Python Power Set of a List [duplicate]

Question

This question already has an answer here:

• How to get all subsets of a set? (powerset) 17 answers

I am trying to implement a function to generate the powerset of a list xs.

The general idea is that we walk through the elements of xs and choose whether to include x or not. The problem I'm facing is that withX ends up being equal to [None] (a singleton list with None) because (I think) s.add(x) returns None.

This isn't a homework assignment, it's an exercise in Cracking the Coding Interview.

def powerSetBF(xs):
    powerSet = [] 
    powerSet.append(set([]))
    for x in xs:
        powerSetCopy = powerSet[:]
        withX = [s.add(x) for s in powerSetCopy] # add x to the list of sets 
        powerSet = powerSet.extend(withX) # append those entries
    return powerSet

Answer 1

Take a look at the powerset example from the itertools recipes:

from itertools import chain, combinations

def powerset(iterable):
    "list(powerset([1,2,3])) --> [(), (1,), (2,), (3,), (1,2), (1,3), (2,3), (1,2,3)]"
    s = list(iterable)
    return chain.from_iterable(combinations(s, r) for r in range(len(s)+1))

For a range of integers up to the length of the given list, make all possible combinations and chain them together as one object.

Answer 2

import itertools

def powerset(L):
  pset = set()
  for n in xrange(len(L) + 1):
    for sset in itertools.combinations(L, n):
      pset.add(sset)
  return pset

powerset([1, 2, 3, 4])

result

set([(1, 2), (1, 3), (1, 2, 3, 4), (1,), (2,), (3,), (1, 4), (4,), (), (2, 3), (1, 2, 4), (1, 3, 4), (2, 3, 4), (1, 2, 3), (3, 4), (2, 4)])

Source code for itertools.combinations can be found here which has a few neat optimizations:

https://docs.python.org/3/library/itertools.html#itertools.combinations

Answer 3

Here's a recursive solution that does not use any modules:

def pset(myset):
  if not myset: # Empty list -> empty set
    return [set()]

  r = []
  for y in myset:
    sy = set((y,))
    for x in pset(myset - sy):
      if x not in r:
        r.extend([x, x|sy])
  return r

print(pset(set((1,2,3,4))))
#[set(), {1}, {2}, {1, 2}, {3}, {1, 3}, {2, 3}, {1, 2, 3}, {4}, 
# {1, 4}, {2, 4}, {1, 2, 4}, {3, 4}, {1, 3, 4}, {2, 3, 4}, {1, 2, 3, 4}]