[FIXED] Efficient algoritm to compare each element in a Numpy array with each other element in the same array

Issue

I have an array (fe: [1,2,3,4,5,...]. I want to compare each element of the array with every other element in the array. So I want to have:

1-2
1-3
1-4
1-5
2-3
2-4
2-5
3-4
3-5
4-5

Currently, this is my code

for el_a in my_array:
    idx_a = np.where(my_array == el_a)[0][0]
    for idx_b in range(idx_a+1, len(my_array)):
        el_b = my_array[idx_b]
        print(el_a,el_b)

I compare every element el_a (first for loop), with every element el_b which comes after el_a (second for loop).

The algoritm is working correctly, however, it is very slow. Is there someone with a better, more efficient solution?

Solution

This will provide the comparison result for all possible combinations as a list:

import numpy as np
from itertools import combinations
my_array = np.array([1, 2, 3, 4, 5])
print([item[0]==item[1] for item in combinations(my_array,2)])

Answered By - Pranta Palit