Multiple nested loops equivalent replacement in python

Multiple nested loops equivalent replacement in python - python

I have a simple dictionary of the form D = {2: 4, 3: 2, 5: 1, 7: 1}. There are 4 keys which are the primes and 4 values which represent the maximum power. Based on that I want to generate all combinations of the form :
p1^range1 * p2^range2 * p3^range3 * p4^range4
where range1 to range4 are the ranges corresponding to the values of the dictionary of each prime. For p1 = 2 the range will be range(0, 4+1)
The most simplistic way would be a nested for loop with four levels :
for i1 in range(0, 4+1):
for i2 in range(0, 2+1):
for i3 in range(0, 1+1):
for i4 in range(0, 1+1):
n = 2**i1 * 3**i2 * 5**i3 * 7**i4
which is what I intend and it is correct but this is a non-elegant and mechanical way.
Another way is to use the itertools.product which works if all the ranges are the same. Basically it is as follows :
for x in itertools.product(range(0, 4+1), repeat=4):
print(x, end=' ')
However this will generate all the possible combinations including (2, 2, 3, 4),
(4, 4, 2, 3) or (2, 3, 4, 4) which are not valid in my case as the power of 5 (which is the third element in the tuple) will never be 3 and the power of 7 will never be 4. I want to limit somehow if possible the range of elements . So in this case the maximum tuple will be (4, 2, 1, 1) corresponding to the powers (values in dictionary D). So basically I need ONLY the range of tuples from (0,0,0,0) to (4,2,1,1).
Is there a way to achieve this without using for loops and do it in a similar approach as with itertools.product?
Many thanks in advance.

Actually,
As miradulo wrote the solution is very simple :
either use :
for x in itertools.product(*map(range, (5, 3, 2, 2))) :
print(x, end=' ')
or :
for x in itertools.product(range(5), range(3), range(2), range(2)) :
print(x, end=' ')
Both ways solve the problem.

While itertools is the way to go, this question is an excellent exercise for practicing generator functions:
def multirange(d):
if len(d) == 1:
for i in range(d[0]):
yield [i]
elif len(d) > 1:
for i in range(d[-1]):
for a in multirange(d[:-1]):
yield a + [i]
def multirange_b(d):
l = len(d)
products = [1]
for k in d:
products.append(products[-1]*k)
n = products[-1]
for i in range(n):
yield [(i%products[j+1])//products[j] for j in range(l)]
for l in multirange([4,3,2]):
print(l)
for l in multirange_b([4,3,2]):
print(l)

Related

Returning a generator inside of a generator

I have a list of integers and and a list of tuples (representing intervals), I want to write a method that, for each tuple, returns the sublists of integers contained in the interval but I want to do it with generators.
For the following input:
l = [1, 2, 3, 4, 5]
intervals = [(1, 2), (2, 4)]
the sublists should be: [1, 2] and [2, 3, 4].
My attempt:
def gen_intervals(l, intervals):
for e in l:
for i in intervals:
if e > i[0] and e < i[1]:
yield e
However, this would give me a list of elements because the generators yield one element at a time. What I want is to yield a generator of the elements in that interval.
Then, I would use it like this:
for interval in gen_intervals(l, intervals):
for e in interval:
print(e)
Important:
The list is sorted and so are the intervals (even though they may overlap). The intervals follow the standard mathematical representation of an interval: [left endpoint, right endpoint] with left endpoint < right endpoint. For any two intervals u, v, they cannot be a subset of the other. By sorted I mean that their right endpoints are sorted in ascending order.
I really want to iterate over elements first and intervals later because the list of elements is likely to be very, very long, so I just want to iterate over that list only once. The length of the list of elements is >> the length of the list of the intervals but the exact lengths are arbitrary.

You can simply use a generator comprehension on the yield line:
def gen_intervals(elements, intervals):
for vmin, vmax in intervals:
yield (elm for elm in elements if (vmin <= elm <= vmax))
Which gives:
l = [1, 2, 3, 4, 5]
intervals = [(2, 4), (1, 2)]
for interval in gen_intervals(l, intervals):
for e in interval:
print(e)
print()
2
3
4
1
2

You can do that like this. Each interval generator has a buffer, when asked for items it flushes its buffer first and then picks the next list element. If this element happens to be "its" element, yield it, otherwise place it in the respective interval's buffer and try with the next element.
def items_by_iterval(lst, intervals):
list_iter = iter(lst)
buffers = [[] for _ in intervals]
def interval_iter(n):
while True:
if buffers[n]:
yield from buffers[n]
buffers[n] = []
try:
k = next(list_iter)
except StopIteration:
return
for m, (a, b) in enumerate(intervals):
if a <= k <= b:
if m == n:
yield k
else:
buffers[m].append(k)
break
return [interval_iter(n) for n, _ in enumerate(intervals)]
##
lst = [1, 9, 2, 8, 3, 7, 4, 6, 5, 1, 9, 2, 8, 3, 7, 4, 6, 5, ]
intervals = [(5, 7), (2, 4), (8, 10)]
for ii in items_by_iterval(lst, intervals):
for k in ii:
print(k, end=' ')
print()
This prints:
7 6 5 7 6 5
2 3 4 2 3 4
9 8 9 8

Instead of any for loops you could use map to create a list with True and False values for each element for each interval. Then you could use itertools.compress to create an iterator object that yields the elements that are in respective interval:
import itertools
l = [1, 2, 3, 4, 5]
intervals = [(2, 4),]
for interval in intervals:
# use range(i+1, j) to check if element is in interval
mapped = map(lambda el: el in range(interval[0] + 1, interval[1]), l)
# mapped = [False, False, True, False, False]
res = itertools.compress(l, mapped) # save this in case of more intervals
print(next(res)) # returns 3

Based on #paime's solution:
def iterate_interval(l, interval):
for e in l:
if e > interval[0]:
if e < interval[1]:
yield e
else: break
def generate_windows(l, intervals):
for interval in intervals:
yield iterate_interval(l, interval)
This is more efficient because we do not need to continue iterating over the elements once an element is larger than the right endpoint of the interval.

Probably faster than yours, as it does two quick binary searches to find the range instead of comparing all elements with the interval bounds. Also, C is faster than Python.
from bisect import bisect_left, bisect_right
from itertools import islice
def generate_windows(l, intervals):
for left, right in intervals:
yield islice(
l,
bisect_left(l, left),
bisect_right(l, right)
)
l = [1, 2, 3, 4, 5]
intervals = [(1, 2), (2, 4)]
for window in generate_windows(l, intervals):
print(*window)
Output (Try it online!):
1 2
2 3 4
I have some other ideas that might be even faster, and usually I'd write a benchmark comparing the various solutions, but for that I'd need to know typical sizes, including average number of list elements per interval, and you didn't answer that.

Python permutations to create combinations using all and partial elements of a list [duplicate]

Given a set
{0, 1, 2, 3}
How can I produce the subsets:
[set(),
{0},
{1},
{2},
{3},
{0, 1},
{0, 2},
{0, 3},
{1, 2},
{1, 3},
{2, 3},
{0, 1, 2},
{0, 1, 3},
{0, 2, 3},
{1, 2, 3},
{0, 1, 2, 3}]

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]]

Use function powerset() from package more_itertools.
Yields all possible subsets of the iterable
>>> list(powerset([1, 2, 3]))
[(), (1,), (2,), (3,), (1, 2), (1, 3), (2, 3), (1, 2, 3)]
If you want sets, use:
list(map(set, powerset(iterable)))

I have found the following algorithm very clear and simple:
def get_powerset(some_list):
"""Returns all subsets of size 0 - len(some_list) for some_list"""
if len(some_list) == 0:
return [[]]
subsets = []
first_element = some_list[0]
remaining_list = some_list[1:]
# Strategy: get all the subsets of remaining_list. For each
# of those subsets, a full subset list will contain both
# the original subset as well as a version of the subset
# that contains first_element
for partial_subset in get_powerset(remaining_list):
subsets.append(partial_subset)
subsets.append(partial_subset[:] + [first_element])
return subsets
Another way one can generate the powerset is by generating all binary numbers that have n bits. As a power set the amount of number with n digits is 2 ^ n. The principle of this algorithm is that an element could be present or not in a subset as a binary digit could be one or zero but not both.
def power_set(items):
N = len(items)
# enumerate the 2 ** N possible combinations
for i in range(2 ** N):
combo = []
for j in range(N):
# test bit jth of integer i
if (i >> j) % 2 == 1:
combo.append(items[j])
yield combo
I found both algorithms when I was taking MITx: 6.00.2x Introduction to Computational Thinking and Data Science, and I consider it is one of the easiest algorithms to understand I have seen.

from functools import reduce
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.

