CODEWITHSUNDEEP
python
Ripster
Ripster

Reputation: 3585

All combinations across multiple lists

Given a multidimensional list (a list of lists) I would like to get all possible combinations of the sub lists items.

For example an input of:

my_list = [
    ['a', 'b'], ['1', '2'], ['@', '&']
]

Would result in:

result = [
['a'],
['b'],
['1'],
['2'],
['@'],
['&'],
['a', '1'],
['a', '2'],
['a', '@'],
['a', '&']
['b', '1'],
['b', '2'],
['b', '@'],
['b', '&'],
['a', '1', '@'],
['a', '1', '&'],
['a', '2', '@'],
['a', '2', '&'],
...]

I tried using itertools.product(*list) but that results in a combination of all items without the smaller sets of combinations. It seems that itertools.combinations, itertools.permutations, etc don't quite give what I am looking for.

Is there a quick way of doing this?

Upvotes: 0

Views: 127

Answers (1)

willeM_ Van Onsem
willeM_ Van Onsem

Reputation: 476503

In that case you first iterate over all possible lengths. For each length you pick all possible combinations of lists, and for each of these combinations you use itertools.product:

def weird_product(*data):
    for i in range(1,len(data)+1):
        for subdata in itertools.combinations(data,i):
            for elem in itertools.product(*subdata):
                yield elem

This generates:

>>> list(weird_product(*data))
[('a',), ('b',), ('1',), ('2',), ('@',), ('&',), ('a', '1'), ('a', '2'), ('b', '1'), ('b', '2'), ('a', '@'), ('a', '&'), ('b', '@'), ('b', '&'), ('1', '@'), ('1', '&'), ('2', '@'), ('2', '&'), ('a', '1', '@'), ('a', '1', '&'), ('a', '2', '@'), ('a', '2', '&'), ('b', '1', '@'), ('b', '1', '&'), ('b', '2', '@'), ('b', '2', '&')]

or more elegantly formatted:

>>> list(weird_product(*data))
[('a',),
 ('b',),
 ('1',),
 ('2',),
 ('@',),
 ('&',),
 ('a', '1'),
 ('a', '2'),
 ('b', '1'),
 ('b', '2'),
 ('a', '@'),
 ('a', '&'),
 ('b', '@'),
 ('b', '&'),
 ('1', '@'),
 ('1', '&'),
 ('2', '@'),
 ('2', '&'),
 ('a', '1', '@'),
 ('a', '1', '&'),
 ('a', '2', '@'),
 ('a', '2', '&'),
 ('b', '1', '@'),
 ('b', '1', '&'),
 ('b', '2', '@'),
 ('b', '2', '&')]

Upvotes: 4

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