Pytorch Tutorial【Chapter 2. Autograd】

Pytorch Tutorial

Chapter 2. Autograd

1. Review Matrix Calculus

1.1 Definition向量对向量求导

​ Define the derivative of a function mapping $f:{\mathbb{R}}^n\to {\mathbb{R}}^m$ as the $n\times m$ matrix of partial derivatives. That is, if $x\in {\mathbb{R}}^n,f(x)\in {\mathbb{R}}^m$, the derivative of $f$ with respect to $x$ is defined as
$${\left[\begin{array}{c}\frac{\partial f}{\partial x}\end{array}\right]}_{ij}=\frac{\partial f_i}{\partial x_i}$$
Let
x = [ x 1 x 2 ⋮ x n ] , f ( x ) = [ f 1 ( x ) f 2 ( x ) ⋮ f m ( x ) ] x = \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix}, f(x) = \begin{bmatrix} f_1(x) \\ f_2(x) \\ \vdots \\ f_m(x) \end{bmatrix} x= ​x1​x2​⋮xn​​ ​,f(x)= ​f1​(x)f2​(x)⋮fm​(x)​ ​

then we define the Jacobian Matrix

∂ f ∂ x = [ ∂ f 1 ∂ x 1 ∂ f 2 ∂ x 1 ⋯ ∂ f m ∂ x 1 ∂ f 1 ∂ x 2 ∂ f 2 ∂ x 2 ⋯ ∂ f m ∂ x 2 ⋮ ⋮ ⋱ ⋮ ∂ f 1 ∂ x n ∂ f 2 ∂ x n ⋯ ∂ f m ∂ x n ] \frac{\partial f}{\partial x} = \begin{bmatrix} \frac{\partial f_1}{\partial x_1} & \frac{\partial f_2}{\partial x_1} & \cdots & \frac{\partial f_m}{\partial x_1} \\ \frac{\partial f_1}{\partial x_2} & \frac{\partial f_2}{\partial x_2} & \cdots & \frac{\partial f_m}{\partial x_2} \\ \vdots & \vdots & \ddots & \vdots \\ \frac{\partial f_1}{\partial x_n} & \frac{\partial f_2}{\partial x_n} & \cdots & \frac{\partial f_m}{\partial x_n} \\ \end{bmatrix} ∂x∂f​= ​∂x1​∂f1​​∂x2​∂f1​​⋮∂xn​∂f1​​​∂x1​∂f2​​∂x2​∂f2​​⋮∂xn​∂f2​​​⋯⋯⋱⋯​∂x1​∂fm​​∂x2​∂fm​​⋮∂xn​∂fm​​​ ​

1.2 Definition标量对向量求导

If $f$ is scalar, one has

∂ f ∂ x = [ ∂ f ∂ x 1 ∂ f ∂ x 2 ⋮ ∂ f ∂ x n ] \frac{\partial f}{\partial x} = \begin{bmatrix} \frac{\partial f}{\partial x_1} \\ \frac{\partial f}{\partial x_2} \\ \vdots \\ \frac{\partial f}{\partial x_n} \\ \end{bmatrix} ∂x∂f​= ​∂x1​∂f​∂x2​∂f​⋮∂xn​∂f​​ ​
这其实是一种分母布局