def get_power_set(s):
power_set=[[]]
for elem in s:
# iterate over the sub sets so far
for sub_set in power_set:
# add a new subset consisting of the subset at hand added elem
power_set=power_set+[list(sub_set)+[elem]]
return power_set
For example:
get_power_set([1,2,3])
yield
[[], [1], [2], [1, 2], [3], [1, 3], [2, 3], [1, 2, 3]]

This can be done very naturally with itertools.product:
import itertools
def powerset(l):
for sl in itertools.product(*[[[], [i]] for i in l]):
yield {j for i in sl for j in i}

I know this is too late
There are many other solutions already but still...
def power_set(lst):
pw_set = [[]]
for i in range(0,len(lst)):
for j in range(0,len(pw_set)):
ele = pw_set[j].copy()
ele = ele + [lst[i]]
pw_set = pw_set + [ele]
return pw_set

I just wanted to provide the most comprehensible solution, the anti code-golf version.
from itertools import combinations
l = ["x", "y", "z", ]
def powerset(items):
combo = []
for r in range(len(items) + 1):
#use a list to coerce a actual list from the combinations generator
combo.append(list(combinations(items,r)))
return combo
l_powerset = powerset(l)
for i, item in enumerate(l_powerset):
print "All sets of length ", i
print item
The results
All sets of length 0
[()]
All sets of length 1
[('x',), ('y',), ('z',)]
All sets of length 2
[('x', 'y'), ('x', 'z'), ('y', 'z')]
All sets of length 3
[('x', 'y', 'z')]
For more see the itertools docs, also the wikipedia entry on power sets

With empty set, which is part of all the subsets, you could use:
def subsets(iterable):
for n in range(len(iterable) + 1):
yield from combinations(iterable, n)

from itertools import combinations
def subsets(arr: set) -> list:
subsets = []
[subsets.extend(list(combinations(arr,n))) for n in range(len(arr))]
return subsets
a = {1,2,3}
print(subsets(a))
Output:
[(), (1,), (2,), (3,), (1, 2), (1, 3), (2, 3)]
For sorted subsets, we can do:
# sorted subsets
print(sorted(subsets(a)))
Output:
[(), (1,), (1, 2), (1, 3), (2,), (2, 3), (3,)]

Just a quick power set refresher !
Power set of a set X, is simply the set of all subsets of X including
the empty set
Example set X = (a,b,c)
Power Set = { { a , b , c } , { a , b } , { a , c } , { b , c } , { a } , { b } , { c } , { } }
Here is another way of finding power set:
def power_set(input):
# returns a list of all subsets of the list a
if (len(input) == 0):
return [[]]
else:
main_subset = [ ]
for small_subset in power_set(input[1:]):
main_subset += [small_subset]
main_subset += [[input[0]] + small_subset]
return main_subset
print(power_set([0,1,2,3]))
full credit to source

If you want any specific length of subsets you can do it like this:
from itertools import combinations
someSet = {0, 1, 2, 3}
([x for i in range(len(someSet)+1) for x in combinations(someSet,i)])
More generally for arbitary length subsets you can modify the range arugment. The output is
[(), (0,), (1,), (2,), (3,), (0, 1), (0, 2), (0, 3), (1, 2), (1, 3), (2, 3), (0, 1, 2), (0, 1, 3), (0, 2, 3), (1, 2, 3), (0, 1, 2, 3)]

You can do it like this:
def powerset(x):
m=[]
if not x:
m.append(x)
else:
A = x[0]
B = x[1:]
for z in powerset(B):
m.append(z)
r = [A] + z
m.append(r)
return m
print(powerset([1, 2, 3, 4]))
Output:
[[], [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]]

A simple way would be to harness the internal representation of integers under 2's complement arithmetic.
Binary representation of integers is as {000, 001, 010, 011, 100, 101, 110, 111} for numbers ranging from 0 to 7. For an integer counter value, considering 1 as inclusion of corresponding element in collection and '0' as exclusion we can generate subsets based on the counting sequence. Numbers have to be generated from 0 to pow(2,n) -1 where n is the length of array i.e. number of bits in binary representation.
A simple Subset Generator Function based on it can be written as below. It basically relies
def subsets(array):
if not array:
return
else:
length = len(array)
for max_int in range(0x1 << length):
subset = []
for i in range(length):
if max_int & (0x1 << i):
subset.append(array[i])
yield subset
and then it can be used as
def get_subsets(array):
powerset = []
for i in subsets(array):
powerser.append(i)
return powerset
Testing
Adding following in local file
if __name__ == '__main__':
sample = ['b', 'd', 'f']
for i in range(len(sample)):
print "Subsets for " , sample[i:], " are ", get_subsets(sample[i:])
gives following output
Subsets for ['b', 'd', 'f'] are [[], ['b'], ['d'], ['b', 'd'], ['f'], ['b', 'f'], ['d', 'f'], ['b', 'd', 'f']]
Subsets for ['d', 'f'] are [[], ['d'], ['f'], ['d', 'f']]
Subsets for ['f'] are [[], ['f']]

Almost all of these answers use list rather than set, which felt like a bit of a cheat to me. So, out of curiosity I tried to do a simple version truly on set and summarize for other "new to Python" folks.
I found there's a couple oddities in dealing with Python's set implementation. The main surprise to me was handling empty sets. This is in contrast to Ruby's Set implementation, where I can simply do Set[Set[]] and get a Set containing one empty Set, so I found it initially a little confusing.
To review, in doing powerset with sets, I encountered two problems:
set() takes an iterable, so set(set()) will return set() because the empty set iterable is empty (duh I guess :))
to get a set of sets, set({set()}) and set.add(set) won't work because set() isn't hashable
To solve both issues, I made use of frozenset(), which means I don't quite get what I want (type is literally set), but makes use of the overall set interace.
def powerset(original_set):
# below gives us a set with one empty set in it
ps = set({frozenset()})
for member in original_set:
subset = set()
for m in ps:
# to be added into subset, needs to be
# frozenset.union(set) so it's hashable
subset.add(m.union(set([member]))
ps = ps.union(subset)
return ps
Below we get 2² (16) frozensets correctly as output:
In [1]: powerset(set([1,2,3,4]))
Out[2]:
{frozenset(),
frozenset({3, 4}),
frozenset({2}),
frozenset({1, 4}),
frozenset({3}),
frozenset({2, 3}),
frozenset({2, 3, 4}),
frozenset({1, 2}),
frozenset({2, 4}),
frozenset({1}),
frozenset({1, 2, 4}),
frozenset({1, 3}),
frozenset({1, 2, 3}),
frozenset({4}),
frozenset({1, 3, 4}),
frozenset({1, 2, 3, 4})}
As there's no way to have a set of sets in Python, if you want to turn these frozensets into sets, you'll have to map them back into a list (list(map(set, powerset(set([1,2,3,4]))))
) or modify the above.

Perhaps the question is getting old, but I hope my code will help someone.
def powSet(set):
if len(set) == 0:
return [[]]
return addtoAll(set[0],powSet(set[1:])) + powSet(set[1:])
def addtoAll(e, set):
for c in set:
c.append(e)
return set

Getting all the subsets with recursion. Crazy-ass one-liner
from typing import List
def subsets(xs: list) -> List[list]:
return subsets(xs[1:]) + [x + [xs[0]] for x in subsets(xs[1:])] if xs else [[]]
Based on a Haskell solution
subsets :: [a] -> [[a]]
subsets [] = [[]]
subsets (x:xs) = map (x:) (subsets xs) ++ subsets xs

def findsubsets(s, n):
return list(itertools.combinations(s, n))
def allsubsets(s) :
a = []
for x in range(1,len(s)+1):
a.append(map(set,findsubsets(s,x)))
return a

def powerSet(s):
sets = [[]]
for i in s:
newsets = []
for k in sets:
newsets.append(k+[i])
sets += newsets
return sets
Code first, for those who want a simple answer.
I have a good explanation here https://leetcode.com/problems/subsets/solutions/3138042/simple-python-solution/
But the short answer is that you start with the set of the empty set, i.e. "sets = [[]]". I recommend to put a print a statement under "for i in s" i.e. "print(sets)" and see that it doubles for each element i

This is wild because none of these answers actually provide the return of an actual Python set. Here is a messy implementation that will give a powerset that actually is a Python set.
test_set = set(['yo', 'whatup', 'money'])
def powerset( base_set ):
""" modified from pydoc's itertools recipe shown above"""
from itertools import chain, combinations
base_list = list( base_set )
combo_list = [ combinations(base_list, r) for r in range(len(base_set)+1) ]
powerset = set([])
for ll in combo_list:
list_of_frozensets = list( map( frozenset, map( list, ll ) ) )
set_of_frozensets = set( list_of_frozensets )
powerset = powerset.union( set_of_frozensets )
return powerset
print powerset( test_set )
# >>> set([ frozenset(['money','whatup']), frozenset(['money','whatup','yo']),
# frozenset(['whatup']), frozenset(['whatup','yo']), frozenset(['yo']),
# frozenset(['money','yo']), frozenset(['money']), frozenset([]) ])
I'd love to see a better implementation, though.

