Re: [Numpy-discussion] nonzero() behaviour has changed

[George N. <gn...@go...>]

This is a very useful thread. Made me think about what the present
arrangement is supposed to do.

My usage of this is to get out indices corresponding to a given condition.

So
E.g.
In [38]: a = arange(12).reshape(3,2,2)
Out[39]:
array([[[ 0,  1],
        [ 2,  3]],

       [[ 4,  5],
        [ 6,  7]],

       [[ 8,  9],
        [10, 11]]])
In [40]: xx = where(a%2==0)

In [41]: xx
Out[41]:
(array([0, 0, 1, 1, 2, 2]),
 array([0, 1, 0, 1, 0, 1]),
 array([0, 0, 0, 0, 0, 0]))

The transpose idea should have been obvious to me...

In [48]: for i,j,k in transpose(xx):
   ....:     print i,j,k
   ....:
   ....:
0 0 0
0 1 0
1 0 0
1 1 0
2 0 0
2 1 0

A 1D array
In [57]: aa = a.ravel()
In [49]: xx = where(aa%2==0)
In [53]: xx
Out[53]: (array([ 0,  2,  4,  6,  8, 10]),)

needs tuple unpacking:
n [52]: for i, in transpose(xx):
   ....:     print i
   ....:
   ....:
0
2
4
6
8
10

Or more simply, the ugly
In [58]: for i in xx[0]:
   ....:     print i
   ....:
   ....:
0
2
4
6
8
10

Alternatively, instead of transpose, we can simply use zip.

E.g. (3D)
In [42]: for i,j,k in zip(*xx):
   ....:     print i,j,k

or (1D, again still needs unpacking)
In [56]: for i, in zip(*xx):
   ....:     print i