1.3 Definition标量对矩阵求导

​ Now we give some results on the derivative of scalar functions of a matrix. Let $X=[x_{ij}]$ be a matrix of order $m\times n$ and let $y=f(X)$ be a scalar function of $X$. The derivative of $y$ with respect to $X$, denoted by $\frac{\partial y}{\partial X}$, is defined as the following matrix of order $m\times n$,
G = ∂ y ∂ X = [ ∂ y ∂ x 11 ∂ y ∂ x 12 ⋯ ∂ y ∂ x 1 n ∂ y ∂ x 21 ∂ y ∂ x 22 ⋯ ∂ y ∂ x 2 n ⋮ ⋮ ⋱ ⋮ ∂ y ∂ x m 1 ∂ y ∂ x m 2 ⋯ ∂ y ∂ x m n ] = [ ∂ y ∂ x i j ] G = \frac{\partial y}{\partial X} = \begin{bmatrix} \frac{\partial y}{\partial x_{11}} & \frac{\partial y}{\partial x_{12}} & \cdots & \frac{\partial y}{\partial x_{1n}} \\ \frac{\partial y}{\partial x_{21}} & \frac{\partial y}{\partial x_{22}} & \cdots & \frac{\partial y}{\partial x_{2n}} \\ \vdots & \vdots & \ddots & \vdots& \\ \frac{\partial y}{\partial x_{m1}} & \frac{\partial y}{\partial x_{m2}} & \cdots & \frac{\partial y}{\partial x_{mn}} \end{bmatrix} = \Big[\frac{\partial y}{\partial x_{ij}} \Big] G=∂X∂y​= ​∂x11​∂y​∂x21​∂y​⋮∂xm1​∂y​​∂x12​∂y​∂x22​∂y​⋮∂xm2​∂y​​⋯⋯⋱⋯​∂x1n​∂y​∂x2n​∂y​⋮∂xmn​∂y​​​ ​=[∂xij​∂y​]

2.关于autograd的说明

torch.Tensor 是包的核心类。如果将其属性 tensor.requires_grad 设置为 True，则会开始跟踪针对 tensor 的所有操作。完成计算后，您可以调用 tensor.backward() 来自动计算所有梯度。该张量的梯度将累积到 tensor.grad 属性中。

要停止 tensor 历史记录的跟踪，您可以调用 tensor.detach()，它将其与计算历史记录分离，并防止将来的计算被跟踪。

要停止跟踪历史记录（和使用内存），您还可以将代码块使用 with torch.no_grad(): 包装起来。在评估模型时，这是特别有用，因为模型在训练阶段具有 requires_grad = True 的可训练参数有利于调参，但在评估阶段我们不需要梯度。

还有一个类对于 autograd 实现非常重要那就是 Function。Tensor 和 Function 互相连接并构建一个非循环图，它保存整个完整的计算过程的历史信息。每个张量都有一个 tensor.grad_fn 属性保存着创建了张量的 Function 的引用，（如果用户自己创建张量，则 grad_fn=None）。

如果你想计算导数，你可以调用 tensor.backward()。如果 Tensor 是标量（即它包含一个元素数据），则不需要指定任何参数backward()，但是如果它有更多元素，则需要指定一个gradient 参数来指定张量的形状。

最后的计算结果保存在tensor.grad属性里

• 使用tensor.requires_grad在初始化时，设置跟踪梯度

import torch
import numpy as np

x = torch.ones(2,2, requires_grad=True)
print(x)

结果如下

tensor([[1., 1.],
        [1., 1.]], requires_grad=True)

• 设置了跟踪梯度的tensor,将会出现tensor.grad_fn的属性，用于记录上次计算的Function

y = torch.add(x, 1)
print(y)
print(y.grad_fn)

结果如下

tensor([[2., 2.],
        [2., 2.]], grad_fn=<AddBackward0>)
<AddBackward0 object at 0x0000020D723EBE80>

• tensor.requires_grad_(True / False) 会改变张量的 requires_grad 标记。 如果没有提供相应的参数输入的标记默认为 False。

a = torch.randn(2,2)
a = (a * 3) / (a-1)
print(a)
a.requires_grad_(True)
print(a)
a = a + 1
print(a)

tensor([[  0.0646, -46.3478],
        [  5.6683,  -0.8896]])
tensor([[  0.0646, -46.3478],
        [  5.6683,  -0.8896]], requires_grad=True)
tensor([[  1.0646, -45.3478],
        [  6.6683,   0.1104]], grad_fn=<AddBackward0>)

3. grad的计算

3.1 Manual手动计算

• 可以使用函数torch.autograd.grad()来手动计算梯度，详细可参考此处

例如计算 $y=x_1^2+x_2^2+x_1{x}_2$的梯度

x1 = torch.tensor(3., requires_grad=True)
x2 = torch.tensor(1., requires_grad=True)
y = x1**2+x2**2+x1*x2

