I was calling nonzero() on a tensor and then getting the mean values, but it turns out that I will need to keep the shape of the original tensor, but just ignore the values that are 0 for the mean calculation, is there a way to do this?
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I don’t understand what do you mean by “keep the shape…”. A mean is an scalar, thus, you can’t “keep the shape”.
You just can filter them out like mean = tensor[tensor !=0].mean()
right, but the original tensor has a shape and I need to preserve the shape which would exist if I were to do…
x = torch.ones((2, 3, 4))
x.mean(dim=0)
which would be a (3,4) tensor of the means along dimension 0. I have found that I could do this by summing the elements along the dimension and then somehow dividing by the count of nonzero elements along the same dimension…
But then I realized my problem is a little bit harder than that because I am dealing with one tensor of values and then one probability distribution tensor with the same shape, which gives the probabilities of those values…
x = torch.rand((2, 3, 4))
y = torch.distributions.Normal(torch.randn((2, 3, 4), torch.randn((2, 3, 4)))
pr = y.log_prob(x)
...
at this point I need to get the mean of dimension 0 of the values in y which are 0 in x but they are not 0 in y because of the log_prob call… very confusing about how to go about it
So then use a masking
If I got it you has a tensor (N,M,P)
mask = x!=0
y_mean = (y*mask).sum(dim=0)/mask.sum(dim=0)
I think this would fit what you want right?
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Yes I believe that will work. thanks for taking the time to show me the way
You could use .nonzero() but just reshape back to the original dimensions
x[x.nonzero(as_tuple=True)].view(-1, x.shape[1])
and then do torch.mean() as normal