Python recursive function to display all subsets of given set

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]]

There is a convenient Python module to help:

import itertoolsdef subs(l):    res = []    for i in range(1, len(l) + 1):        for combo in itertools.combinations(l, i):            res.append(list(combo))    return res

The results are:

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

Actually there is no problem with Python call by reference as I originally thought. In that case l[-1] would be 8 in all recursive calls. But l[-1] is 3, 5, 8 respectively in the recursive calls. This modified function solves the issue:

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

returns:

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