Numpy slice of arbitrary dimensions

Question

I would like to slice a numpy array to obtain the i-th index in the last dimension. For a 3D array, this would be:

slice = myarray[:,:,i]

B

Answer 1

Actually, just found the answer. As stated in numpy's documentation this can be done with the slice object. In my particular case, this would do it:

idx = [slice(None)] * (myarray.ndim - 1) + [i] 
my_slice = myarray[idx]

The slice(None) is equivalent to choosing all elements in that index, and the last [i] selects a specific index for the last dimension.

Answer 2

There is ... or Ellipsis, which does exactly this:

slice = myarray[..., i]

Ellipsis is the python object, if you should want to use it outside the square bracket notation.

Answer 3

In terms of slicing an arbitrary dimension, the previous excellent answers can be extended to:

indx = [slice(None)]*myarray.ndim
indx[slice_dim] = i
sliced = myarray[indx]

This returns the slice from any dimension slice_dim - slice_dim = -1 reproduces the previous answers. For completeness - the first two lines of the above listing can be condensed to:

indx = [slice(None)]*(slice_dim) + [i] + [slice(None)]*(myarray.ndim-slice_dim-1)

though I find the previous version more readable.