Here is my quick implementation utilizing combinations but using only built-ins.
def powerSet(array):
length = str(len(array))
formatter = '{:0' + length + 'b}'
combinations = []
for i in xrange(2**int(length)):
combinations.append(formatter.format(i))
sets = set()
currentSet = []
for combo in combinations:
for i,val in enumerate(combo):
if val=='1':
currentSet.append(array[i])
sets.add(tuple(sorted(currentSet)))
currentSet = []
return sets

All subsets in range n as set:
n = int(input())
l = [i for i in range (1, n + 1)]
for number in range(2 ** n) :
binary = bin(number)[: 1 : -1]
subset = [l[i] for i in range(len(binary)) if binary[i] == "1"]
print(set(sorted(subset)) if number > 0 else "{}")

import math
def printPowerSet(set,set_size):
pow_set_size =int(math.pow(2, set_size))
for counter in range(pow_set_size):
for j in range(set_size):
if((counter & (1 << j)) > 0):
print(set[j], end = "")
print("")
set = ['a', 'b', 'c']
printPowerSet(set,3)

A variation of the question, is an exercise I see on the book "Discovering Computer Science: Interdisciplinary Problems, Principles, and Python Programming. 2015 edition". In that exercise 10.2.11, the input is just an integer number, and the output should be the power sets. Here is my recursive solution (not using anything else but basic python3 )
def powerSetR(n):
assert n >= 0
if n == 0:
return [[]]
else:
input_set = list(range(1, n+1)) # [1,2,...n]
main_subset = [ ]
for small_subset in powerSetR(n-1):
main_subset += [small_subset]
main_subset += [ [input_set[-1]] + small_subset]
return main_subset
superset = powerSetR(4)
print(superset)
print("Number of sublists:", len(superset))
And the output is
[[], [4], [3], [4, 3], [2], [4, 2], [3, 2], [4, 3, 2], [1], [4, 1], [3, 1], [4, 3, 1], [2, 1], [4, 2, 1], [3, 2, 1], [4, 3, 2, 1]]
Number of sublists: 16

I hadn't come across the more_itertools.powerset function and would recommend using that. I also recommend not using the default ordering of the output from itertools.combinations, often instead you want to minimise the distance between the positions and sort the subsets of items with shorter distance between them above/before the items with larger distance between them.
The itertools recipes page shows it uses chain.from_iterable
Note that the r here matches the standard notation for the lower part of a binomial coefficient, the s is usually referred to as n in mathematics texts and on calculators (“n Choose r”)
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))
The other examples here give the powerset of [1,2,3,4] in such a way that the 2-tuples are listed in "lexicographic" order (when we print the numbers as integers). If I write the distance between the numbers alongside it (i.e. the difference), it shows my point:
12 ⇒ 1
13 ⇒ 2
14 ⇒ 3
23 ⇒ 1
24 ⇒ 2
34 ⇒ 1
The correct order for subsets should be the order which 'exhausts' the minimal distance first, like so:
12 ⇒ 1
23 ⇒ 1
34 ⇒ 1
13 ⇒ 2
24 ⇒ 2
14 ⇒ 3
Using numbers here makes this ordering look 'wrong', but consider for example the letters ["a","b","c","d"] it is clearer why this might be useful to obtain the powerset in this order:
ab ⇒ 1
bc ⇒ 1
cd ⇒ 1
ac ⇒ 2
bd ⇒ 2
ad ⇒ 3
This effect is more pronounced with more items, and for my purposes it makes the difference between being able to describe the ranges of the indexes of the powerset meaningfully.
(There is a lot written on Gray codes etc. for the output order of algorithms in combinatorics, I don't see it as a side issue).
I actually just wrote a fairly involved program which used this fast integer partition code to output the values in the proper order, but then I discovered more_itertools.powerset and for most uses it's probably fine to just use that function like so:
from more_itertools import powerset
from numpy import ediff1d
def ps_sorter(tup):
l = len(tup)
d = ediff1d(tup).tolist()
return l, d
ps = powerset([1,2,3,4])
ps = sorted(ps, key=ps_sorter)
for x in ps:
print(x)
⇣
()
(1,)
(2,)
(3,)
(4,)
(1, 2)
(2, 3)
(3, 4)
(1, 3)
(2, 4)
(1, 4)
(1, 2, 3)
(2, 3, 4)
(1, 2, 4)
(1, 3, 4)
(1, 2, 3, 4)
I wrote some more involved code which will print the powerset nicely (see the repo for pretty printing functions I've not included here: print_partitions, print_partitions_by_length, and pprint_tuple).
Repo: ordered-powerset, specifically pset_partitions.py
This is all pretty simple, but still might be useful if you want some code that'll let you get straight to accessing the different levels of the powerset:
from itertools import permutations as permute
from numpy import cumsum
# http://jeromekelleher.net/generating-integer-partitions.html
# via
# https://stackoverflow.com/questions/10035752/elegant-python-code-for-integer-partitioning#comment25080713_10036764
def asc_int_partitions(n):
a = [0 for i in range(n + 1)]
k = 1
y = n - 1
while k != 0:
x = a[k - 1] + 1
k -= 1
while 2 * x <= y:
a[k] = x
y -= x
k += 1
l = k + 1
while x <= y:
a[k] = x
a[l] = y
yield tuple(a[:k + 2])
x += 1
y -= 1
a[k] = x + y
y = x + y - 1
yield tuple(a[:k + 1])
# https://stackoverflow.com/a/6285330/2668831
def uniquely_permute(iterable, enforce_sort=False, r=None):
previous = tuple()
if enforce_sort: # potential waste of effort (default: False)
iterable = sorted(iterable)
for p in permute(iterable, r):
if p > previous:
previous = p
yield p
def sum_min(p):
return sum(p), min(p)
def partitions_by_length(max_n, sorting=True, permuting=False):
partition_dict = {0: ()}
for n in range(1,max_n+1):
partition_dict.setdefault(n, [])
partitions = list(asc_int_partitions(n))
for p in partitions:
if permuting:
perms = uniquely_permute(p)
for perm in perms:
partition_dict.get(len(p)).append(perm)
else:
partition_dict.get(len(p)).append(p)
if not sorting:
return partition_dict
for k in partition_dict:
partition_dict.update({k: sorted(partition_dict.get(k), key=sum_min)})
return partition_dict
def print_partitions_by_length(max_n, sorting=True, permuting=True):
partition_dict = partitions_by_length(max_n, sorting=sorting, permuting=permuting)
for k in partition_dict:
if k == 0:
print(tuple(partition_dict.get(k)), end="")
for p in partition_dict.get(k):
print(pprint_tuple(p), end=" ")
print()
return
def generate_powerset(items, subset_handler=tuple, verbose=False):
"""
Generate the powerset of an iterable `items`.
Handling of the elements of the iterable is by whichever function is passed as
`subset_handler`, which must be able to handle the `None` value for the
empty set. The function `string_handler` will join the elements of the subset
with the empty string (useful when `items` is an iterable of `str` variables).
"""
ps = {0: [subset_handler()]}
n = len(items)
p_dict = partitions_by_length(n-1, sorting=True, permuting=True)
for p_len, parts in p_dict.items():
ps.setdefault(p_len, [])
if p_len == 0:
# singletons
for offset in range(n):
subset = subset_handler([items[offset]])
if verbose:
if offset > 0:
print(end=" ")
if offset == n - 1:
print(subset, end="\n")
else:
print(subset, end=",")
ps.get(p_len).append(subset)
for pcount, partition in enumerate(parts):
distance = sum(partition)
indices = (cumsum(partition)).tolist()
for offset in range(n - distance):
subset = subset_handler([items[offset]] + [items[offset:][i] for i in indices])
if verbose:
if offset > 0:
print(end=" ")
if offset == n - distance - 1:
print(subset, end="\n")
else:
print(subset, end=",")
ps.get(p_len).append(subset)
if verbose and p_len < n-1:
print()
return ps
As an example, I wrote a CLI demo program which takes a string as a command line argument:
python string_powerset.py abcdef
⇣
a, b, c, d, e, f
ab, bc, cd, de, ef
ac, bd, ce, df
ad, be, cf
ae, bf
af
abc, bcd, cde, def
abd, bce, cdf
acd, bde, cef
abe, bcf
ade, bef
ace, bdf
abf
aef
acf
adf
abcd, bcde, cdef
abce, bcdf
abde, bcef
acde, bdef
abcf
abef
adef
abdf
acdf
acef
abcde, bcdef
abcdf
abcef
abdef
acdef
abcdef

