Python Power Set of a List

Question

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

could you provide more context? exemplaric input and expected vs. actual output? — MSeifert, Jan 13 '17 at 02:21
capturing return value of `[s.add(x) for s in powerSetCopy]` definitely wrong. It will always be a list of `None`s. `s.add(x)` returns `None`. — Paul Rooney, Jan 13 '17 at 02:26
maybe related [print powerset of a string](http://stackoverflow.com/questions/12632421/print-powerset-of-a-string) — chickity china chinese chicken, Jan 13 '17 at 02:45
@user2993016 there are quite a few mistakes in this code. e.g. `powerSet = powerSet.extend(withX)` will assign `None` to `powerSet`, as `extend` modifies in place and returns `None`. I recommend learning more about list manipulation in python. — Paul Rooney, Jan 13 '17 at 02:50

pylang · Answer 1 · 2017-01-13T03:18:53.080

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.

score 3 · Answer 2 · answered Jan 13 '17 at 03:14

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}]

OregonTrail · Answer 3 · 2017-01-13T03:11:21.893

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

edited Jan 13 '17 at 03:11

answered Jan 13 '17 at 03:02

OregonTrail

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Note that that code in the docs is "roughly equivalent" to the source code of `itertools. combinations`, which is implemented in C, so it's more efficient than doing it in pure Python code. – PM 2Ring Jan 13 '17 at 03:59

Python Power Set of a List

3 Answers3

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