# 求一阶导数
# torch.autograd.grad(y, x1,retain_graph=True, create_graph=True)
x1_1 = torch.autograd.grad(y, x1, retain_graph=True, create_graph=True)[0]
x2_1 = torch.autograd.grad(y, x2, retain_graph=True, create_graph=True)[0]
print(x1_1,x2_1)

# 求二阶混合偏导数
x1_11 = torch.autograd.grad(x1_1, x1)[0]
x1_12 = torch.autograd.grad(x1_1, x2)[0]
x2_21 = torch.autograd.grad(x2_1, x1)[0]
x2_22 = torch.autograd.grad(x2_1, x2)[0]
print(x1_11,x1_12,x2_21,x2_22)

结果如下

tensor(7., grad_fn=<AddBackward0>) tensor(5., grad_fn=<AddBackward0>)
tensor(2.) tensor(1.) tensor(1.) tensor(2.)

3.2 backward()自动计算

当输出是标量scalar函数时
考虑如下的计算问题

x = torch.ones(2,2, requires_grad=True)
y = x + 2
print(y)
z = y * y * 3
out = z.mean()
print(z, out)  #输出out是一个标量
out.backward()
print(x.grad)

输出是

tensor([[3., 3.],
        [3., 3.]], grad_fn=<AddBackward0>)
tensor([[27., 27.],
        [27., 27.]], grad_fn=<MulBackward0>) tensor(27., grad_fn=<MeanBackward0>)
tensor([[4.5000, 4.5000],
        [4.5000, 4.5000]])

$$X=\left[\begin{array}{cc}{x}_1& x_2\\ x_3& x_4\end{array}\right]=\left[\begin{array}{cc}1& 1\\ 1& 1\end{array}\right]$$

中间变量是

$$Z=\left[\begin{array}{cc}{z}_1& z_2\\ z_3& z_4\end{array}\right]=\left[\begin{array}{cc}3(x_1+2{)}^2& 3(x_1+2{)}^2\\ 3(x_1+2{)}^2& 3(x_1+2{)}^2\end{array}\right]$$

最后获得是输出是

$$\begin{array}{rl}\text{out}& =\frac{1}{4}\sum _{i=1}{z}_i=\frac{1}{4}(z_1+z_2+z_3+z_4)\\ & =\frac{1}{4}(3(x_1+2{)}^2+3(x_2+2{)}^2+3(x_3+2{)}^2+3(x_4+2{)}^2)\\ & =f(\mathrm{x})\end{array}$$

其中将矩阵$X$和矩阵$Z$中的所有元素拼接为向量

$$\begin{gathered} \mathrm{x}=[x_1,x_2,x_3,x_4{]}^T \\ \mathrm{z}=[z_1,z_2,z_3,z_4{]}^T \end{gathered}$$

我们利用矩阵求导的链式法则

$$\frac{\partial f}{\partial \mathrm{x}}=\frac{f(\mathrm{x})}{\partial \mathrm{x}}=\frac{\partial \mathrm{z}}{\partial \mathrm{x}}\frac{\partial f(\mathrm{x})}{\partial \mathrm{z}}$$

