Re: [Numpy-discussion] repmat equivalent?

[Christopher B. <Chr...@no...>]

Albert Strasheim wrote:
> There are other (unexpected, for me at least) differences between
> MATLAB/Octave and NumPy too.

First: numpy is not, and was never intended to be, a MATLAB clone, 
work-alike, whatever. You should *expect* there to be differences.

 > For a 3D array in MATLAB, only indexing
> on the last dimension yields a 2D array, where NumPy always returns a
> 2D array.

I think the key here is that MATLAB's core data type is a matrix, which 
is 2-d. The ability to do 3-d arrays was added later, and it looks like 
they are still preserving the core matrix concept, so that a 3-d array 
is not really a 3-d array; it is, as someone on this thread mentioned, a 
"stack" of matrices.

In numpy, the core data type is an n-d array. That means that there is 
nothing special about 2-d vs 4-d vs whatever, except 0-d (scalars). So a 
3-d array is a cube shape, that you might want to pull a 2-d array out 
of it in any orientation. There's nothing special about which axis 
you're indexing. For that reason, it's very important that indexing any 
axis will give you the same rank array.

Here's the rule:

-- indexing reduces the rank by 1, regardless of which axis is being 
indexed.
 >>> import numpy as N
 >>> a = N.zeros((2,3,4))
 >>> a
array([[[0, 0, 0, 0],
         [0, 0, 0, 0],
         [0, 0, 0, 0]],

        [[0, 0, 0, 0],
         [0, 0, 0, 0],
         [0, 0, 0, 0]]])
 >>> a[0,:,:]
array([[0, 0, 0, 0],
        [0, 0, 0, 0],
        [0, 0, 0, 0]])
 >>> a[:,0,:]
array([[0, 0, 0, 0],
        [0, 0, 0, 0]])
 >>> a[:,:,0]
array([[0, 0, 0],
        [0, 0, 0]])

-- slicing does not reduce the rank:
 >>> a[:,0:1,:]
array([[[0, 0, 0, 0]],

        [[0, 0, 0, 0]]])
 >>> a[:,0:1,:].shape
(2, 1, 4)

It's actually very clean, logical, and useful.

-Chris







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Christopher Barker, Ph.D.
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Chr...@no...