python numpy ndarray index permutations in all dimensions (variable dimension number)

Question

'M' is a numpy ndarray, which dimension 'dim' number is variable (previously generated) but each dimension is of equal size 'size'. In my code it will be more like dim = 5, size = 7.

ex: (dim = 3,size = 4).

M = np.arange(size**dim).reshape((size,)*dim)

[[[ 0  1  2  3]
  [ 4  5  6  7]
  [ 8  9 10 11]
  [12 13 14 15]]

 [[16 17 18 19]
  [20 21 22 23]
  [24 25 26 27]
  [28 29 30 31]]

 [[32 33 34 35]
  [36 37 38 39]
  [40 41 42 43]
  [44 45 46 47]]

 [[48 49 50 51]
  [52 53 54 55]
  [56 57 58 59]
  [60 61 62 63]]]

And I have a permutation generator of my own, 'per', that generates specific permutations (not random) of range(size).

print(next(per))
(1,0,3,2)

My need: transform M, by moving its elements according to as many permutations as I need. In the example: 21 (1 permutation for first dimension, 4 for second, 16 for third - generalised: size**d for d in range(dim) permutations). My permutations are not random, but they are independant, different from each other.

A result may be:

[[[36 39 37 38]
  [33 34 32 35]
  [46 44 45 47]]
  [41 43 40 42]
 [[9  10 11 8]
  [2  1  3  0]
  [6  7  5  4]
  [13 12 14 15]]
 [[56 59 57 58]
  [63 61 62 60]
  [53 54 52 55]
  [51 50 49 48]]
 [[28 30 29 31]
  [27 25 24 26]
  [17 18 16 19]
  [23 21 20 22]]]

How can I do it directly from M as numpy array, whith my code remaining dynamic?

What do you mean by "by moving its elements according to as many permutations as I need"? What is the exact relationship supposed to be between the input array, the output array, and the permutation(s?) used? — user2357112
– user2357112, Commented Jul 2, 2018 at 23:36
If you see my example, each line content has been permuted, but the lines in each 4*4 grid have been also, and the grid order also. All this needs to be done according to individual permutations (pertmuted index) from my generator. — Cépagrave
– Cépagrave, Commented Jul 2, 2018 at 23:49

Paul Panzer · Accepted Answer · 2018-07-05 01:23:39Z

1

This is an exercise in broadcasting:

>>> import numpy as np
>>> from itertools import islice
>>> 
>>> A = np.arange(27).reshape(3,3,3)
>>> def per(n):
...     while True:
...         yield np.random.permutation(n)
... 
>>> pg = per(3)
>>> 
>>> p0 = next(pg)[..., None, None]
>>> p1 = np.array([p for p in islice(pg, 3)])[..., None]
>>> p2 = np.array([[p for p in islice(pg, 3)] for _ in range(3)])
>>> 
>>> p0
array([[[2]],

       [[1]],

       [[0]]])
>>> p1
array([[[1],
        [0],
        [2]],

       [[1],
        [0],
        [2]],

       [[0],
        [2],
        [1]]])
>>> p2
array([[[1, 0, 2],
        [0, 2, 1],
        [0, 1, 2]],

       [[2, 1, 0],
        [1, 2, 0],
        [2, 1, 0]],

       [[1, 2, 0],
        [2, 1, 0],
        [2, 1, 0]]])
>>> A[p0, p1, p2]
array([[[22, 21, 23],
        [18, 20, 19],
        [24, 25, 26]],

       [[14, 13, 12],
        [10, 11,  9],
        [17, 16, 15]],

       [[ 1,  2,  0],
        [ 8,  7,  6],
        [ 5,  4,  3]]])

General soln:

import numpy as np
from itertools import islice

def per(n=None):
    while True:
        n = (yield n if n is None else np.random.permutation(n)) or n

def sweep(shp):
    pg = per()
    pg.send(None)
    redshp = len(shp)*[1]
    sz = 1
    for j, k in enumerate(shp):
        pg.send(k)
        redshp[j] = k
        yield np.reshape((*islice(pg, sz),), redshp)
        sz *= k

# example
a = np.arange(24).reshape((2,3,4))
idx = *sweep(a.shape),
for i in idx:
    print(i)
print(a[idx])

edited Jul 5, 2018 at 1:23

answered Jul 3, 2018 at 0:45

Paul Panzer

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2 Comments

Cépagrave Over a year ago

As I'm new here, I would like to understand why my question is downvoted. Is there something wrong with it ?

Paul Panzer Over a year ago

@Cépagrave I can only guess - but some people here have very strong opinions on what a question should look like - look up "mcve" in the help section. If they suspect you to be lazy either in not trying hard enough yourself or in not making a good effort at explaining your problem, they get annoyed. For example, the line that user2357112 threw back at you---while I find it clear enough from context---some may find it a bit cavalier. But as I said, only guessing.

Cépagrave · Accepted Answer · 2018-07-11 17:42:52Z

0

Thanks to Paul Panzer's brilliant answers, I managed to solve this question. Paul's last answer is great and totally generalized (even with non-square matrixes!). My initial question was a little less general, as my goal was to have my all-dimensions permutation working for square matrices. I simplified and shortened the code, so I'm sharing it here:

import numpy as np
from itertools import islice,count

size,dim = 4,3

per = (np.random.permutation(size) for _ in count())
idx = *(np.reshape((*islice(per, size**d),),[size]*(d+1)+[1]*(dim-d-1)) for d in range(dim)),

a = np.arange(size**dim).reshape((size,)*dim)
print(a[idx])

answered Jul 11, 2018 at 17:42

Cépagrave

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