再利用标量函数对矩阵导数的定义，则有

∂ f ∂ x = [ ∂ z 1 ∂ x 1 ∂ z 2 ∂ x 1 ∂ z 3 ∂ x 1 ∂ z 4 ∂ x 1 ∂ z 1 ∂ x 2 ∂ z 2 ∂ x 2 ∂ z 3 ∂ x 2 ∂ z 4 ∂ x 2 ∂ z 1 ∂ x 3 ∂ z 2 ∂ x 3 ∂ z 3 ∂ x 3 ∂ z 4 ∂ x 3 ∂ z 1 ∂ x 4 ∂ z 2 ∂ x 4 ∂ z 3 ∂ x 4 ∂ z 4 ∂ x 4 ] [ ∂ f ∂ z 1 ∂ f ∂ z 2 ∂ f ∂ z 3 ∂ f ∂ z 4 ] = [ 6 ( x 1 + 2 ) 0 0 0 0 6 ( x 2 + 2 ) 0 0 0 0 6 ( x 3 + 2 ) 0 0 0 0 6 ( x 4 + 2 ) ] [ 1 4 1 4 1 4 1 4 ] = [ 4.5 4.5 4.5 4.5 ] \frac{\partial f}{\partial \mathrm{x}} = \begin{bmatrix} \frac{\partial z_1}{\partial x_1} & \frac{\partial z_2}{\partial x_1} & \frac{\partial z_3}{\partial x_1} & \frac{\partial z_4}{\partial x_1} \\ \frac{\partial z_1}{\partial x_2} & \frac{\partial z_2}{\partial x_2} & \frac{\partial z_3}{\partial x_2} & \frac{\partial z_4}{\partial x_2} \\ \frac{\partial z_1}{\partial x_3} & \frac{\partial z_2}{\partial x_3} & \frac{\partial z_3}{\partial x_3} & \frac{\partial z_4}{\partial x_3} \\ \frac{\partial z_1}{\partial x_4} & \frac{\partial z_2}{\partial x_4} & \frac{\partial z_3}{\partial x_4} & \frac{\partial z_4}{\partial x_4} \end{bmatrix} \begin{bmatrix} \frac{\partial f}{\partial z_1} \\ \frac{\partial f}{\partial z_2} \\ \frac{\partial f}{\partial z_3} \\ \frac{\partial f}{\partial z_4} \end{bmatrix}= \begin{bmatrix} 6(x_1+2) & 0 & 0 & 0 \\ 0 & 6(x_2+2) & 0 & 0 \\ 0 & 0 & 6(x_3+2) & 0 \\ 0 & 0 & 0 & 6(x_4+2) \end{bmatrix} \begin{bmatrix} \frac{1}{4} \\ \frac{1}{4} \\ \frac{1}{4} \\ \frac{1}{4} \end{bmatrix} = \begin{bmatrix} 4.5 \\ 4.5 \\ 4.5 \\ 4.5 \\ \end{bmatrix} ∂x∂f​= ​∂x1​∂z1​​∂x2​∂z1​​∂x3​∂z1​​∂x4​∂z1​​​∂x1​∂z2​​∂x2​∂z2​​∂x3​∂z2​​∂x4​∂z2​​​∂x1​∂z3​​∂x2​∂z3​​∂x3​∂z3​​∂x4​∂z3​​​∂x1​∂z4​​∂x2​∂z4​​∂x3​∂z4​​∂x4​∂z4​​​ ​ ​∂z1​∂f​∂z2​∂f​∂z3​∂f​∂z4​∂f​​ ​= ​6(x1​+2)000​06(x2​+2)00​006(x3​+2)0​0006(x4​+2)​ ​ ​41​41​41​41​​ ​= ​4.54.54.54.5​ ​

所以最后获得关于的矩阵$X$的导数为

$$\frac{\partial f}{\partial X}=\left[\begin{array}{cc}\frac{\partial f}{\partial x_1}& \frac{\partial f}{\partial x_2}\\ \frac{\partial f}{\partial x_3}& \frac{\partial f}{\partial x_4}\end{array}\right]=\left[\begin{array}{cc}4.5& 4.5\\ 4.5& 4.5\end{array}\right]$$

当输出是张量tensor函数时

x = torch.tensor([[1.0, 2, 3],[4, 5, 6],[7, 8, 9]], requires_grad=True)
w = torch.tensor([[1.0, 2, 3],[4, 5, 6]], requires_grad=True)

y = torch.matmul(x,w.T)
print(y)
print(torch.ones_like(y))
y.backward(gradient = torch.ones_like(y))
print(x.grad)

输出是

tensor([[ 14.,  32.],
        [ 32.,  77.],
        [ 50., 122.]], grad_fn=<MmBackward0>)
tensor([[1., 1.],
        [1., 1.],
        [1., 1.]])
tensor([[5., 7., 9.],
        [5., 7., 9.],
        [5., 7., 9.]])

