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def get_jacobian(net, x, noutputs): | |
x = x.squeeze() | |
n = x.size()[0] | |
x = x.repeat(noutputs, 1) | |
x.requires_grad_(True) | |
y = net(x) | |
y.backward(torch.eye(noutputs)) | |
return x.grad.data |
how about this experimental api for jacobian: https://pytorch.org/docs/stable/_modules/torch/autograd/functional.html#jacobian
is it good?
how about this experimental api for jacobian: https://pytorch.org/docs/stable/_modules/torch/autograd/functional.html#jacobian
is it good?
I took a look and:
for j in range(out.nelement()):
vj = _autograd_grad((out.reshape(-1)[j],), inputs, retain_graph=True, create_graph=create_graph)
It's just for-looping over the output and computing the gradient one by one (i.e. each row of the jacobian one by one). This will for sure be slow as hell if you have a lot of outputs. I actually think it's a tad bit deceiving that they advertise this functionality, because really the functionality just isn't there.
And actually, to be honest I wanted the jacobian earlier to do some gauss newton type optimization, but I've actually since discovered that the optim.LBFGS
optimizer (now built into pytorch) might work well for my problem. I think it even has some backtracking type stuff built into it. So for now I don't think I even need the jacobian anymore.
@justinblaber , autodiff either computes matrix-vector products or vector-matrix products (depending on forward mode / reverse mode). The Jacobian is a matrix - there's no easy way to recover this by itself. Either you perform multiple backwards passes, using different elementary basis vector on each pass, or you blow the batch size up and do one massive backwards pass. There's no way around this.