Removing a sublist from a list that meets a certain condition

Question

I create all three-element permutations without mirroring, using itertools.product():

import itertools

list_1 = [list(i) for i in itertools.product(tuple(range(4)), repeat=3) if tuple(reversed(i)) >= tuple(i)]

Output:

[[0, 0, 0], [0, 0, 1], [0, 0, 2], [0, 0, 3], [0, 1, 0], [0, 1, 1], [0, 1, 2], [0, 1, 3], [0, 2, 0], [0, 2, 1], [0, 2, 2], [0, 2, 3], [0, 3, 0], [0, 3, 1], [0, 3, 2], [0, 3, 3], [1, 0, 1], [1, 0, 2], [1, 0, 3], [1, 1, 1], [1, 1, 2], [1, 1, 3], [1, 2, 1], [1, 2, 2], [1, 2, 3], [1, 3, 1], [1, 3, 2], [1, 3, 3], [2, 0, 2], [2, 0, 3], [2, 1, 2], [2, 1, 3], [2, 2, 2], [2, 2, 3], [2, 3, 2], [2, 3, 3], [3, 0, 3], [3, 1, 3], [3, 2, 3], [3, 3, 3]]

How do I delete these sublisters from the list list_1, which have the same number of corresponding values and then leave only one of them?

For example, in sublists [1,1,2], [1,2,1] the number of given values is the same in all, that is, in each sub-list there are two 1 and one 2, that's why I consider the sublisters to be the same and that's why I want to leave only the first one, namely [1,1,2]. How can this be done?

I was thinking about counting the number of corresponding values in each sub-list and creating a list with the occurring feature regarding the amount of given values, and then checking each element from the list list_1 in the loop or the element with the given feature has not occurred before. But it seems to me to be very complicated.

Answer 1

Rather than using product from the itertools module, use combinations_with_replacement. That does what you want in one line without any massaging afterward:

list1 = [list(i) for i in combinations_with_replacement(range(4),3)]

The result of print(list1) after that is

[[0, 0, 0], [0, 0, 1], [0, 0, 2], [0, 0, 3], [0, 1, 1], [0, 1, 2], [0, 1, 3], [0, 2, 2], [0, 2, 3], [0, 3, 3], [1, 1, 1], [1, 1, 2], [1, 1, 3], [1, 2, 2], [1, 2, 3], [1, 3, 3], [2, 2, 2], [2, 2, 3], [2, 3, 3], [3, 3, 3]]

Note that your conversion of the range object to a tuple is not necessary.

Answer 2

This might do the trick:

import itertools

def uniqifyList(list):
    indexToReturn = []
    sortedUniqueItems = []

    for idx, value in enumerate(list):
        # check if exists in unique_list or not
        value.sort()
        if value not in sortedUniqueItems:
            sortedUniqueItems.append(value)
            indexToReturn.append(idx)

    return [list[i] for i in indexToReturn]

list1 = [list(i) for i in itertools.product(tuple(range(4)), repeat=3) if tuple(reversed(i)) >= tuple(i)]
print(list1)

list2 = uniqifyList(list1)
print(list2)

Which outputs:

[[0, 0, 0], [0, 0, 1], [0, 0, 2], [0, 0, 3], [0, 1, 1], [0, 1, 2], [0, 1, 3], [0, 2, 2], [0, 2, 3], [0, 3, 3], [1, 1, 1], [1, 1, 2], [1, 1, 3], [1, 2, 2], [1, 2, 3], [1, 3, 3], [2, 2, 2], [2, 2, 3], [2, 3, 3], [3, 3, 3]]