Y = X W T [ y 11 y 21 y 12 y 22 y 13 y 23 ] = [ x 11 x 12 x 13 x 21 x 22 x 23 x 31 x 32 x 33 ] [ w 11 w 21 w 12 w 22 w 13 w 23 ] [ y 11 y 21 y 12 y 22 y 13 y 23 ] = [ ( x 11 w 11 + x 12 w 12 + x 13 w 13 ) ( x 11 w 21 + x 12 w 22 + x 13 w 23 ) ( x 21 w 11 + x 22 w 12 + x 23 w 13 ) ( x 21 w 21 + x 22 w 22 + x 23 w 23 ) ( x 31 w 11 + x 32 w 12 + x 33 w 13 ) ( x 31 w 21 + x 32 w 22 + x 33 w 23 ) ] Y = XW^T \\ \begin{bmatrix} y_{11} & y_{21} \\ y_{12} & y_{22} \\ y_{13} & y_{23} \\ \end{bmatrix} = \begin{bmatrix} x_{11} & x_{12} & x_{13} \\ x_{21} & x_{22} & x_{23} \\ x_{31} & x_{32} & x_{33} \\ \end{bmatrix} \begin{bmatrix} w_{11} & w_{21} \\ w_{12} & w_{22} \\ w_{13} & w_{23} \\ \end{bmatrix} \\ \begin{bmatrix} y_{11} & y_{21} \\ y_{12} & y_{22} \\ y_{13} & y_{23} \\ \end{bmatrix} = \begin{bmatrix} (x_{11}w_{11} + x_{12}w_{12} + x_{13}w_{13}) & (x_{11}w_{21} + x_{12}w_{22} + x_{13}w_{23}) \\ (x_{21}w_{11} + x_{22}w_{12} + x_{23}w_{13}) & (x_{21}w_{21} + x_{22}w_{22} + x_{23}w_{23}) \\ (x_{31}w_{11} + x_{32}w_{12} + x_{33}w_{13}) & (x_{31}w_{21} + x_{32}w_{22} + x_{33}w_{23}) \\ \end{bmatrix} Y=XWT ​y11​y12​y13​​y21​y22​y23​​ ​= ​x11​x21​x31​​x12​x22​x32​​x13​x23​x33​​ ​ ​w11​w12​w13​​w21​w22​w23​​ ​ ​y11​y12​y13​​y21​y22​y23​​ ​= ​(x11​w11​+x12​w12​+x13​w13​)(x21​w11​+x22​w12​+x23​w13​)(x31​w11​+x32​w12​+x33​w13​)​(x11​w21​+x12​w22​+x13​w23​)(x21​w21​+x22​w22​+x23​w23​)(x31​w21​+x32​w22​+x33​w23​)​ ​

gradient=torch.ones_like(y)用于指定矩阵$Y$中每一项的权重都为1，由矩阵$Y$中元素加权得到的scalar函数为

$$\begin{array}{rl}f(\mathrm{x},\mathrm{w})& =1\times y_{11}+1\times y_{12}+1\times y_{13}+1\times y_{21}+1\times y_{22}+1\times y_{23},\\ & \mathrm{x}=[x_{11},x_{12},x_{13},x_{21},x_{22},x_{23},x_{31},x_{32},x_{33}{]}^T\\ & \mathrm{w}=[w_{11},w_{12},w_{13},w_{21},w_{22},w_{23},w_{31},w_{32},w_{33}{]}^T\end{array}$$

这里不包括复合求导，可以直接计算

$$\begin{array}{rl}\frac{\partial f}{\partial \mathrm{x}}& =[\frac{\partial f}{\partial x_{11}},\frac{\partial f}{\partial x_{12}},\frac{\partial f}{\partial x_{13}},\frac{\partial f}{\partial x_{21}},\frac{\partial f}{\partial x_{22}},\frac{\partial f}{\partial x_{23}},\frac{\partial f}{\partial x_{31}},\frac{\partial f}{\partial x_{32}},\frac{\partial f}{\partial x_{33}}{]}^T\\ \frac{\partial f}{\partial \mathrm{x}}& =[w_{11}+w_{21},w_{12}+w_{22},w_{13}+w_{23},w_{11}+w_{21},w_{12}+w_{22},w_{13}+w_{23},w_{11}+w_{21},w_{12}+w_{22},w_{13}+w_{23}{]}^T\\ & =[5,7,9,5,7,9,5,7,9{]}^T\end{array}$$
再写成矩阵的形式则有

