在pytorch中获取矢量化函数的梯度

Getting gradient of vectorized function in pytorch(在pytorch中获取矢量化函数的梯度)

问题描述

我是 PyTorch 的新手，想做我认为非常简单的事情，但遇到了很多困难.

I am brand new to PyTorch and want to do what I assume is a very simple thing but am having a lot of difficulty.

我有函数 sin(x) * cos(x) + x^2 并且我想在任何时候获得该函数的导数.

I have the function sin(x) * cos(x) + x^2 and I want to get the derivative of that function at any point.

如果我做到这一点，它就像

If I do this with one point it works perfectly as

x = torch.autograd.Variable(torch.Tensor([4]),requires_grad=True)
y = torch.sin(x)*torch.cos(x)+torch.pow(x,2)
y.backward()
print(x.grad) # outputs tensor([7.8545])

但是，我希望能够将向量作为 x 传入并让它按元素计算导数.例如:

However, I want to be able to pass in a vector as x and for it to evaluate the derivative element-wise. For example:

Input: [4., 4., 4.,]
Output: tensor([7.8545, 7.8545, 7.8545])

但我似乎无法使其正常工作.

But I can't seem to get this working.

我试着简单地做

x = torch.tensor([4., 4., 4., 4.], requires_grad=True)
out = torch.sin(x)*torch.cos(x)+x.pow(2)
out.backward()
print(x.grad)

但是我收到错误消息RuntimeError:grad 只能为标量输出隐式创建"

But I get the error "RuntimeError: grad can be implicitly created only for scalar outputs"

如何为向量调整此代码?

How do I adjust this code for vectors?

提前致谢，

推荐答案

在这里你可以找到关于你的错误的相关讨论.

Here you can find relevant discussion about your error.

本质上，当你不带参数调用backward()时，它被隐式转换为backward(torch.Tensor([1]))，其中torch.Tensor([1]) 是计算梯度的输出值.

In essence, when you call backward() without arguments it is implicitly converted to backward(torch.Tensor([1])), where torch.Tensor([1]) is the output value with respect to which gradients are calculated.

如果您传递 4(或更多)个输入，则每个输入都需要一个值来计算梯度.您可以像这样将 torch.ones_like 显式传递给 backward:

If you pass 4 (or more) inputs, each needs a value with respect to which you calculate gradient. You can pass torch.ones_like explicitly to backward like this:

import torch

x = torch.tensor([4.0, 2.0, 1.5, 0.5], requires_grad=True)
out = torch.sin(x) * torch.cos(x) + x.pow(2)
# Pass tensor of ones, each for each item in x
out.backward(torch.ones_like(x))
print(x.grad)

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