Python recursive function to display all subsets of given set

Question

I have the following python function to print all subsets of a list of numbers:

def subs(l):
    if len(l) == 1:
        return [l]
    res = []
    for sub in subs(l[0:-1]):
        res.append(sub)
        res.append([l[-1]])
        res.append(sub+[l[-1]])
    return res

li = [2, 3, 5, 8]
print(subs(li))

This returns:

[[2], [8], [2, 8], [5], [8], [5, 8], [2, 5], [8], [2, 5, 8], [3], [8], [3, 8], [5], [8], [5, 8], [3, 5], [8], [3, 5, 8], [2, 3], [8], [2, 3, 8], [5], [8], [5, 8], [2, 3, 5], [8], [2, 3, 5, 8]]

Which is not the expected answer. It looks like python takes the list l into the function by reference. So when I append l[-1], it appends the last element of original list, not the smaller list sent into the recursive method. Is there any way to solve this?

This could possibly be solved using tuples but I'm wondering if there is a solution using lists.

Answer 1

def subs(l):
    if l == []:
        return [[]]

    x = subs(l[1:])

    return x + [[l[0]] + y for y in x]

Results:

>>> print (subs([1, 2, 3]))
[[], [3], [2], [2, 3], [1], [1, 3], [1, 2], [1, 2, 3]]