Pytorch. Can autograd be used when the final tensor has more than a single value in it?
Can autograd be used when the final tensor has more than a single value in it?
I tried the following.
x = torch.tensor([4.0, 5.0], requires_grad=True)
y = x ** 2
print(y)
y.backward()
Throws an error
RuntimeError: grad can be implicitly created only for scalar outputs
The following however works.
x = torch.tensor([4.0, 5.0], requires_grad=True)
y = x ** 2
y = torch.sum(y)
print(y)
y.backward()
print(x.grad)
The output is as
tensor(41., grad_fn=<SumBackward0>)
tensor([ 8., 10.])
Am I missing something here or can I proceed with the assumption that autograd only works when the final tensor has a single value in it?
python pytorch autograd
edited Nov 13 '18 at 11:01
Milo Lu
1,60511327
asked Nov 13 '18 at 4:08
sbetagerisbetageri
1815
add a comment |
Can autograd be used when the final tensor has more than a single value in it?
I tried the following.
x = torch.tensor([4.0, 5.0], requires_grad=True)
y = x ** 2
print(y)
y.backward()
Throws an error
RuntimeError: grad can be implicitly created only for scalar outputs
The following however works.
x = torch.tensor([4.0, 5.0], requires_grad=True)
y = x ** 2
y = torch.sum(y)
print(y)
y.backward()
print(x.grad)
The output is as
tensor(41., grad_fn=<SumBackward0>)
tensor([ 8., 10.])
Am I missing something here or can I proceed with the assumption that autograd only works when the final tensor has a single value in it?
python pytorch autograd
edited Nov 13 '18 at 11:01
Milo Lu
1,60511327
asked Nov 13 '18 at 4:08
sbetagerisbetageri
1815
add a comment |
Can autograd be used when the final tensor has more than a single value in it?
I tried the following.
x = torch.tensor([4.0, 5.0], requires_grad=True)
y = x ** 2
print(y)
y.backward()
Throws an error
RuntimeError: grad can be implicitly created only for scalar outputs
The following however works.
x = torch.tensor([4.0, 5.0], requires_grad=True)
y = x ** 2
y = torch.sum(y)
print(y)
y.backward()
print(x.grad)
The output is as
tensor(41., grad_fn=<SumBackward0>)
tensor([ 8., 10.])
Am I missing something here or can I proceed with the assumption that autograd only works when the final tensor has a single value in it?
python pytorch autograd
edited Nov 13 '18 at 11:01
Milo Lu
1,60511327
asked Nov 13 '18 at 4:08
sbetagerisbetageri
1815
Can autograd be used when the final tensor has more than a single value in it?
I tried the following.
x = torch.tensor([4.0, 5.0], requires_grad=True)
y = x ** 2
print(y)
y.backward()
Throws an error
RuntimeError: grad can be implicitly created only for scalar outputs
The following however works.
x = torch.tensor([4.0, 5.0], requires_grad=True)
y = x ** 2
y = torch.sum(y)
print(y)
y.backward()
print(x.grad)
The output is as
tensor(41., grad_fn=<SumBackward0>)
tensor([ 8., 10.])
Am I missing something here or can I proceed with the assumption that autograd only works when the final tensor has a single value in it?
python pytorch autograd
python pytorch autograd
edited Nov 13 '18 at 11:01
Milo Lu
1,60511327
asked Nov 13 '18 at 4:08
sbetagerisbetageri
1815
edited Nov 13 '18 at 11:01
Milo Lu
1,60511327
asked Nov 13 '18 at 4:08
sbetagerisbetageri
1815
edited Nov 13 '18 at 11:01
Milo Lu
1,60511327
edited Nov 13 '18 at 11:01
Milo Lu
1,60511327
edited Nov 13 '18 at 11:01
Milo Lu
1,60511327
1,60511327
asked Nov 13 '18 at 4:08
sbetagerisbetageri
1815
asked Nov 13 '18 at 4:08
sbetagerisbetageri
1815
asked Nov 13 '18 at 4:08
sbetagerisbetageri
1815
1815
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
See https://pytorch.org/tutorials/beginner/blitz/autograd_tutorial.html#gradients
y.backward() is same as y.backward(torch.tensor(1.0))
Usually, the output is scalar and hence the scalar is passed to backward as a default choice. However, since your output is two dimensional you should call
y.backward(torch.tensor([1.0,1.0]))
This will give expected results with x.grad being tensor([ 8., 10.])
answered Nov 13 '18 at 6:02
Umang GuptaUmang Gupta
3,17511535
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
See https://pytorch.org/tutorials/beginner/blitz/autograd_tutorial.html#gradients
y.backward() is same as y.backward(torch.tensor(1.0))
Usually, the output is scalar and hence the scalar is passed to backward as a default choice. However, since your output is two dimensional you should call
y.backward(torch.tensor([1.0,1.0]))
This will give expected results with x.grad being tensor([ 8., 10.])
answered Nov 13 '18 at 6:02
Umang GuptaUmang Gupta
3,17511535
add a comment |
See https://pytorch.org/tutorials/beginner/blitz/autograd_tutorial.html#gradients
y.backward() is same as y.backward(torch.tensor(1.0))
Usually, the output is scalar and hence the scalar is passed to backward as a default choice. However, since your output is two dimensional you should call
y.backward(torch.tensor([1.0,1.0]))
This will give expected results with x.grad being tensor([ 8., 10.])
answered Nov 13 '18 at 6:02
Umang GuptaUmang Gupta
3,17511535
add a comment |
See https://pytorch.org/tutorials/beginner/blitz/autograd_tutorial.html#gradients
y.backward() is same as y.backward(torch.tensor(1.0))
Usually, the output is scalar and hence the scalar is passed to backward as a default choice. However, since your output is two dimensional you should call
y.backward(torch.tensor([1.0,1.0]))
This will give expected results with x.grad being tensor([ 8., 10.])
answered Nov 13 '18 at 6:02
Umang GuptaUmang Gupta
3,17511535
See https://pytorch.org/tutorials/beginner/blitz/autograd_tutorial.html#gradients
y.backward() is same as y.backward(torch.tensor(1.0))
Usually, the output is scalar and hence the scalar is passed to backward as a default choice. However, since your output is two dimensional you should call
y.backward(torch.tensor([1.0,1.0]))
This will give expected results with x.grad being tensor([ 8., 10.])
answered Nov 13 '18 at 6:02
Umang GuptaUmang Gupta
3,17511535
edited Nov 13 '18 at 8:07
answered Nov 13 '18 at 6:02
Umang GuptaUmang Gupta
3,17511535
answered Nov 13 '18 at 6:02
Umang GuptaUmang Gupta
3,17511535
answered Nov 13 '18 at 6:02
Umang GuptaUmang Gupta
3,17511535
3,17511535
add a comment |
add a comment |
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