Python-有遗漏的排列新清单

Question

Lets say I have a list of values

my_list = [1,2,3,4]

I user itertools.permutations to find all combinations of this list

perms = itertools.permutations(my_list)

Which creates

[(1, 2, 3, 4),
(1, 2, 4, 3),
(1, 3, 2, 4),
(1, 3, 4, 2),
(1, 4, 2, 3),
(1, 4, 3, 2),
(2, 1, 3, 4),
(2, 1, 4, 3),
(2, 3, 1, 4),
(2, 3, 4, 1),
(2, 4, 1, 3),
(2, 4, 3, 1),
(3, 1, 2, 4),
(3, 1, 4, 2),
(3, 2, 1, 4),
(3, 2, 4, 1),
(3, 4, 1, 2),
(3, 4, 2, 1),
(4, 1, 2, 3),
(4, 1, 3, 2),
(4, 2, 1, 3),
(4, 2, 3, 1),
(4, 3, 1, 2),
(4, 3, 2, 1)]

I begin iterating through this and find that I no longer need any items in perms which begin with (4,1... or (3,1... .

How can I recreate this list with those specific omissions? Iterating through and removing items is not viable as this needs to scale to very large sizes.

EDIT: For clarification (4,1,2,3) should be removed as it starts with (4,1...) , but not (4,2,1,3) as it starts with (4,2...) .

Answer 1

>>> from itertools import permutations
>>> my_list = [1,2,3,4]
>>> perms = permutations(my_list)
>>> perms
<itertools.permutations object at 0x107a63ad0>
>>> perms = filter(lambda x: x[:2] != (4,1) and x[:2] != (3,1), perms)
>>> perms
[(1, 2, 3, 4), (1, 2, 4, 3), (1, 3, 2, 4), (1, 3, 4, 2), (1, 4, 2, 3), (1, 4, 3, 2), (2, 1, 3, 4), (2, 1, 4, 3), (2, 3, 1, 4), (2, 3, 4, 1), (2, 4, 1, 3), (2, 4, 3, 1), (3, 2, 1, 4), (3, 2, 4, 1), (3, 4, 1, 2), (3, 4, 2, 1), (4, 2, 1, 3), (4, 2, 3, 1), (4, 3, 1, 2), (4, 3, 2, 1)]

Answer 2

Since you state

Iterating through and removing items is not viable as this needs to scale to very large sizes.

The best is to wrap the interator produced by permutations that will generate the tuples you want and skip the tuples you do not want:

my_list = [1,2,3,4]

def my_perms(my_list, f):
    for e in permutations(my_list):
        if f(e):
            yield e 

>>> list(my_perms(my_list, lambda t: t[:2] not in {(4,1), (3,1)}))
[(1, 2, 3, 4), (1, 2, 4, 3), (1, 3, 2, 4), (1, 3, 4, 2), (1, 4, 2, 3), (1, 4, 3, 2), (2, 1, 3, 4), (2, 1, 4, 3), (2, 3, 1, 4), (2, 3, 4, 1), (2, 4, 1, 3), (2, 4, 3, 1), (3, 2, 1, 4), (3, 2, 4, 1), (3, 4, 1, 2), (3, 4, 2, 1), (4, 2, 1, 3), (4, 2, 3, 1), (4, 3, 1, 2), (4, 3, 2, 1)]

Or, use ifilter from itertools :

>>> list(ifilter(lambda t: t[:2] not in {(4,1), (3,1)}, permutations(my_list)))

Or, filter directly on Python 3+ since that also creates an iterator vs creating a list as filter on Python 2 does.

Answer 3

You can use a list comprehension to get the expected permutations :

>>> [i for i in perms if i[0] not in {3,4}]
[(1, 2, 3, 4), (1, 2, 4, 3), (1, 3, 2, 4), (1, 3, 4, 2), (1, 4, 2, 3), (1, 4, 3, 2), (2, 1, 3, 4), (2, 1, 4, 3), (2, 3, 1, 4), (2, 3, 4, 1), (2, 4, 1, 3), (2, 4, 3, 1)]

Note that using a set container has O(1) for check the membership.and it would be more efficient if you have more filter numbers!

If you don't want the tuples that starts with 3,1 and 4,1 you can just simply do :

>>> perms = it.permutations(my_list)
>>> [i for i in perms if i[:2] !=[4,1] and i[:2] !=[4,1]]
[(1, 2, 3, 4), (1, 2, 4, 3), (1, 3, 2, 4), (1, 3, 4, 2), (1, 4, 2, 3), (1, 4, 3, 2), (2, 1, 3, 4), (2, 1, 4, 3), (2, 3, 1, 4), (2, 3, 4, 1), (2, 4, 1, 3), (2, 4, 3, 1), (3, 1, 2, 4), (3, 1, 4, 2), (3, 2, 1, 4), (3, 2, 4, 1), (3, 4, 1, 2), (3, 4, 2, 1), (4, 1, 2, 3), (4, 1, 3, 2), (4, 2, 1, 3), (4, 2, 3, 1), (4, 3, 1, 2), (4, 3, 2, 1)]
>>>