I want to compute the gradient between two tensors in a net. The input X tensor (batch size x m) is sent through a set of convolutional layers which give me back and output Y tensor(batch size x n).
I’m creating a new loss and I would like to know the gradient of Y w.r.t. X. Something that in tensorflow would be like:
tf.gradients(ys=Y, xs=X)
Unfortunately, I’ve been making tests with torch.autograd.grad()
, but I could not figure out how to do it. I get errors like: “RunTimeerror: grad can be implicitly created only for scalar outputs”
.
What should be the inputs in torch.autograd.grad()
if I want to know the gradient of Y w.r.t. X?
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 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}')
Output:
Loss: 42.25
dloss/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 + 2
y = v ** 2
try:
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 + 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}')
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))
.
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