Here it is my solutions, it is similar (conceptually) with the solution of lmiguelvargasf.
Let me say that
-[math item] by defintion the powerset do contain the empty set
-[personal taste] and also that I don't like using frozenset.
So the input is a list and the output will be a list of lists. The function could close earlier, but I like the element of the power set to be order lexicographically, that essentially means nicely.
def power_set(L):
"""
L is a list.
The function returns the power set, but as a list of lists.
"""
cardinality=len(L)
n=2 ** cardinality
powerset = []
for i in range(n):
a=bin(i)[2:]
subset=[]
for j in range(len(a)):
if a[-j-1]=='1':
subset.append(L[j])
powerset.append(subset)
#the function could stop here closing with
#return powerset
powerset_orderred=[]
for k in range(cardinality+1):
for w in powerset:
if len(w)==k:
powerset_orderred.append(w)
return powerset_orderred

Powerset of a python list [duplicate]

Given a set
{0, 1, 2, 3}
How can I produce the subsets:
[set(),
{0},
{1},
{2},
{3},
{0, 1},
{0, 2},
{0, 3},
{1, 2},
{1, 3},
{2, 3},
{0, 1, 2},
{0, 1, 3},
{0, 2, 3},
{1, 2, 3},
{0, 1, 2, 3}]

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]]

Use function powerset() from package more_itertools.
Yields all possible subsets of the iterable
>>> list(powerset([1, 2, 3]))
[(), (1,), (2,), (3,), (1, 2), (1, 3), (2, 3), (1, 2, 3)]
If you want sets, use:
list(map(set, powerset(iterable)))

I have found the following algorithm very clear and simple:
def get_powerset(some_list):
"""Returns all subsets of size 0 - len(some_list) for some_list"""
if len(some_list) == 0:
return [[]]
subsets = []
first_element = some_list[0]
remaining_list = some_list[1:]
# Strategy: get all the subsets of remaining_list. For each
# of those subsets, a full subset list will contain both
# the original subset as well as a version of the subset
# that contains first_element
for partial_subset in get_powerset(remaining_list):
subsets.append(partial_subset)
subsets.append(partial_subset[:] + [first_element])
return subsets
Another way one can generate the powerset is by generating all binary numbers that have n bits. As a power set the amount of number with n digits is 2 ^ n. The principle of this algorithm is that an element could be present or not in a subset as a binary digit could be one or zero but not both.
def power_set(items):
N = len(items)
# enumerate the 2 ** N possible combinations
for i in range(2 ** N):
combo = []
for j in range(N):
# test bit jth of integer i
if (i >> j) % 2 == 1:
combo.append(items[j])
yield combo
I found both algorithms when I was taking MITx: 6.00.2x Introduction to Computational Thinking and Data Science, and I consider it is one of the easiest algorithms to understand I have seen.

from functools import reduce
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.

def get_power_set(s):
power_set=[[]]
for elem in s:
# iterate over the sub sets so far
for sub_set in power_set:
# add a new subset consisting of the subset at hand added elem to it
# effectively doubling the sets resulting in the 2^n sets in the powerset of s.
power_set=power_set+[list(sub_set)+[elem]]
return power_set
For example:
get_power_set([1,2,3])
yield
[[], [1], [2], [1, 2], [3], [1, 3], [2, 3], [1, 2, 3]]

This can be done very naturally with itertools.product:
import itertools
def powerset(l):
for sl in itertools.product(*[[[], [i]] for i in l]):
yield {j for i in sl for j in i}

I know this is too late
There are many other solutions already but still...
def power_set(lst):
pw_set = [[]]
for i in range(0,len(lst)):
for j in range(0,len(pw_set)):
ele = pw_set[j].copy()
ele = ele + [lst[i]]
pw_set = pw_set + [ele]
return pw_set

I just wanted to provide the most comprehensible solution, the anti code-golf version.
from itertools import combinations
l = ["x", "y", "z", ]
def powerset(items):
combo = []
for r in range(len(items) + 1):
#use a list to coerce a actual list from the combinations generator
combo.append(list(combinations(items,r)))
return combo
l_powerset = powerset(l)
for i, item in enumerate(l_powerset):
print "All sets of length ", i
print item
The results
All sets of length 0
[()]
All sets of length 1
[('x',), ('y',), ('z',)]
All sets of length 2
[('x', 'y'), ('x', 'z'), ('y', 'z')]
All sets of length 3
[('x', 'y', 'z')]
For more see the itertools docs, also the wikipedia entry on power sets

With empty set, which is part of all the subsets, you could use:
def subsets(iterable):
for n in range(len(iterable) + 1):
yield from combinations(iterable, n)

from itertools import combinations
def subsets(arr: set) -> list:
subsets = []
[subsets.extend(list(combinations(arr,n))) for n in range(len(arr))]
return subsets
a = {1,2,3}
print(subsets(a))
Output:
[(), (1,), (2,), (3,), (1, 2), (1, 3), (2, 3)]
For sorted subsets, we can do:
# sorted subsets
print(sorted(subsets(a)))
Output:
[(), (1,), (1, 2), (1, 3), (2,), (2, 3), (3,)]

Just a quick power set refresher !
Power set of a set X, is simply the set of all subsets of X including
the empty set
Example set X = (a,b,c)
Power Set = { { a , b , c } , { a , b } , { a , c } , { b , c } , { a } , { b } , { c } , { } }
Here is another way of finding power set:
def power_set(input):
# returns a list of all subsets of the list a
if (len(input) == 0):
return [[]]
else:
main_subset = [ ]
for small_subset in power_set(input[1:]):
main_subset += [small_subset]
main_subset += [[input[0]] + small_subset]
return main_subset
print(power_set([0,1,2,3]))
full credit to source

If you want any specific length of subsets you can do it like this:
from itertools import combinations
someSet = {0, 1, 2, 3}
([x for i in range(len(someSet)+1) for x in combinations(someSet,i)])
More generally for arbitary length subsets you can modify the range arugment. The output is
[(), (0,), (1,), (2,), (3,), (0, 1), (0, 2), (0, 3), (1, 2), (1, 3), (2, 3), (0, 1, 2), (0, 1, 3), (0, 2, 3), (1, 2, 3), (0, 1, 2, 3)]

You can do it like this:
def powerset(x):
m=[]
if not x:
m.append(x)
else:
A = x[0]
B = x[1:]
for z in powerset(B):
m.append(z)
r = [A] + z
m.append(r)
return m
print(powerset([1, 2, 3, 4]))
Output:
[[], [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]]

A simple way would be to harness the internal representation of integers under 2's complement arithmetic.
Binary representation of integers is as {000, 001, 010, 011, 100, 101, 110, 111} for numbers ranging from 0 to 7. For an integer counter value, considering 1 as inclusion of corresponding element in collection and '0' as exclusion we can generate subsets based on the counting sequence. Numbers have to be generated from 0 to pow(2,n) -1 where n is the length of array i.e. number of bits in binary representation.
A simple Subset Generator Function based on it can be written as below. It basically relies
def subsets(array):
if not array:
return
else:
length = len(array)
for max_int in range(0x1 << length):
subset = []
for i in range(length):
if max_int & (0x1 << i):
subset.append(array[i])
yield subset
and then it can be used as
def get_subsets(array):
powerset = []
for i in subsets(array):
powerser.append(i)
return powerset
Testing
Adding following in local file
if __name__ == '__main__':
sample = ['b', 'd', 'f']
for i in range(len(sample)):
print "Subsets for " , sample[i:], " are ", get_subsets(sample[i:])
gives following output
Subsets for ['b', 'd', 'f'] are [[], ['b'], ['d'], ['b', 'd'], ['f'], ['b', 'f'], ['d', 'f'], ['b', 'd', 'f']]
Subsets for ['d', 'f'] are [[], ['d'], ['f'], ['d', 'f']]
Subsets for ['f'] are [[], ['f']]

Almost all of these answers use list rather than set, which felt like a bit of a cheat to me. So, out of curiosity I tried to do a simple version truly on set and summarize for other "new to Python" folks.
I found there's a couple oddities in dealing with Python's set implementation. The main surprise to me was handling empty sets. This is in contrast to Ruby's Set implementation, where I can simply do Set[Set[]] and get a Set containing one empty Set, so I found it initially a little confusing.
To review, in doing powerset with sets, I encountered two problems:
set() takes an iterable, so set(set()) will return set() because the empty set iterable is empty (duh I guess :))
to get a set of sets, set({set()}) and set.add(set) won't work because set() isn't hashable
To solve both issues, I made use of frozenset(), which means I don't quite get what I want (type is literally set), but makes use of the overall set interace.
def powerset(original_set):
# below gives us a set with one empty set in it
ps = set({frozenset()})
for member in original_set:
subset = set()
for m in ps:
# to be added into subset, needs to be
# frozenset.union(set) so it's hashable
subset.add(m.union(set([member]))
ps = ps.union(subset)
return ps
Below we get 2² (16) frozensets correctly as output:
In [1]: powerset(set([1,2,3,4]))
Out[2]:
{frozenset(),
frozenset({3, 4}),
frozenset({2}),
frozenset({1, 4}),
frozenset({3}),
frozenset({2, 3}),
frozenset({2, 3, 4}),
frozenset({1, 2}),
frozenset({2, 4}),
frozenset({1}),
frozenset({1, 2, 4}),
frozenset({1, 3}),
frozenset({1, 2, 3}),
frozenset({4}),
frozenset({1, 3, 4}),
frozenset({1, 2, 3, 4})}
As there's no way to have a set of sets in Python, if you want to turn these frozensets into sets, you'll have to map them back into a list (list(map(set, powerset(set([1,2,3,4]))))
) or modify the above.