∂ f ∂ X = [ 5 7 9 5 7 9 5 7 9 ] \frac{\partial f}{\partial X} = \begin{bmatrix} 5 & 7 & 9 \\ 5 & 7 & 9 \\ 5 & 7 & 9 \\ \end{bmatrix} ∂X∂f​= ​555​777​999​ ​

再考虑一个更一般求二阶导的情况

x = torch.ones(3, requires_grad=True)
print(x)
y = x * 2
print(y)
z = y * 2
print(z)
v = torch.tensor([0.1, 1.0, 0.0001], dtype=torch.float)
z.backward(gradient=v)
print(x.grad)

结果如下

tensor([1., 1., 1.], requires_grad=True)
tensor([2., 2., 2.], grad_fn=<MulBackward0>)
tensor([4., 4., 4.], grad_fn=<MulBackward0>)
tensor([4.0000e-01, 4.0000e+00, 4.0000e-04])

其中
$$\begin{array}{rl}\mathrm{x}& =[x_1,x_2,x_3{]}^T=[1,1,1{]}^T\\ \mathrm{y}=2\mathrm{x}& =[y_1,y_2,y_3{]}^T=[2,2,2{]}^T\\ \mathrm{z}=2\mathrm{y}& =[z_1,z_2,z_3{]}^T=[4,4,4{]}^T\end{array}$$
若考虑gradient=torch.tensor([a1,a2,a3], dtype=torch.folat)，那么最终加权得到的scalar函数为
$$f=a_1{z}_1+a_2{z}_2+a_3{z}_3$$
那么对$\mathrm{x}$求偏导则有
∂ f ∂ x = ∂ y ∂ x ∂ f ∂ y = [ ∂ y 1 ∂ x 1 ∂ y 2 ∂ x 1 ∂ y 3 ∂ x 1 ∂ y 1 ∂ x 2 ∂ y 2 ∂ x 2 ∂ y 3 ∂ x 2 ∂ y 1 ∂ x 3 ∂ y 2 ∂ x 3 ∂ y 3 ∂ x 3 ] [ ∂ f ∂ y 1 ∂ f ∂ y 2 ∂ f ∂ y 3 ] = [ 2 2 2 ] [ 2 a 1 2 a 2 2 a 3 ] = [ 4 a 1 , 4 a 2 , 4 a 3 ] T \begin{aligned} \frac{\partial f}{\partial \mathrm{x}} & = \frac{\partial \mathrm{y}}{\partial \mathrm{x}} \frac{\partial f}{\partial \mathrm{y}} \\ & = \begin{bmatrix} \frac{\partial y_1}{\partial x_1} & \frac{\partial y_2}{\partial x_1} & \frac{\partial y_3}{\partial x_1} \\ \frac{\partial y_1}{\partial x_2} & \frac{\partial y_2}{\partial x_2} & \frac{\partial y_3}{\partial x_2} \\ \frac{\partial y_1}{\partial x_3} & \frac{\partial y_2}{\partial x_3} & \frac{\partial y_3}{\partial x_3} \\ \end{bmatrix} \begin{bmatrix} \frac{\partial f}{\partial y_1} \\ \frac{\partial f}{\partial y_2} \\ \frac{\partial f}{\partial y_3} \\ \end{bmatrix} \\ & = \begin{bmatrix} 2 & & \\ & 2 & \\ & & 2 \end{bmatrix} \begin{bmatrix} 2 a_1 \\ 2 a_2 \\ 2 a_3 \\ \end{bmatrix} \\ & = [4a_1, 4a_2, 4a_3]^T \end{aligned} ∂x∂f​​=∂x∂y​∂y∂f​= ​∂x1​∂y1​​∂x2​∂y1​​∂x3​∂y1​​​∂x1​∂y2​​∂x2​∂y2​​∂x3​∂y2​​​∂x1​∂y3​​∂x2​∂y3​​∂x3​∂y3​​​ ​ ​∂y1​∂f​∂y2​∂f​∂y3​∂f​​ ​= ​2​2​2​ ​ ​2a1​2a2​2a3​​ ​=[4a1​,4a2​,4a3​]T​

一些启示

Reference

参考教程1
参考教程2文章来源地址https://www.toymoban.com/news/detail-624954.html

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