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

Question

The Python itertools page has exactly a powerset recipe for this:

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))

Output:

>>> list(powerset("abcd"))
[(), ('a',), ('b',), ('c',), ('d',), ('a', 'b'), ('a', 'c'), ('a', 'd'), ('b', 'c'), ('b', 'd'), ('c', 'd'), ('a', 'b', 'c'), ('a', 'b', 'd'), ('a', 'c', 'd'), ('b', 'c', 'd'), ('a', 'b', 'c', 'd')]

If you don't like that empty tuple at the beginning, you can just change the range statement to range(1, len(s)+1) to avoid a 0-length combination.

Here is more code for a powerset. This is written from scratch:

>>> def powerset(s):
...     x = len(s)
...     for i in range(1 << x):
...         print [s[j] for j in range(x) if (i & (1 << j))]
...
>>> powerset([4,5,6])
[]
[4]
[5]
[4, 5]
[6]
[4, 6]
[5, 6]
[4, 5, 6]

Mark Rushakoff's comment is applicable here: "If you don't like that empty tuple at the beginning, on."you can just change the range statement to range(1, len(s)+1) to avoid a 0-length combination", except in my case you change for i in range(1 << x) to for i in range(1, 1 << x).

---

Returning to this years later, I'd now write it like this:

def powerset(s):
    x = len(s)
    masks = [1 << i for i in range(x)]
    for i in range(1 << x):
        yield [ss for mask, ss in zip(masks, s) if i & mask]

And then the test code would look like this, say:

print(list(powerset([4, 5, 6])))

Using yield means that you do not need to calculate all results in a single piece of memory. Precalculating the masks outside the main loop is assumed to be a worthwhile optimization.

If you're looking for a quick answer, I just searched "python power set" on google and came up with this: Python Power Set Generator

Here's a copy-paste from the code in that page:

def powerset(seq):
    """
    Returns all the subsets of this set. This is a generator.
    """
    if len(seq) <= 1:
        yield seq
        yield []
    else:
        for item in powerset(seq[1:]):
            yield [seq[0]]+item
            yield item

This can be used like this:

 l = [1, 2, 3, 4]
 r = [x for x in powerset(l)]

Now r is a list of all the elements you wanted, and can be sorted and printed:

r.sort()
print r
[[], [1], [1, 2], [1, 2, 3], [1, 2, 3, 4], [1, 2, 4], [1, 3], [1, 3, 4], [1, 4], [2], [2, 3], [2, 3, 4], [2, 4], [3], [3, 4], [4]]

def powerset(lst):
    return reduce(lambda result, x: result + [subset + [x] for subset in result],
                  lst, [[]])

There is a refinement of powerset:

def powerset(seq):
    """
    Returns all the subsets of this set. This is a generator.
    """
    if len(seq) <= 0:
        yield []
    else:
        for item in powerset(seq[1:]):
            yield [seq[0]]+item
            yield item

TL;DR (go directly to Simplification)

I know I have previously added an answer, but I really like my new implementation. I am taking a set as input, but it actually could be any iterable, and I am returning a set of sets which is the power set of the input. I like this approach because it is more aligned with the mathematical definition of power set (set of all subsets).

def power_set(A):
    """A is an iterable (list, tuple, set, str, etc)
    returns a set which is the power set of A."""
    length = len(A)
    l = [a for a in A]
    ps = set()

    for i in range(2 ** length):
        selector = f'{i:0{length}b}'
        subset = {l[j] for j, bit in enumerate(selector) if bit == '1'}
        ps.add(frozenset(subset))

    return ps

If you want exactly the output you posted in your answer use this:

>>> [set(s) for s in power_set({1, 2, 3, 4})]
[{3, 4},
 {2},
 {1, 4},
 {2, 3, 4},
 {2, 3},
 {1, 2, 4},
 {1, 2},
 {1, 2, 3},
 {3},
 {2, 4},
 {1},
 {1, 2, 3, 4},
 set(),
 {1, 3},
 {1, 3, 4},
 {4}]

Explanation

It is known that the number of elements of the power set is 2 ** len(A), so that could clearly be seen in the for loop.

I need to convert the input (ideally a set) into a list because by a set is a data structure of unique unordered elements, and the order will be crucial to generate the subsets.

selector is key in this algorithm. Note that selector has the same length as the input set, and to make this possible it is using an f-string with padding. Basically, this allows me to select the elements that will be added to each subset during each iteration. Let's say the input set has 3 elements {0, 1, 2}, so selector will take values between 0 and 7 (inclusive), which in binary are:

000 # 0
001 # 1
010 # 2
011 # 3
100 # 4
101 # 5
110 # 6
111 # 7

So, each bit could serve as an indicator if an element of the original set should be added or not. Look at the binary numbers, and just think of each number as an element of the super set in which 1 means that an element at index j should be added, and 0 means that this element should not be added.

I am using a set comprehension to generate a subset at each iteration, and I convert this subset into a frozenset so I can add it to ps (power set). Otherwise, I won't be able to add it because a set in Python consists only of immutable objects.

Simplification

You can simplify the code using some python comprehensions, so you can get rid of those for loops. You can also use zip to avoid using j index and the code will end up as the following:

def power_set(A):
    length = len(A)
    return {
        frozenset({e for e, b in zip(A, f'{i:{length}b}') if b == '1'})
        for i in range(2 ** length)
    }

That's it. What I like of this algorithm is that is clearer and more intuitive than others because it looks quite magical to rely on itertools even though it works as expected.