[FIXED] Replace numpy.array values with randomly selected values from other rows in the array only

Issue

I have the following array of shape (A, B) where A=4 and B=2 here: [[1 1] [2 2] [3 3] [4 4]] I want to modify the array so each row consists of randomly selected values from the other rows, with no repetition within the same row. An example result would be: [[3 2] [1 3] [2 4] [3 1]]

I tried to use np.random.shuffle and np.random.choice but I can't figure out how to exclude the row itself and how to replace values per value instead of per row. np.random.shuffle leads to: [[4 4] [2 2] [3 3] [1 1]] and np.random.choice gives me errors because my array is 2D and not 1D
I'm a beginner and I feel like this should be obvious but I've been racking my brain all day... Any help would be very much appreciated

Solution

Assuming that the values for an A=N matrix are always to be the values 1 to N, the following will generate a random matrix where the values come from that set but no row contains the one-based index of that row.

import random
N = 4
nums = list(range(1, N+1))
a = []
for irow in range(1, N+1):
    nums.remove(irow)
    a.append([random.choice(nums), random.choice(nums)])
    nums.append(irow)
print(a)

generates

[[3, 4], [4, 4], [4, 1], [1, 2]]

If you want to be sure that each column contains all of the possible values, but again with no value appearing in the row with the value equaling its one-based index. The following (based of @BRivera's solution described in Find the permutations where no element stays in place) should do the trick

import random

def generate_all_moved_permutation_tree(level, nums):
    if len(nums) == 0:
        raise RunTimeError('generate_permutation_tree must be called with a non-empty nums list')
    if len(nums) == 1:
        if level == nums[0]:
            return None
        else:
            return {nums[0]: {}}
    allowed_nums = list(nums)
    if level in allowed_nums:
        allowed_nums.remove(level)
    result = {}
    for n in allowed_nums:
        sublevel_nums = list(nums)
        if n in sublevel_nums:
            sublevel_nums.remove(n)
        subtree = generate_all_moved_permutation_tree(level+1, sublevel_nums)
        if subtree is not None:
            result[n] = subtree
    if len(result) == 0:
        return None
    return result

def pick_an_all_moved_permutation(all_moved_permutation_tree):
    n = random.choice(list(all_moved_permutation_tree))
    l = [n]
    sub_tree = all_moved_permutation_tree[n]
    if len(sub_tree) > 0:
        l.extend(pick_an_all_moved_permutation(sub_tree))
    return l

Examples

>>> t = generate_all_moved_permutation_tree(1, range(1,4))
>>> print(t)
{2: {3: {1: {}}}, 3: {1: {2: {}}}}
>>> t = generate_all_moved_permutation_tree(1, range(1,5))
>>> print(list(zip(pick_an_all_moved_permutation(t), pick_an_all_moved_permutation(t))))
[(2, 4), (3, 1), (4, 2), (1, 3)]
>>> print(list(zip(pick_an_all_moved_permutation(t), pick_an_all_moved_permutation(t))))
[(3, 4), (1, 1), (4, 2), (2, 3)]
>>> print(list(zip(pick_an_all_moved_permutation(t), pick_an_all_moved_permutation(t))))
[(3, 2), (4, 1), (1, 4), (2, 3)]

Answered By - Malcolm