Perhaps the question is getting old, but I hope my code will help someone.
def powSet(set):
if len(set) == 0:
return [[]]
return addtoAll(set[0],powSet(set[1:])) + powSet(set[1:])
def addtoAll(e, set):
for c in set:
c.append(e)
return set

Getting all the subsets with recursion. Crazy-ass one-liner
from typing import List
def subsets(xs: list) -> List[list]:
return subsets(xs[1:]) + [x + [xs[0]] for x in subsets(xs[1:])] if xs else [[]]
Based on a Haskell solution
subsets :: [a] -> [[a]]
subsets [] = [[]]
subsets (x:xs) = map (x:) (subsets xs) ++ subsets xs

def findsubsets(s, n):
return list(itertools.combinations(s, n))
def allsubsets(s) :
a = []
for x in range(1,len(s)+1):
a.append(map(set,findsubsets(s,x)))
return a

def powerSet(s):
sets = [[]]
for i in s:
newsets = []
for k in sets:
newsets.append(k+[i])
sets += newsets
return sets
Code first, for those who want a simple answer.
I have a good explanation here https://leetcode.com/problems/subsets/solutions/3138042/simple-python-solution/
But the short answer is that you start with the set of the empty set, i.e. "sets = [[]]". I recommend to put a print a statement under "for i in s" i.e. "print(sets)" and see that it doubles for each element i

This is wild because none of these answers actually provide the return of an actual Python set. Here is a messy implementation that will give a powerset that actually is a Python set.
test_set = set(['yo', 'whatup', 'money'])
def powerset( base_set ):
""" modified from pydoc's itertools recipe shown above"""
from itertools import chain, combinations
base_list = list( base_set )
combo_list = [ combinations(base_list, r) for r in range(len(base_set)+1) ]
powerset = set([])
for ll in combo_list:
list_of_frozensets = list( map( frozenset, map( list, ll ) ) )
set_of_frozensets = set( list_of_frozensets )
powerset = powerset.union( set_of_frozensets )
return powerset
print powerset( test_set )
# >>> set([ frozenset(['money','whatup']), frozenset(['money','whatup','yo']),
# frozenset(['whatup']), frozenset(['whatup','yo']), frozenset(['yo']),
# frozenset(['money','yo']), frozenset(['money']), frozenset([]) ])
I'd love to see a better implementation, though.

Here is my quick implementation utilizing combinations but using only built-ins.
def powerSet(array):
length = str(len(array))
formatter = '{:0' + length + 'b}'
combinations = []
for i in xrange(2**int(length)):
combinations.append(formatter.format(i))
sets = set()
currentSet = []
for combo in combinations:
for i,val in enumerate(combo):
if val=='1':
currentSet.append(array[i])
sets.add(tuple(sorted(currentSet)))
currentSet = []
return sets

All subsets in range n as set:
n = int(input())
l = [i for i in range (1, n + 1)]
for number in range(2 ** n) :
binary = bin(number)[: 1 : -1]
subset = [l[i] for i in range(len(binary)) if binary[i] == "1"]
print(set(sorted(subset)) if number > 0 else "{}")

import math
def printPowerSet(set,set_size):
pow_set_size =int(math.pow(2, set_size))
for counter in range(pow_set_size):
for j in range(set_size):
if((counter & (1 << j)) > 0):
print(set[j], end = "")
print("")
set = ['a', 'b', 'c']
printPowerSet(set,3)

A variation of the question, is an exercise I see on the book "Discovering Computer Science: Interdisciplinary Problems, Principles, and Python Programming. 2015 edition". In that exercise 10.2.11, the input is just an integer number, and the output should be the power sets. Here is my recursive solution (not using anything else but basic python3 )
def powerSetR(n):
assert n >= 0
if n == 0:
return [[]]
else:
input_set = list(range(1, n+1)) # [1,2,...n]
main_subset = [ ]
for small_subset in powerSetR(n-1):
main_subset += [small_subset]
main_subset += [ [input_set[-1]] + small_subset]
return main_subset
superset = powerSetR(4)
print(superset)
print("Number of sublists:", len(superset))
And the output is
[[], [4], [3], [4, 3], [2], [4, 2], [3, 2], [4, 3, 2], [1], [4, 1], [3, 1], [4, 3, 1], [2, 1], [4, 2, 1], [3, 2, 1], [4, 3, 2, 1]]
Number of sublists: 16

I hadn't come across the more_itertools.powerset function and would recommend using that. I also recommend not using the default ordering of the output from itertools.combinations, often instead you want to minimise the distance between the positions and sort the subsets of items with shorter distance between them above/before the items with larger distance between them.
The itertools recipes page shows it uses chain.from_iterable
Note that the r here matches the standard notation for the lower part of a binomial coefficient, the s is usually referred to as n in mathematics texts and on calculators (“n Choose r”)
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))
The other examples here give the powerset of [1,2,3,4] in such a way that the 2-tuples are listed in "lexicographic" order (when we print the numbers as integers). If I write the distance between the numbers alongside it (i.e. the difference), it shows my point:
12 ⇒ 1
13 ⇒ 2
14 ⇒ 3
23 ⇒ 1
24 ⇒ 2
34 ⇒ 1
The correct order for subsets should be the order which 'exhausts' the minimal distance first, like so:
12 ⇒ 1
23 ⇒ 1
34 ⇒ 1
13 ⇒ 2
24 ⇒ 2
14 ⇒ 3
Using numbers here makes this ordering look 'wrong', but consider for example the letters ["a","b","c","d"] it is clearer why this might be useful to obtain the powerset in this order:
ab ⇒ 1
bc ⇒ 1
cd ⇒ 1
ac ⇒ 2
bd ⇒ 2
ad ⇒ 3
This effect is more pronounced with more items, and for my purposes it makes the difference between being able to describe the ranges of the indexes of the powerset meaningfully.
(There is a lot written on Gray codes etc. for the output order of algorithms in combinatorics, I don't see it as a side issue).
I actually just wrote a fairly involved program which used this fast integer partition code to output the values in the proper order, but then I discovered more_itertools.powerset and for most uses it's probably fine to just use that function like so:
from more_itertools import powerset
from numpy import ediff1d
def ps_sorter(tup):
l = len(tup)
d = ediff1d(tup).tolist()
return l, d
ps = powerset([1,2,3,4])
ps = sorted(ps, key=ps_sorter)
for x in ps:
print(x)
⇣
()
(1,)
(2,)
(3,)
(4,)
(1, 2)
(2, 3)
(3, 4)
(1, 3)
(2, 4)
(1, 4)
(1, 2, 3)
(2, 3, 4)
(1, 2, 4)
(1, 3, 4)
(1, 2, 3, 4)
I wrote some more involved code which will print the powerset nicely (see the repo for pretty printing functions I've not included here: print_partitions, print_partitions_by_length, and pprint_tuple).
Repo: ordered-powerset, specifically pset_partitions.py
This is all pretty simple, but still might be useful if you want some code that'll let you get straight to accessing the different levels of the powerset:
from itertools import permutations as permute
from numpy import cumsum
# http://jeromekelleher.net/generating-integer-partitions.html
# via
# https://stackoverflow.com/questions/10035752/elegant-python-code-for-integer-partitioning#comment25080713_10036764
def asc_int_partitions(n):
a = [0 for i in range(n + 1)]
k = 1
y = n - 1
while k != 0:
x = a[k - 1] + 1
k -= 1
while 2 * x <= y:
a[k] = x
y -= x
k += 1
l = k + 1
while x <= y:
a[k] = x
a[l] = y
yield tuple(a[:k + 2])
x += 1
y -= 1
a[k] = x + y
y = x + y - 1
yield tuple(a[:k + 1])
# https://stackoverflow.com/a/6285330/2668831
def uniquely_permute(iterable, enforce_sort=False, r=None):
previous = tuple()
if enforce_sort: # potential waste of effort (default: False)
iterable = sorted(iterable)
for p in permute(iterable, r):
if p > previous:
previous = p
yield p
def sum_min(p):
return sum(p), min(p)
def partitions_by_length(max_n, sorting=True, permuting=False):
partition_dict = {0: ()}
for n in range(1,max_n+1):
partition_dict.setdefault(n, [])
partitions = list(asc_int_partitions(n))
for p in partitions:
if permuting:
perms = uniquely_permute(p)
for perm in perms:
partition_dict.get(len(p)).append(perm)
else:
partition_dict.get(len(p)).append(p)
if not sorting:
return partition_dict
for k in partition_dict:
partition_dict.update({k: sorted(partition_dict.get(k), key=sum_min)})
return partition_dict
def print_partitions_by_length(max_n, sorting=True, permuting=True):
partition_dict = partitions_by_length(max_n, sorting=sorting, permuting=permuting)
for k in partition_dict:
if k == 0:
print(tuple(partition_dict.get(k)), end="")
for p in partition_dict.get(k):
print(pprint_tuple(p), end=" ")
print()
return
def generate_powerset(items, subset_handler=tuple, verbose=False):
"""
Generate the powerset of an iterable `items`.
Handling of the elements of the iterable is by whichever function is passed as
`subset_handler`, which must be able to handle the `None` value for the
empty set. The function `string_handler` will join the elements of the subset
with the empty string (useful when `items` is an iterable of `str` variables).
"""
ps = {0: [subset_handler()]}
n = len(items)
p_dict = partitions_by_length(n-1, sorting=True, permuting=True)
for p_len, parts in p_dict.items():
ps.setdefault(p_len, [])
if p_len == 0:
# singletons
for offset in range(n):
subset = subset_handler([items[offset]])
if verbose:
if offset > 0:
print(end=" ")
if offset == n - 1:
print(subset, end="\n")
else:
print(subset, end=",")
ps.get(p_len).append(subset)
for pcount, partition in enumerate(parts):
distance = sum(partition)
indices = (cumsum(partition)).tolist()
for offset in range(n - distance):
subset = subset_handler([items[offset]] + [items[offset:][i] for i in indices])
if verbose:
if offset > 0:
print(end=" ")
if offset == n - distance - 1:
print(subset, end="\n")
else:
print(subset, end=",")
ps.get(p_len).append(subset)
if verbose and p_len < n-1:
print()
return ps
As an example, I wrote a CLI demo program which takes a string as a command line argument:
python string_powerset.py abcdef
⇣
a, b, c, d, e, f
ab, bc, cd, de, ef
ac, bd, ce, df
ad, be, cf
ae, bf
af
abc, bcd, cde, def
abd, bce, cdf
acd, bde, cef
abe, bcf
ade, bef
ace, bdf
abf
aef
acf
adf
abcd, bcde, cdef
abce, bcdf
abde, bcef
acde, bdef
abcf
abef
adef
abdf
acdf
acef
abcde, bcdef
abcdf
abcef
abdef
acdef
abcdef

