获取python中列表的所有排列而没有重复？

Question

I am trying to write a script that gets a set of strings-

["ab", "ls", "u"]

Then creates every possible combination of them, but doesn't necessarily use all of them. I want possible outputs for the above example to be:

ab
ab ls
ab ls u
ab u ls
ab u

ls
ls ab
ls ab u
ls u ab
ls u

u
u ls
u ls ab
u ab ls
u ab

My script, having removed the other things it does:

stuff = ["ab", "ls", "u"]

for subset in itertools.permutations(stuff):
    concat = ""
    for part in subset:
        concat = concat + part

    #the rest of my script now uses this data

It returns:

ablsu
abuls
lsabu
lsuab
uabls
ulsab

How would I make it return what I want?

Answer 1

You can use combinations and permutations together. This should be able to get you going

a = ["ab", "ls", "u"]
for i in range(1, len(a)+1):
    for comb in combinations(a, i):
        for perm in permutations(comb):
            print(perm)

Output:

('ab',)
('ls',)
('u',)
('ab', 'ls')
('ls', 'ab')
('ab', 'u')
('u', 'ab')
('ls', 'u')
('u', 'ls')
('ab', 'ls', 'u')
('ab', 'u', 'ls')
('ls', 'ab', 'u')
('ls', 'u', 'ab')
('u', 'ab', 'ls')
('u', 'ls', 'ab')

You can handle comb how ever you see fit

Answer 2

As you are giving list with 3 elements permutations is giving you back result with all 3 elements. You need to supply 1 element to get your ab / ls / u in output. You need to supply 2 element to get your ab ls / ab u in output.

So same program you can use by calling it with 1/2 elements in list.

stuff = ["ab", "ls", "u"]

for subset in itertools.permutations(stuff):
    concat = ""
    for part in subset:
        concat = concat + part

    #the rest of my script now uses this data

stuff = ["ab", "ls"]

for subset in itertools.permutations(stuff):
    concat = ""
    for part in subset:
        concat = concat + part


stuff = ["ls", "u"]

for subset in itertools.permutations(stuff):
    concat = ""
    for part in subset:
        concat = concat + part

Answer 3

stuff = ["ab", "ls", "u"]
final_list = []
for subset in itertools.permutations(stuff):
    concat = ""
    for part in subset:
        concat = concat + part
        final_list.append(concat)

print(final_list)

['ab',
 'abls',
 'ablsu',
 'ab',
 'abu',
 'abuls',
 'ls',
 'lsab',
 'lsabu',
 'ls',
 'lsu',
 'lsuab',
 'u',
 'uab',
 'uabls',
 'u',
 'uls',
 'ulsab']

Answer 4

import itertools

stuff = ["ab", "ls", "u"]

for i in range(len(stuff) + 1):
    for x in itertools.permutations(stuff[:i]):
        print(x)

But this solution show all permutations::

()
('ab',)
('ab', 'ls')
('ls', 'ab')
('ab', 'ls', 'u')
('ab', 'u', 'ls')
('ls', 'ab', 'u')
('ls', 'u', 'ab')
('u', 'ab', 'ls')
('u', 'ls', 'ab')