CODEWITHSUNDEEP
python-3.xpython-itertools
Jared
Jared

Reputation: 21

Permutations in python3

I can't figure out how to get permutations to return the actual permutation and not I tried a lot of different things to no avail. The code I used was from itertools import permutations and then permutations([1,2,3]). Thanks!

Upvotes: 2

Views: 9238

Answers (3)

kimxons
kimxons

Reputation: 1

from itertools import permutations

#iteration
for p in permutations([1,2,3]):
    print(p)

This should work perfectly.

Upvotes: 0

Michael F
Michael F

Reputation: 40879

itertools.permutations is a generator, which means you have to retrieve results from it by using it like this:

for permutation in itertools.permutations([1,2,3]):
   do_stuff_with(permutation)

or alternatively put all of them in a list:

list(itertools.permutations([1,2,3]))

or, less conveniently:

generator = itertools.permutations([1,2,3])
generator.__next__()

Upvotes: 1

RocketDonkey
RocketDonkey

Reputation: 37279

This may not be answering your question (it appears to be missing the part after 'and not'), but from your code, what you are likely seeing is the repr of the itertools.permutations iterator. You can iterate through this object just as you would a normal list in order to access all of the items. If you want to convert it to a list, you can wrap it in list:

>>> from itertools import permutations
>>> permutations([1, 2, 3])
<itertools.permutations object at 0x1e67890>
>>> list(permutations([1, 2, 3]))
[(1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2), (3, 2, 1)]

However as mentioned above, the iterator can be iterated over just like you would a normal list (the benefit of returning an iterator is that the entire sequence is not loaded into memory right away - it is instead loaded 'as needed'):

>>> for perm in permutations([1, 2, 3]):
...     print(perm)
... 
(1, 2, 3)
(1, 3, 2)
(2, 1, 3)
(2, 3, 1)
(3, 1, 2)
(3, 2, 1)

Upvotes: 2

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