Here it is my solutions, it is similar (conceptually) with the solution of lmiguelvargasf.
Let me say that
-[math item] by defintion the powerset do contain the empty set
-[personal taste] and also that I don't like using frozenset.
So the input is a list and the output will be a list of lists. The function could close earlier, but I like the element of the power set to be order lexicographically, that essentially means nicely.
def power_set(L):
"""
L is a list.
The function returns the power set, but as a list of lists.
"""
cardinality=len(L)
n=2 ** cardinality
powerset = []
for i in range(n):
a=bin(i)[2:]
subset=[]
for j in range(len(a)):
if a[-j-1]=='1':
subset.append(L[j])
powerset.append(subset)
#the function could stop here closing with
#return powerset
powerset_orderred=[]
for k in range(cardinality+1):
for w in powerset:
if len(w)==k:
powerset_orderred.append(w)
return powerset_orderred

The fastest way to find all pairs of numbers in a list that differ not more than x

What I have now:
d = 0
res = 0
newlist = []
l = [4, 1, 6, 1, 1, 1]
for el in range(len(l)):
for j in range(len(l)):
if abs(l[el] - l[j]) <= d and el != j and el not in newlist and j not in newlist:
newlist.append(el)
newlist.append(j)
res += 1
print(res)
It works well and returns 2 which is correct(1,1; 1,1) but takes too much time. How can I make it work faster ? Thanks.
For example if list = [1, 1, 1, 1] and d = 0 there will be 2 pairs because you can use each number only once. Using (a, b) and (b, c) is not allowed and (a, b) with (b, a) is the same pair...

Sort the list, then walk through it.
Once you have the list sorted, you can just be greedy: take the earliest pair that works, then the next, then the next... and you will end up with the maximum number of valid pairs.
def get_pairs(lst, maxdiff):
sl = sorted(lst) # may want to do lst.sort() if you don't mind changing lst
count = 0
i = 1
N = len(sl)
while i < N:
# no need for abs -- we know the previous value is not bigger.
if sl[i] - sl[i-1] <= maxdiff:
count += 1
i += 2 # these two values are now used
else:
i += 1
return count
And here's some code to benchmark it:
print('generating list...')
from random import randrange, seed
seed(0) # always same contents
l = []
for i in range(1000000):
l.append(randrange(0,5000))
print('ok, measuring...')
from time import time
start = time();
print(get_pairs(l, 0))
print('took', time()-start, 'seconds')
And the result (with 1 million values in list):
tmp$ ./test.py
generating list...
ok, measuring...
498784
took 0.6729779243469238 seconds

You may want to compute all the pairs separately and then collect the pairs you want.
def get_pairs(l, difference):
pairs = []
# first compute all pairs: n choose 2 which is O(n^2)
for i in xrange(len(l)):
for j in xrange(i+1, len(l)):
pairs.append((l[i], l[j]))
# collect pairs you want: O(n^2)
res = []
for pair in pairs:
if abs(pair[0] - pair[1]) <= difference:
res.append(pair)
return res
>>> get_pairs([1,2,3,4,2], 0)
>>> [(2, 2)]
>>> get_pairs([1,2,3,4,2], 1)
>>> [(1, 2), (1, 2), (2, 3), (2, 2), (3, 4), (3, 2)]
If you want to remove duplicates from you result, you can convert the res list to a set before you return it with set(res).

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

Given a set
{0, 1, 2, 3}
How can I produce the subsets:
[set(),
{0},
{1},
{2},
{3},
{0, 1},
{0, 2},
{0, 3},
{1, 2},
{1, 3},
{2, 3},
{0, 1, 2},
{0, 1, 3},
{0, 2, 3},
{1, 2, 3},
{0, 1, 2, 3}]

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]]

Use function powerset() from package more_itertools.
Yields all possible subsets of the iterable
>>> list(powerset([1, 2, 3]))
[(), (1,), (2,), (3,), (1, 2), (1, 3), (2, 3), (1, 2, 3)]
If you want sets, use:
list(map(set, powerset(iterable)))

I have found the following algorithm very clear and simple:
def get_powerset(some_list):
"""Returns all subsets of size 0 - len(some_list) for some_list"""
if len(some_list) == 0:
return [[]]
subsets = []
first_element = some_list[0]
remaining_list = some_list[1:]
# Strategy: get all the subsets of remaining_list. For each
# of those subsets, a full subset list will contain both
# the original subset as well as a version of the subset
# that contains first_element
for partial_subset in get_powerset(remaining_list):
subsets.append(partial_subset)
subsets.append(partial_subset[:] + [first_element])
return subsets
Another way one can generate the powerset is by generating all binary numbers that have n bits. As a power set the amount of number with n digits is 2 ^ n. The principle of this algorithm is that an element could be present or not in a subset as a binary digit could be one or zero but not both.
def power_set(items):
N = len(items)
# enumerate the 2 ** N possible combinations
for i in range(2 ** N):
combo = []
for j in range(N):
# test bit jth of integer i
if (i >> j) % 2 == 1:
combo.append(items[j])
yield combo
I found both algorithms when I was taking MITx: 6.00.2x Introduction to Computational Thinking and Data Science, and I consider it is one of the easiest algorithms to understand I have seen.

from functools import reduce
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.

def get_power_set(s):
power_set=[[]]
for elem in s:
# iterate over the sub sets so far
for sub_set in power_set:
# add a new subset consisting of the subset at hand added elem to it
# effectively doubling the sets resulting in the 2^n sets in the powerset of s.
power_set=power_set+[list(sub_set)+[elem]]
return power_set
For example:
get_power_set([1,2,3])
yield
[[], [1], [2], [1, 2], [3], [1, 3], [2, 3], [1, 2, 3]]

This can be done very naturally with itertools.product:
import itertools
def powerset(l):
for sl in itertools.product(*[[[], [i]] for i in l]):
yield {j for i in sl for j in i}

I know this is too late
There are many other solutions already but still...
def power_set(lst):
pw_set = [[]]
for i in range(0,len(lst)):
for j in range(0,len(pw_set)):
ele = pw_set[j].copy()
ele = ele + [lst[i]]
pw_set = pw_set + [ele]
return pw_set

