How to resize a Numpy array to add/replace rows with combinations determined by the values in each r-CodePudding

So I have a program that at some point creates random arrays and I have perform an operation which is to add rows while replacing other rows based on the values found in the rows. One of the random arrays will look something like this but keep in mind that it could randomly vary in size ranging from 3x3 up to 10x10:

For every row that has at least one value equal to 2 I need to remove/replace the row and add some more rows. The number of rows added will depend on the number of combinations possible of 0s and 1s where the number of digits is equal to the number of 2s counted in each row. Each added row will introduce one of these combinations in the positions where the 2s are located. The result that I'm looking for will look like this:

0 1 0 1 # First combination to replace 0 2 0 1
0 0 0 1 # Second combination to replace 0 2 0 1 (Only 2 combinations, only one 2)
1 0 0 1 # Stays the same
1 0 1 1 # First combination to replace 1 0 2 1 
1 0 0 1 # Second combination to replace 1 0 2 1 (Only 2 combinations, only one 2)
0 0 1 0 # First combination to replace 2 0 1 2
0 0 1 1 # Second combination to replace 2 0 1 2
1 0 1 1 # Third combination to replace 2 0 1 2
1 0 1 0 # Fourth combination to replace 2 0 1 2 (4 combinations, there are two 2s)

If you know a Numpy way of accomplishing this I will be grateful.

CodePudding user response：

Not the prettiest code but it does the job. You could clean up the itertools calls but this lets you see how it works.

import numpy as np
import itertools

X = np.array([[0, 2, 0, 1],
              [1, 0, 0, 1],
              [1, 0, 2, 1],
              [2, 0, 1, 2]])


def add(X_,Y):
    if Y.size == 0:
        Y = X_
    else:
        Y = np.vstack((Y, X_))
    return(Y)

Y = np.array([])
for i in range(len(X)):
    if 2 not in X[i,:]:
        Y = add(X[i,:], Y)
    else:
        a = np.where(X[i,:]==2)[0]
        n = [[i for i in itertools.chain([1, 0])] for _ in range(len(a))]
        m = list(itertools.product(*n))
        for j in range(len(m)):            
            M = 1 * X[i,:]
            u = list(m[j])
            for k in range(len(a)):
                M[a[k]] = u[k]
            Y = add(M, Y)

print(Y)
#[[0 1 0 1]
# [0 0 0 1]
# [1 0 0 1]
# [1 0 1 1]
# [1 0 0 1]
# [1 0 1 1]
# [1 0 1 0]
# [0 0 1 1]
# [0 0 1 0]]

CodePudding user response：

You can try the following. Create a sample array:

import numpy as np

np.random.seed(5)
a = np.random.randint(0, 3, (4, 4))
print(a)

This gives:

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

Compute the output array:

ts = (a == 2).sum(axis=1)
r = np.hstack([np.array(np.meshgrid(*[[0, 1]] * t)).reshape(t, -1).T.ravel() for t in ts if t])
out = np.repeat(a, 2**ts, axis=0)
out[out == 2] = r
print(out)

Result: