Getting subsets of a set in Python

Question

Suppose we need to write a function that gives a list of all subsets of a set. Function and doctrine are given below. And we need to complete the entire function definition

def subsets(s):
   """Return a list of the subsets of s.

   >>> subsets({True, False})
   [{False, True}, {False}, {True}, set()]
   >>> counts = {x for x in range(10)} # A set comprehension
   >>> subs = subsets(counts)
   >>> len(subs)
   1024
   >>> counts in subs
   True
   >>> len(counts)
   10
   """
   assert type(s) == set, str(s) + ' is not a set.'
   if not s:
       return [set()]
   element = s.pop() 
   rest = subsets(s)
   s.add(element)    

---

It should not use the built-in function

My approach is to add the "element" to rest and return them all, but I don't know very well how to use set, list in Python.

Answer 1

See poweret () in the itertools docs .

from itertools import chain, combinations

def powerset(iterable):
    "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))

def subsets(s):
    return map(set, powerset(s))

Answer 2

>>> s=set(range(10))
>>> L=list(s)
>>> subs = [{L[j] for j in range(len(L)) if 1<<j&i} for i in range(1,1<<len(L))]
>>> s in subs
True
>>> set() in subs
False

Answer 3

>>> from itertools import combinations
>>> s=set(range(10))
>>> subs = [set(j) for i in range(len(s)) for j in combinations(s, i+1)]
>>> len(subs)
1023

Answer 4

A typical poweret implementation doesn't listlook something like this:

def powerset(elements):
    if len(elements) > 0:
        head = elements[0]
        for tail in powerset(elements[1:]):
            yield [head] + tail
            yield tail
    else:
        yield []

You just need to adapt a little to set.

Answer 5

Slightly more efficient (less copy than previous answers):

# Generate all subsets of the list v of length l.
def subsets(v, l):
  return _subsets(v, 0, l, [])

def _subsets(v, k, l, acc):
  if l == 0:
    return [acc]
  else:
    r = []
    for i in range(k, len(v)):
      # Take i-th position and continue with subsets of length l - 1:
      r.extend(_subsets(v, i + 1, l - 1, acc + [v[i]]))
    return r

Answer 6

>>> from itertools import combinations
>>> s=set([1,2,3])
>>> sum(map(lambda r: list(combinations(s, r)), range(1, len(s)+1)), [])
[(1,), (2,), (3,), (1, 2), (1, 3), (2, 3), (1, 2, 3)]

creates tuples, but for you it's close enough