I just wanted to provide the most comprehensible solution, the anti code-golf version.
from itertools import combinations
l = ["x", "y", "z", ]
def powerset(items):
combo = []
for r in range(len(items) + 1):
#use a list to coerce a actual list from the combinations generator
combo.append(list(combinations(items,r)))
return combo
l_powerset = powerset(l)
for i, item in enumerate(l_powerset):
print "All sets of length ", i
print item
The results
All sets of length 0
[()]
All sets of length 1
[('x',), ('y',), ('z',)]
All sets of length 2
[('x', 'y'), ('x', 'z'), ('y', 'z')]
All sets of length 3
[('x', 'y', 'z')]
For more see the itertools docs, also the wikipedia entry on power sets

With empty set, which is part of all the subsets, you could use:
def subsets(iterable):
for n in range(len(iterable) + 1):
yield from combinations(iterable, n)

from itertools import combinations
def subsets(arr: set) -> list:
subsets = []
[subsets.extend(list(combinations(arr,n))) for n in range(len(arr))]
return subsets
a = {1,2,3}
print(subsets(a))
Output:
[(), (1,), (2,), (3,), (1, 2), (1, 3), (2, 3)]
For sorted subsets, we can do:
# sorted subsets
print(sorted(subsets(a)))
Output:
[(), (1,), (1, 2), (1, 3), (2,), (2, 3), (3,)]

Just a quick power set refresher !
Power set of a set X, is simply the set of all subsets of X including
the empty set
Example set X = (a,b,c)
Power Set = { { a , b , c } , { a , b } , { a , c } , { b , c } , { a } , { b } , { c } , { } }
Here is another way of finding power set:
def power_set(input):
# returns a list of all subsets of the list a
if (len(input) == 0):
return [[]]
else:
main_subset = [ ]
for small_subset in power_set(input[1:]):
main_subset += [small_subset]
main_subset += [[input[0]] + small_subset]
return main_subset
print(power_set([0,1,2,3]))
full credit to source

If you want any specific length of subsets you can do it like this:
from itertools import combinations
someSet = {0, 1, 2, 3}
([x for i in range(len(someSet)+1) for x in combinations(someSet,i)])
More generally for arbitary length subsets you can modify the range arugment. The output is
[(), (0,), (1,), (2,), (3,), (0, 1), (0, 2), (0, 3), (1, 2), (1, 3), (2, 3), (0, 1, 2), (0, 1, 3), (0, 2, 3), (1, 2, 3), (0, 1, 2, 3)]

You can do it like this:
def powerset(x):
m=[]
if not x:
m.append(x)
else:
A = x[0]
B = x[1:]
for z in powerset(B):
m.append(z)
r = [A] + z
m.append(r)
return m
print(powerset([1, 2, 3, 4]))
Output:
[[], [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]]

A simple way would be to harness the internal representation of integers under 2's complement arithmetic.
Binary representation of integers is as {000, 001, 010, 011, 100, 101, 110, 111} for numbers ranging from 0 to 7. For an integer counter value, considering 1 as inclusion of corresponding element in collection and '0' as exclusion we can generate subsets based on the counting sequence. Numbers have to be generated from 0 to pow(2,n) -1 where n is the length of array i.e. number of bits in binary representation.
A simple Subset Generator Function based on it can be written as below. It basically relies
def subsets(array):
if not array:
return
else:
length = len(array)
for max_int in range(0x1 << length):
subset = []
for i in range(length):
if max_int & (0x1 << i):
subset.append(array[i])
yield subset
and then it can be used as
def get_subsets(array):
powerset = []
for i in subsets(array):
powerser.append(i)
return powerset
Testing
Adding following in local file
if __name__ == '__main__':
sample = ['b', 'd', 'f']
for i in range(len(sample)):
print "Subsets for " , sample[i:], " are ", get_subsets(sample[i:])
gives following output
Subsets for ['b', 'd', 'f'] are [[], ['b'], ['d'], ['b', 'd'], ['f'], ['b', 'f'], ['d', 'f'], ['b', 'd', 'f']]
Subsets for ['d', 'f'] are [[], ['d'], ['f'], ['d', 'f']]
Subsets for ['f'] are [[], ['f']]

Almost all of these answers use list rather than set, which felt like a bit of a cheat to me. So, out of curiosity I tried to do a simple version truly on set and summarize for other "new to Python" folks.
I found there's a couple oddities in dealing with Python's set implementation. The main surprise to me was handling empty sets. This is in contrast to Ruby's Set implementation, where I can simply do Set[Set[]] and get a Set containing one empty Set, so I found it initially a little confusing.
To review, in doing powerset with sets, I encountered two problems:
set() takes an iterable, so set(set()) will return set() because the empty set iterable is empty (duh I guess :))
to get a set of sets, set({set()}) and set.add(set) won't work because set() isn't hashable
To solve both issues, I made use of frozenset(), which means I don't quite get what I want (type is literally set), but makes use of the overall set interace.
def powerset(original_set):
# below gives us a set with one empty set in it
ps = set({frozenset()})
for member in original_set:
subset = set()
for m in ps:
# to be added into subset, needs to be
# frozenset.union(set) so it's hashable
subset.add(m.union(set([member]))
ps = ps.union(subset)
return ps
Below we get 2² (16) frozensets correctly as output:
In [1]: powerset(set([1,2,3,4]))
Out[2]:
{frozenset(),
frozenset({3, 4}),
frozenset({2}),
frozenset({1, 4}),
frozenset({3}),
frozenset({2, 3}),
frozenset({2, 3, 4}),
frozenset({1, 2}),
frozenset({2, 4}),
frozenset({1}),
frozenset({1, 2, 4}),
frozenset({1, 3}),
frozenset({1, 2, 3}),
frozenset({4}),
frozenset({1, 3, 4}),
frozenset({1, 2, 3, 4})}
As there's no way to have a set of sets in Python, if you want to turn these frozensets into sets, you'll have to map them back into a list (list(map(set, powerset(set([1,2,3,4]))))
) or modify the above.

Perhaps the question is getting old, but I hope my code will help someone.
def powSet(set):
if len(set) == 0:
return [[]]
return addtoAll(set[0],powSet(set[1:])) + powSet(set[1:])
def addtoAll(e, set):
for c in set:
c.append(e)
return set

Getting all the subsets with recursion. Crazy-ass one-liner
from typing import List
def subsets(xs: list) -> List[list]:
return subsets(xs[1:]) + [x + [xs[0]] for x in subsets(xs[1:])] if xs else [[]]
Based on a Haskell solution
subsets :: [a] -> [[a]]
subsets [] = [[]]
subsets (x:xs) = map (x:) (subsets xs) ++ subsets xs

def findsubsets(s, n):
return list(itertools.combinations(s, n))
def allsubsets(s) :
a = []
for x in range(1,len(s)+1):
a.append(map(set,findsubsets(s,x)))
return a

def powerSet(s):
sets = [[]]
for i in s:
newsets = []
for k in sets:
newsets.append(k+[i])
sets += newsets
return sets
Code first, for those who want a simple answer.
I have a good explanation here https://leetcode.com/problems/subsets/solutions/3138042/simple-python-solution/
But the short answer is that you start with the set of the empty set, i.e. "sets = [[]]". I recommend to put a print a statement under "for i in s" i.e. "print(sets)" and see that it doubles for each element i

This is wild because none of these answers actually provide the return of an actual Python set. Here is a messy implementation that will give a powerset that actually is a Python set.
test_set = set(['yo', 'whatup', 'money'])
def powerset( base_set ):
""" modified from pydoc's itertools recipe shown above"""
from itertools import chain, combinations
base_list = list( base_set )
combo_list = [ combinations(base_list, r) for r in range(len(base_set)+1) ]
powerset = set([])
for ll in combo_list:
list_of_frozensets = list( map( frozenset, map( list, ll ) ) )
set_of_frozensets = set( list_of_frozensets )
powerset = powerset.union( set_of_frozensets )
return powerset
print powerset( test_set )
# >>> set([ frozenset(['money','whatup']), frozenset(['money','whatup','yo']),
# frozenset(['whatup']), frozenset(['whatup','yo']), frozenset(['yo']),
# frozenset(['money','yo']), frozenset(['money']), frozenset([]) ])
I'd love to see a better implementation, though.

