CODEWITHSUNDEEP
pythonlistpermutation
Xeoncross
Xeoncross

Reputation: 57184

All possible permutations of multiple lists and sizes

It's easy in python to calculate simple permutations using itertools.permutations().

You can even find some possible permutations of multiple lists.

import itertools
s=[ [ 'a', 'b', 'c'], ['d'], ['e', 'f'] ]
for l in list(itertools.product(*s)):
    print(l)


('a', 'd', 'e')
('a', 'd', 'f')
('b', 'd', 'e')
('b', 'd', 'f')
('c', 'd', 'e')
('c', 'd', 'f')

It's also possible to find permutations of different lengths.

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

()
(1,)
(2,)
(3,)
(1, 2)
(1, 3)
(2, 3)
(1, 2, 3)

How would you find permutations of all possible 1) lengths, 2) orders, and 3) from multiple lists?

I would assume the first step would be to combine the lists into one. A list will not de-dup items like a set would.

s=[ [ 'a', 'b', 'c'], ['d'], ['e', 'f'] ]

('a', 'b')
('a', 'c')
('a', 'd')
('a', 'e')
('a', 'f')
...
('b', 'a')
('c', 'a')
...
('a', 'b', 'c', 'd', 'e')
...
('a', 'b', 'c', 'd', 'e', 'f')
...
('f', 'a', 'b', 'c', 'd', 'e')

Upvotes: 5

Views: 2406

Answers (2)

A H
A H

Reputation: 2570

Here's a simple one liner (You can replace feature_cols instead of s)

Combinations:

[combo for i in range(1, len(feature_cols) + 1) for combo in itertools.combinations(feature_cols, i) ]

Permutations:

[combo for i in range(1, len(feature_cols) + 1) for combo in itertools.permutations(feature_cols, i) ]

See my answer here for more details

Upvotes: 2

fferri
fferri

Reputation: 18940

Like you suggested, do:

s = [x for y in s for x in y]

and then use your solution for finding permutations of different lengths:

for L in range(0, len(s)+1):
    for subset in itertools.combinations(s, L):
        print(subset)

would find:

()
('a',)
('b',)
('c',)
('d',)
('e',)
('f',)
('a', 'b')
('a', 'c')
('a', 'd')
('a', 'e')
('a', 'f')
('b', 'c')
('b', 'd')
('b', 'e')
('b', 'f')
('c', 'd')
('c', 'e')
('c', 'f')
('d', 'e')
('d', 'f')
('e', 'f')
('a', 'b', 'c')
('a', 'b', 'd')
('a', 'b', 'e')
('a', 'b', 'f')
('a', 'c', 'd')
('a', 'c', 'e')
('a', 'c', 'f')
('a', 'd', 'e')
('a', 'd', 'f')
('a', 'e', 'f')
('b', 'c', 'd')
('b', 'c', 'e')
('b', 'c', 'f')
('b', 'd', 'e')
('b', 'd', 'f')
('b', 'e', 'f')
('c', 'd', 'e')
('c', 'd', 'f')
('c', 'e', 'f')
('d', 'e', 'f')
('a', 'b', 'c', 'd')
('a', 'b', 'c', 'e')
('a', 'b', 'c', 'f')
('a', 'b', 'd', 'e')
('a', 'b', 'd', 'f')
('a', 'b', 'e', 'f')
('a', 'c', 'd', 'e')
('a', 'c', 'd', 'f')
('a', 'c', 'e', 'f')
('a', 'd', 'e', 'f')
('b', 'c', 'd', 'e')
('b', 'c', 'd', 'f')
('b', 'c', 'e', 'f')
('b', 'd', 'e', 'f')
('c', 'd', 'e', 'f')
('a', 'b', 'c', 'd', 'e')
('a', 'b', 'c', 'd', 'f')
('a', 'b', 'c', 'e', 'f')
('a', 'b', 'd', 'e', 'f')
('a', 'c', 'd', 'e', 'f')
('b', 'c', 'd', 'e', 'f')
('a', 'b', 'c', 'd', 'e', 'f')

If you want to distinguish e.g. ('d', 'e', 'f') from ('f', 'e', 'd') (thanks @Kefeng91 for pointing this out) and others, replace itertools.combinations with itertools.permutations, like @YakymPirozhenko suggests.

Upvotes: 3

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