Return all labels that contain repeated elements of array

Question

I have data that looks like this in python: Notice that the first and last two elements have repeats of that array.

 for _, m in enumerate(array)
    print(_,m)

#Output:   
    1 [2,3]
    2 [8,8]
    2 [9,7]
    3 [1,1]       
    3 [2,3]
    4 [8,8]

I want a function that outputs the labels only if there are repeats of that array. Something like this:

    [2,3]: 1,3
    [8,8]: 2,4

The code I want should be something like this?

    for _, m in enumerate(array):
      if m repeats:
        print(m, _.all())

Answer 1

hope this will answer to your needs, still try to follow the advice about MRE next time

lets start defining a simple function to list all positions of an element into a list

the next step would be to actually print an element one time only if it is contained 2 or more times into the "array". this is achievable using a list to keep the items already checked (found)

def find_positions(n_array, to_find):
        return [i for i in [i for i in range(len(n_array))] if to_find == n_array[i]]

if __name__ == "__main__":
        array = [[2,3],[8,8],[9,7],[1,1],[2,3],[8,8]]
        found = []
        for element in array:
            if element not in found:
                found.append(element)
                position = find_positions(array, element)
                if len(position)>1:
                    print(element,":",position)

this above is an example of an MRE of a solution ;)

output:
[2,3] : [0,4]
[8,8] : [1,5]

Note that the indexing of the positions starts with 0, a choice to be in line with python indexing ;)