Htykuut

Many similar questions has been asked before, but I have not found one that is the same use case as me.

So let's say I have the following list:

List = [1,2,3]

How would I print all combinations of this list, where the same number doesn't repeat twice. The output I would want is:

1

2

3

12

13

21

23

31

32

123

132

213

232

312

321

How would I achieve this?

Most solutions give outputs like:

1

2

3

12

13

32

123

132

213

232

312

321

Here 21,23 and 31 is missing, as I think it counts 12, 32 and 13 as the same combinations, which I don't want the code to do.

Thanks in advance

Pretty sure this is a duplicate, but I cannot find a good one. itertools.permutations does what you want when you iterate over all the desired lengthes:

from itertools import permutations, chain



lst = [1,2,3]

list(chain(*(permutations(lst, n) for n in range(1, len(lst)+1))))

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

or:

for p in chain(*(permutations(lst, n) for n in range(1, len(lst)+1))):

    print(''.join(map(str, p)))



1

2

3

12

13

21

23

31

32

123

132

213

231

312

321

Horribly disgusting one-liner which seems to do the trick:

[

    int(''.join(map(str, number)))

    for n in range(1,4) 

    for number in itertools.product([1,2,3], repeat = n) 

    if not any([first == second for first, second in zip(number, number[1:])])

]

Outputs:

[1, 2, 3, 12, 13, 21, 23, 31, 32, 121, 123, 131, 132, 212, 213, 231, 232, 312, 313, 321, 323]