Python 3 list of all possible cohesive combinations

Question

Hi I'm trying to make a list of all possible cohesive combinations of another list, so from [0, 1, 2, 3] I'd like to get [[0], [0, 1], [0, 1, 2], [0, 1, 2, 3], [1], [1, 2], [1, 2, 3], [2], [2, 3], [3]]. So far I've got this:

def expandArray(arr):
    result = []
    for x in range(0, len(arr)):
        subArray = [arr[x]]
        result.append(subArray)
        for y in range(x + 1, len(arr)):
            subArray.append(arr[y])
            result.append(subArray)
    return(result)

But this returns: [[0, 1, 2, 3], [0, 1, 2, 3], [0, 1, 2, 3], [0, 1, 2, 3], [1, 2, 3], [1, 2, 3], [1, 2, 3], [2, 3], [2, 3], [3]].

What am I doing wrong ?

possible duplicate of [What's a good way to get all subsets of a set?](https://stackoverflow.com/questions/1482308/whats-a-good-way-to-get-all-subsets-of-a-set-powerset) — D. Jones, Oct 18 '18 at 14:59
It looks like you are trying to create a 'Power Set' https://stackoverflow.com/questions/1482308/whats-a-good-way-to-get-all-subsets-of-a-set-powerset — user78090, Oct 18 '18 at 14:59
No Snakey does not seem to get sublists with "holes" in them. So [0, 3] should not be included in the result. — Moberg, Oct 18 '18 at 15:01

score 0 · Accepted Answer · answered Oct 18 '18 at 15:00

0

subArray is a list that you modify in your for loop. When you append to it, you do not create a new list, but you modify it, and then put it in the list again, so will in the end get a result with several copies of the same list. Compare this code:

a = []
b = [5]
a.append(b)
b.append(1)
a.append(b)

print(a)

would output:

[[5, 1], [5, 1]]

answered Oct 18 '18 at 15:00

Moberg

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That makes sense, but how can i get around that ? – Snakey Oct 18 '18 at 15:04
There are many ways: https://stackoverflow.com/questions/2612802/how-to-clone-or-copy-a-list – Moberg Oct 18 '18 at 15:11

Chiheb Nexus · Answer 2 · 2018-10-18T15:27:38.260

Here is a way to have your desired output using list slicing:

def get_combs(iterable):
    for k, _ in enumerate(iterable):
        elm = k
        while elm <= len(iterable):
            data = iterable[k:elm]
            elm += 1
            if data:
                yield data

combs = list(get_combs([0, 1, 2, 3]))
print(combs)

Output:

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

Python 3 list of all possible cohesive combinations

2 Answers2