How to generate all permutations of a list but also add strings that can be present only once in any permutation

Question

itertools.product is very convenient to generate all permutations of a list, but how would one proceed to implicate another list of elements where only one element of that list could populate any permutation at a time ?

In pseudocode :

from itertools import product
product('ABC', repeat=1) + ['.', '%3A']
->A, B, C, .A, .B, .C, A., B., C., %3AA, %3AB, %3AC, A%3A, B%3A, C%3A

I suppose it would be non trivial and one would have to tweak the permutation building algorithm.

Sometimes it is easier to ask about your problem instead of asking how to implement a solution you devised (see The X/Y Problem) — Paulo Scardine

Martijn Pieters Martijn Pieters · Accepted Answer · 2017-01-02T17:54:51

You have 3 series. The original input string as a list, and then you prepend and append. Generate those three separately:

from itertools import chain, product

inputstring = 'ABC'
extra = ['.', '%3A']
for combo in chain(inputstring, 
        product(extra, inputstring), product(inputstring, extra)):
    combo = ''.join(combo)
    print(combo)

Demo with a list comprehension:

>>> from itertools import chain, product
>>> inputstring = 'ABC'
>>> extra = ['.', '%3A']
>>> [''.join(combo) for combo in chain(inputstring, product(extra, inputstring), product(inputstring, extra))]
['A', 'B', 'C', '.A', '.B', '.C', '%3AA', '%3AB', '%3AC', 'A.', 'A%3A', 'B.', 'B%3A', 'C.', 'C%3A']

How to generate all permutations of a list but also add strings that can be present only once in any permutation

1 Answers