如何将形状（3、1、2）的3D张量重塑为（1、2、3）

Question

I intended

(Pdb) aa = torch.tensor([[[1,2]], [[3,4]], [[5,6]]])
(Pdb) aa.shape
torch.Size([3, 1, 2])
(Pdb) aa
tensor([[[ 1,  2]],

        [[ 3,  4]],

        [[ 5,  6]]])
(Pdb) aa.view(1, 2, 3)
tensor([[[ 1,  2,  3],
         [ 4,  5,  6]]])

But what I really want is

tensor([[[ 1,  3,  5],
         [ 2,  4,  6]]])

How?

In my application, I am trying to transform my input data of shape (L, N, C_in) to (N, C_in, L) in order to use Conv1d , where

• L: sequence length
• N: batch size
• C_in: number of channels in the input, I also understand it as the dimensionality of the input at each position of a sequence.

I am also wondering the input of Conv1d doesn't have the same input shape as GRU ?

Answer 1

You can permute the axes to the desired shape. (This is in some sense similar to np.rollaxis operation).

In [90]: aa
Out[90]: 
tensor([[[ 1,  2]],

        [[ 3,  4]],

        [[ 5,  6]]])

In [91]: aa.shape
Out[91]: torch.Size([3, 1, 2])

# pass the desired ordering of the axes as argument
# assign the result back to some tensor since permute returns a "view"
In [97]: permuted = aa.permute(1, 2, 0)

In [98]: permuted.shape
Out[98]: torch.Size([1, 2, 3])

In [99]: permuted
Out[99]: 
tensor([[[ 1,  3,  5],
         [ 2,  4,  6]]])

Answer 2

This is one way to do it, still hope to see a solution with a single operation.

(Pdb) torch.transpose(aa, 0, 2).t()
tensor([[[ 1,  3,  5],
         [ 2,  4,  6]]])
(Pdb) torch.transpose(aa, 0, 2).t().shape
torch.Size([1, 2, 3])