How to get all possible combinations of a list’s elements? How to get all possible combinations of a list’s elements? python python

How to get all possible combinations of a list’s elements?

This answer missed one aspect: the OP asked for ALL combinations... not just combinations of length "r".

So you'd either have to loop through all lengths "L":

import itertoolsstuff = [1, 2, 3]for L in range(0, len(stuff)+1):    for subset in itertools.combinations(stuff, L):        print(subset)

Or -- if you want to get snazzy (or bend the brain of whoever reads your code after you) -- you can generate the chain of "combinations()" generators, and iterate through that:

from itertools import chain, combinationsdef all_subsets(ss):    return chain(*map(lambda x: combinations(ss, x), range(0, len(ss)+1)))for subset in all_subsets(stuff):    print(subset)

Have a look at itertools.combinations:

itertools.combinations(iterable, r)

Return r length subsequences of elements from the input iterable.

Combinations are emitted in lexicographic sort order. So, if the input iterable is sorted, the combination tuples will be produced in sorted order.

Since 2.6, batteries are included!

Here's a lazy one-liner, also using itertools:

from itertools import compress, productdef combinations(items):    return ( set(compress(items,mask)) for mask in product(*[[0,1]]*len(items)) )    # alternative:             product([0,1], repeat=len(items)) )

Main idea behind this answer: there are 2^N combinations -- same as the number of binary strings of length N. For each binary string, you pick all elements corresponding to a "1".

items=abc * mask=### | V000 -> 001 ->   c010 ->  b011 ->  bc100 -> a101 -> a c110 -> ab111 -> abc

Things to consider:

  • This requires that you can call len(...) on items (workaround: if items is something like an iterable like a generator, turn it into a list first with items=list(_itemsArg))
  • This requires that the order of iteration on items is not random (workaround: don't be insane)
  • This requires that the items are unique, or else {2,2,1} and {2,1,1} will both collapse to {2,1} (workaround: use collections.Counter as a drop-in replacement for set; it's basically a multiset... though you may need to later use tuple(sorted(Counter(...).elements())) if you need it to be hashable)


>>> list(combinations(range(4)))[set(), {3}, {2}, {2, 3}, {1}, {1, 3}, {1, 2}, {1, 2, 3}, {0}, {0, 3}, {0, 2}, {0, 2, 3}, {0, 1}, {0, 1, 3}, {0, 1, 2}, {0, 1, 2, 3}]>>> list(combinations('abcd'))[set(), {'d'}, {'c'}, {'c', 'd'}, {'b'}, {'b', 'd'}, {'c', 'b'}, {'c', 'b', 'd'}, {'a'}, {'a', 'd'}, {'a', 'c'}, {'a', 'c', 'd'}, {'a', 'b'}, {'a', 'b', 'd'}, {'a', 'c', 'b'}, {'a', 'c', 'b', 'd'}]