numpy.ndenumerate to return indices in Fortran order?

arrays, iteration, iterator, numpy, python

Solution

You can do it with `np.nditer` as follows:

it = np.nditer(a, flags=['multi_index'], order='F')
while not it.finished:
    print it.multi_index, it[0]
    it.iternext()

`np.nditer` is a very powerful beast that exposes some of the internal C iterator in Python, take a look at Iterating Over Arrays in the docs.

Problem

When using `numpy.ndenumerate` the indices are returned following for a `C-contiguous` array order, for example: ``` import numpy as np a = np.array([[11, 12], [21, 22], [31, 32]]) for (i,j),v in np.ndenumerate(a): print i, j, v ``` No mather if the `order` in `a` is `'F'` or `'C'`, this gives: ``` 0 0 11 0 1 12 1 0 21 1 1 22 2 0 31 2 1 32 ``` Is there any built-in iterator in `numpy` like `ndenumerate` to give this (following the array `order='F'`): ``` 0 0 11 1 0 21 2 0 31 0 1 12 1 1 22 2 1 32 ```

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