Applying a condition to rows of a 2d array-CodePudding

Consider a = np.array([0, 0, 1, 1, 2, 2, 3, 3])

In the multiset a, there are exactly 2 instances each of 0,1,2, and 3.

I want to find all permutations of a that meet a condition as we move through each row from left to right:

condition: the 1st instance of 0,1,2, and 3 must appear in that order, though they do not need to be consecutive.

[0, 1, , 2, , 3, , ] is ok, [0, 1, , 3, , 2, , ] is not ok

The 2nd instance of each number may appear anywhere in the row as long as it is after (to the right of) the 1st instance.

[0, 1, 0, 2, 2, 3, 1, 3] is ok

I've started by finding all 8!/2**4 = 2525 permutations of the multiset a:

from sympy.utilities.iterables import multiset_permutations
import numpy as np

a = np.array([0, 0, 1, 1, 2, 2, 3, 3])

resultList = []
for p in multiset_permutations(a):
    resultList.append(p)
    
out = np.array(resultList)

My difficulty is that I'm drowning in the details when I try to set the condition. To compound the problem, the actual array a could have up to 5 pairs of values. QUESTION: How can the condition be written so that I can eliminate, from array out, all permutation rows that do not satisfy the condition?

CodePudding user response：

Since you know your array consists of exactly pairs of the elements in np.arange(4), you can use np.argmax to check:

max_values = np.max(a)
uniques = np.arange(max_values   1)
# or you can just do
# uniques = np.unique(a)

resultList = []
for p in multiset_permutations(a):
    idx = np.argmax(p==uniques[:,None], axis=1)
    if (idx[:-1] < idx[1:]).all():
        resultList.append(p)

Then resultList would contains 420 permutations for 4 pairs; and 4725 for 5 pairs.