Here is my quick implementation utilizing combinations but using only built-ins.
def powerSet(array):
length = str(len(array))
formatter = '{:0' + length + 'b}'
combinations = []
for i in xrange(2**int(length)):
combinations.append(formatter.format(i))
sets = set()
currentSet = []
for combo in combinations:
for i,val in enumerate(combo):
if val=='1':
currentSet.append(array[i])
sets.add(tuple(sorted(currentSet)))
currentSet = []
return sets

All subsets in range n as set:
n = int(input())
l = [i for i in range (1, n + 1)]
for number in range(2 ** n) :
binary = bin(number)[: 1 : -1]
subset = [l[i] for i in range(len(binary)) if binary[i] == "1"]
print(set(sorted(subset)) if number > 0 else "{}")

import math
def printPowerSet(set,set_size):
pow_set_size =int(math.pow(2, set_size))
for counter in range(pow_set_size):
for j in range(set_size):
if((counter & (1 << j)) > 0):
print(set[j], end = "")
print("")
set = ['a', 'b', 'c']
printPowerSet(set,3)

A variation of the question, is an exercise I see on the book "Discovering Computer Science: Interdisciplinary Problems, Principles, and Python Programming. 2015 edition". In that exercise 10.2.11, the input is just an integer number, and the output should be the power sets. Here is my recursive solution (not using anything else but basic python3 )
def powerSetR(n):
assert n >= 0
if n == 0:
return [[]]
else:
input_set = list(range(1, n+1)) # [1,2,...n]
main_subset = [ ]
for small_subset in powerSetR(n-1):
main_subset += [small_subset]
main_subset += [ [input_set[-1]] + small_subset]
return main_subset
superset = powerSetR(4)
print(superset)
print("Number of sublists:", len(superset))
And the output is
[[], [4], [3], [4, 3], [2], [4, 2], [3, 2], [4, 3, 2], [1], [4, 1], [3, 1], [4, 3, 1], [2, 1], [4, 2, 1], [3, 2, 1], [4, 3, 2, 1]]
Number of sublists: 16

I hadn't come across the more_itertools.powerset function and would recommend using that. I also recommend not using the default ordering of the output from itertools.combinations, often instead you want to minimise the distance between the positions and sort the subsets of items with shorter distance between them above/before the items with larger distance between them.
The itertools recipes page shows it uses chain.from_iterable
Note that the r here matches the standard notation for the lower part of a binomial coefficient, the s is usually referred to as n in mathematics texts and on calculators (“n Choose r”)
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))
The other examples here give the powerset of [1,2,3,4] in such a way that the 2-tuples are listed in "lexicographic" order (when we print the numbers as integers). If I write the distance between the numbers alongside it (i.e. the difference), it shows my point:
12 ⇒ 1
13 ⇒ 2
14 ⇒ 3
23 ⇒ 1
24 ⇒ 2
34 ⇒ 1
The correct order for subsets should be the order which 'exhausts' the minimal distance first, like so:
12 ⇒ 1
23 ⇒ 1
34 ⇒ 1
13 ⇒ 2
24 ⇒ 2
14 ⇒ 3
Using numbers here makes this ordering look 'wrong', but consider for example the letters ["a","b","c","d"] it is clearer why this might be useful to obtain the powerset in this order:
ab ⇒ 1
bc ⇒ 1
cd ⇒ 1
ac ⇒ 2
bd ⇒ 2
ad ⇒ 3
This effect is more pronounced with more items, and for my purposes it makes the difference between being able to describe the ranges of the indexes of the powerset meaningfully.
(There is a lot written on Gray codes etc. for the output order of algorithms in combinatorics, I don't see it as a side issue).
I actually just wrote a fairly involved program which used this fast integer partition code to output the values in the proper order, but then I discovered more_itertools.powerset and for most uses it's probably fine to just use that function like so:
from more_itertools import powerset
from numpy import ediff1d
def ps_sorter(tup):
l = len(tup)
d = ediff1d(tup).tolist()
return l, d
ps = powerset([1,2,3,4])
ps = sorted(ps, key=ps_sorter)
for x in ps:
print(x)
⇣
()
(1,)
(2,)
(3,)
(4,)
(1, 2)
(2, 3)
(3, 4)
(1, 3)
(2, 4)
(1, 4)
(1, 2, 3)
(2, 3, 4)
(1, 2, 4)
(1, 3, 4)
(1, 2, 3, 4)
I wrote some more involved code which will print the powerset nicely (see the repo for pretty printing functions I've not included here: print_partitions, print_partitions_by_length, and pprint_tuple).
Repo: ordered-powerset, specifically pset_partitions.py
This is all pretty simple, but still might be useful if you want some code that'll let you get straight to accessing the different levels of the powerset:
from itertools import permutations as permute
from numpy import cumsum
# http://jeromekelleher.net/generating-integer-partitions.html
# via
# https://stackoverflow.com/questions/10035752/elegant-python-code-for-integer-partitioning#comment25080713_10036764
def asc_int_partitions(n):
a = [0 for i in range(n + 1)]
k = 1
y = n - 1
while k != 0:
x = a[k - 1] + 1
k -= 1
while 2 * x <= y:
a[k] = x
y -= x
k += 1
l = k + 1
while x <= y:
a[k] = x
a[l] = y
yield tuple(a[:k + 2])
x += 1
y -= 1
a[k] = x + y
y = x + y - 1
yield tuple(a[:k + 1])
# https://stackoverflow.com/a/6285330/2668831
def uniquely_permute(iterable, enforce_sort=False, r=None):
previous = tuple()
if enforce_sort: # potential waste of effort (default: False)
iterable = sorted(iterable)
for p in permute(iterable, r):
if p > previous:
previous = p
yield p
def sum_min(p):
return sum(p), min(p)
def partitions_by_length(max_n, sorting=True, permuting=False):
partition_dict = {0: ()}
for n in range(1,max_n+1):
partition_dict.setdefault(n, [])
partitions = list(asc_int_partitions(n))
for p in partitions:
if permuting:
perms = uniquely_permute(p)
for perm in perms:
partition_dict.get(len(p)).append(perm)
else:
partition_dict.get(len(p)).append(p)
if not sorting:
return partition_dict
for k in partition_dict:
partition_dict.update({k: sorted(partition_dict.get(k), key=sum_min)})
return partition_dict
def print_partitions_by_length(max_n, sorting=True, permuting=True):
partition_dict = partitions_by_length(max_n, sorting=sorting, permuting=permuting)
for k in partition_dict:
if k == 0:
print(tuple(partition_dict.get(k)), end="")
for p in partition_dict.get(k):
print(pprint_tuple(p), end=" ")
print()
return
def generate_powerset(items, subset_handler=tuple, verbose=False):
"""
Generate the powerset of an iterable `items`.
Handling of the elements of the iterable is by whichever function is passed as
`subset_handler`, which must be able to handle the `None` value for the
empty set. The function `string_handler` will join the elements of the subset
with the empty string (useful when `items` is an iterable of `str` variables).
"""
ps = {0: [subset_handler()]}
n = len(items)
p_dict = partitions_by_length(n-1, sorting=True, permuting=True)
for p_len, parts in p_dict.items():
ps.setdefault(p_len, [])
if p_len == 0:
# singletons
for offset in range(n):
subset = subset_handler([items[offset]])
if verbose:
if offset > 0:
print(end=" ")
if offset == n - 1:
print(subset, end="\n")
else:
print(subset, end=",")
ps.get(p_len).append(subset)
for pcount, partition in enumerate(parts):
distance = sum(partition)
indices = (cumsum(partition)).tolist()
for offset in range(n - distance):
subset = subset_handler([items[offset]] + [items[offset:][i] for i in indices])
if verbose:
if offset > 0:
print(end=" ")
if offset == n - distance - 1:
print(subset, end="\n")
else:
print(subset, end=",")
ps.get(p_len).append(subset)
if verbose and p_len < n-1:
print()
return ps
As an example, I wrote a CLI demo program which takes a string as a command line argument:
python string_powerset.py abcdef
⇣
a, b, c, d, e, f
ab, bc, cd, de, ef
ac, bd, ce, df
ad, be, cf
ae, bf
af
abc, bcd, cde, def
abd, bce, cdf
acd, bde, cef
abe, bcf
ade, bef
ace, bdf
abf
aef
acf
adf
abcd, bcde, cdef
abce, bcdf
abde, bcef
acde, bdef
abcf
abef
adef
abdf
acdf
acef
abcde, bcdef
abcdf
abcef
abdef
acdef
abcdef

Here it is my solutions, it is similar (conceptually) with the solution of lmiguelvargasf.
Let me say that
-[math item] by defintion the powerset do contain the empty set
-[personal taste] and also that I don't like using frozenset.
So the input is a list and the output will be a list of lists. The function could close earlier, but I like the element of the power set to be order lexicographically, that essentially means nicely.
def power_set(L):
"""
L is a list.
The function returns the power set, but as a list of lists.
"""
cardinality=len(L)
n=2 ** cardinality
powerset = []
for i in range(n):
a=bin(i)[2:]
subset=[]
for j in range(len(a)):
if a[-j-1]=='1':
subset.append(L[j])
powerset.append(subset)
#the function could stop here closing with
#return powerset
powerset_orderred=[]
for k in range(cardinality+1):
for w in powerset:
if len(w)==k:
powerset_orderred.append(w)
return powerset_orderred

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Multiple nested loops equivalent replacement in python - python

Actually, As miradulo wrote the solution is very simple : either use : for x in itertools.product(*map(range, (5, 3, 2, 2))) : print(x, end=' ') or : for x in itertools.product(range(5), range(3), range(2), range(2)) : print(x, end=' ') Both ways solve the problem.

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