[v_1] v=[v1],则:
v ⋅ J = [ v 1 ] ⋅ [ ∂ y 1 ∂ x 1 ∂ y 1 ∂ x 2 ∂ y 1 ∂ x 3 ] = [ v 1 ] ⋅ [ 2 x 1 2 x 2 2 x 3 ] = [ 1 ] ⋅ [ 2 4 6 ] = [ 2 4 6 ] \begin{aligned} \bold{v} \cdot J &= [v_1]\cdot \begin{bmatrix} \frac {\partial y_1} {\partial x_1} & \frac {\partial y_1} {\partial x_2} & \frac {\partial y_1} {\partial x_3} \end{bmatrix}\\ &= [v_1]\cdot \begin{bmatrix} 2x_1 & 2x_2 & 2x_3 \end{bmatrix}\\ &= [1]\cdot \begin{bmatrix} 2&4&6 \end{bmatrix}\\ &= \begin{bmatrix} 2&4&6 \end{bmatrix}\\ \end{aligned} vJ=[v1][x1y1x2y1x3y1]=[v1][2x12x22x3]=[1][246]=[246]

输入为标量,输出为向量

参见上一节 vector-Jacobian product的例子理解

额外例子:输出为标量,gradient为向量

在上一节的例子中,gradient的维度与输出维度保持一致,本节探索gradient的维度与输出维度不一致的情况。

输入为标量,输出为标量,gradient为向量

x = torch.tensor(2., requires_grad=True)
y = x**2+x
y.backward(gradient=torch.tensor([1.,10]))
print(x.grad)

结果:报错

RuntimeError: Mismatch in shape: grad_output[0] has a shape of torch.Size([4]) and output[0] has a shape of torch.Size([]).

输入为向量,输出为标量,gradient为向量

x = torch.tensor([1.,2,3], requires_grad=True)
y = torch.sum(x**2)
y.backward(gradient=torch.tensor([1,2.]))
print(x.grad)

结果:报错

RuntimeError: Mismatch in shape: grad_output[0] has a shape of torch.Size([2]) and output[0] has a shape of torch.Size([]).

总结

  • gradient默认为1,其维度应与输出维度保持一致;
  • gradient类似于在梯度前面乘以一个动量,类似于学习率,不过可正可负(个人理解);

■ \blacksquare

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