Swapping the dimensions of a numpy array

I would like to do the following:

for i in dimension1:
  for j in dimension2:
    for k in dimension3:
      for l in dimension4:
        B[k,l,i,j] = A[i,j,k,l]

without the use of loops. In the end both A and B contain the same information but indexed differently.

I must point out that the dimension 1,2,3 and 4 can be the same or different. So a numpy.reshape() seems difficult.

The canonical way of doing this in numpy would be to use np.transpose's optional permutation argument. In your case, to go from ijkl to klij, the permutation is (2, 3, 0, 1), e.g.:

In [16]: a = np.empty((2, 3, 4, 5))

In [17]: b = np.transpose(a, (2, 3, 0, 1))

In [18]: b.shape
Out[18]: (4, 5, 2, 3)

Please note: Jaime's answer is better. NumPy provides np.transpose precisely for this purpose.

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Or use np.einsum; this is perhaps a perversion of its intended purpose, but the syntax is quite nice:

In [195]: A = np.random.random((2,4,3,5))

In [196]: B = np.einsum('klij->ijkl', A)

In [197]: A.shape
Out[197]: (2, 4, 3, 5)

In [198]: B.shape
Out[198]: (3, 5, 2, 4)

In [199]: import itertools as IT    
In [200]: all(B[k,l,i,j] == A[i,j,k,l] for i,j,k,l in IT.product(*map(range, A.shape)))
Out[200]: True

You could rollaxis twice:

>>> A = np.random.random((2,4,3,5))
>>> B = np.rollaxis(np.rollaxis(A, 2), 3, 1)
>>> A.shape
(2, 4, 3, 5)
>>> B.shape
(3, 5, 2, 4)
>>> from itertools import product
>>> all(B[k,l,i,j] == A[i,j,k,l] for i,j,k,l in product(*map(range, A.shape)))
True

or maybe swapaxes twice is easier to follow:

>>> A = np.random.random((2,4,3,5))
>>> C = A.swapaxes(0, 2).swapaxes(1,3)
>>> C.shape
(3, 5, 2, 4)
>>> all(C[k,l,i,j] == A[i,j,k,l] for i,j,k,l in product(*map(range, A.shape)))
True

I would look at numpy.ndarray.shape and itertools.product:

import numpy, itertools
A = numpy.ones((10,10,10,10))
B = numpy.zeros((10,10,10,10))

for i, j, k, l in itertools.product(*map(xrange, A.shape)):
    B[k,l,i,j] = A[i,j,k,l]

By "without the use of loops" I'm assuming you mean "without the use of nested loops", of course. Unless there's some numpy built-in that does this, I think this is your best bet.

One can also leverage numpy.moveaxis() for moving the required axes to desired locations. Here is an illustration, stealing the example from Jaime's answer:

In [160]: a = np.empty((2, 3, 4, 5))

# move the axes that are originally at positions [0, 1] to [2, 3]
In [161]: np.moveaxis(a, [0, 1], [2, 3]).shape 
Out[161]: (4, 5, 2, 3)