pytorch - Autograd.grad() for Tensor in pytorch

Answer 1

让我们从简单的工作示例开始，该示例具有普通的损失函数和常规的后向。我们将构建简短的计算图并对其进行一些梯度计算。

代码：

import torch
from torch.autograd import grad
import torch.nn as nn


# Create some dummy data.
x = torch.ones(2, 2, requires_grad=True)
gt = torch.ones_like(x) * 16 - 0.5  # "ground-truths" 

# We will use MSELoss as an example.
loss_fn = nn.MSELoss()

# Do some computations.
v = x + 2
y = v ** 2

# Compute loss.
loss = loss_fn(y, gt)

print(f'Loss: {loss}')

# Now compute gradients:
d_loss_dx = grad(outputs=loss, inputs=x)
print(f'dloss/dx:\n {d_loss_dx}')

输出：

Loss: 42.25
dloss/dx:
(tensor([[-19.5000, -19.5000], [-19.5000, -19.5000]]),)

好的，这行得通！现在让我们尝试重现错误“grad can be implicitly created only for scalar outputs”。如您所见，前面示例中的损失是一个标量。backward()并且grad()默认处理单个标量值：loss.backward(torch.tensor(1.)). 如果您尝试传递具有更多值的张量，您将收到错误消息。

代码：

v = x + 2
y = v ** 2

try:
    dy_hat_dx = grad(outputs=y, inputs=x)
except RuntimeError as err:
    print(err)

输出：

grad can be implicitly created only for scalar outputs

因此，使用grad()时需要指定grad_outputs参数如下：

代码：

v = x + 2
y = v ** 2

dy_dx = grad(outputs=y, inputs=x, grad_outputs=torch.ones_like(y))
print(f'dy/dx:\n {dy_dx}')

dv_dx = grad(outputs=v, inputs=x, grad_outputs=torch.ones_like(v))
print(f'dv/dx:\n {dv_dx}')

输出：

dy/dx:
(tensor([[6., 6.],[6., 6.]]),)

dv/dx:
(tensor([[1., 1.], [1., 1.]]),)

注意：如果您backward()改用，只需执行y.backward(torch.ones_like(y)).