Autograd.grad() for Tensor in pytorch

✔ Recommended Answer

Let's start from simple working example with plain loss function and regular backward. We will build short computational graph and do some grad computations on it.

Code:

import torchfrom torch.autograd import gradimport 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 + 2y = 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: {d_loss_dx}')

Output:

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

Ok, this works! Now let's try to reproduce error "grad can be implicitly created only for scalar outputs". As you can notice, loss in previous example is a scalar. backward() and grad() by defaults deals with single scalar value: loss.backward(torch.tensor(1.)). If you try to pass tensor with more values you will get an error.

Code:

v = x + 2y = v ** 2try:    dy_hat_dx = grad(outputs=y, inputs=x)except RuntimeError as err:    print(err)

Output:

grad can be implicitly created only for scalar outputs

Therefore, when using grad() you need to specify grad_outputs parameter as follows:

Code:

v = x + 2y = v ** 2dy_dx = grad(outputs=y, inputs=x, grad_outputs=torch.ones_like(y))print(f'dy/dx: {dy_dx}')dv_dx = grad(outputs=v, inputs=x, grad_outputs=torch.ones_like(v))print(f'dv/dx: {dv_dx}')

Output:

dy/dx:(tensor([[6., 6.],[6., 6.]]),)dv/dx:(tensor([[1., 1.], [1., 1.]]),)

NOTE: If you are using backward() instead, simply do y.backward(torch.ones_like(y)).

Source: stackoverflow.com

Answered By: trsvchn