Create variations of a string

Question

I generated random strings of 16-characters each and added them to a list using the following:

import random
strings = []
for x in range(0, 99): 
 strings.append(''.join(random.choice('0123456789ABCDEF') for i in range(16)))

This worked as expected. Now, for each generated string, I want to find all possible combinations such that at least two characters remain same as the original string and the order of characters does not change. For example, if we have CDD733665417E3F1, then I want to generate all CDXXXXXXXXXXXXXX where X could be anything (0-9 or A-F). Similarly XXD7XXXXXXXXXXXX and so on. Previous similar questions hint towards using itertools.product but I am not sure how it can be used to generate permutations and not fixed replacements. Any help will be appreciated. Thank you

haha considering each could be 0-9 or A-F, it will be humongous. Just interested to know how its done and then I might limit a few variations — user2061944
– user2061944, Commented Aug 3, 2016 at 14:40
Since you use 0-9A-F it's just hex numbers, why not generate random numbers between 0 and 0xffffffffffffffff then modify some bits ? — polku
– polku, Commented Aug 3, 2016 at 14:44

R Nar · Accepted Answer · 2016-08-03 15:26:42Z

2

Create an iterator for the indices of the two characters you want to stay the same using itertools.combinations

>>> from itertools import combinations
>>> s = 'ABC123'
>>> for indices in combinations(range(len(s)), 2):
...     print ''.join([s[x] if x in indices else 'X' for x in range(len(s))])
...
ABXXXX
AXCXXX
AXX1XX
AXXX2X
AXXXX3
XBCXXX
XBX1XX
XBXX2X
XBXXX3
XXC1XX
XXCX2X
XXCXX3
XXX12X
XXX1X3
XXXX23

Creates all the variable strings.

You can then do a nested loop to replace the X's.

Then you can use product to get all the letters that you would need to replace the X's with:

>>> for letters in product('ABCDEF0123456789', repeat = 4):
...     print letters
...
('A', 'A', 'A', 'A')
('A', 'A', 'A', 'B')
('A', 'A', 'A', 'C')
('A', 'A', 'A', 'D')
('A', 'A', 'A', 'E')
('A', 'A', 'A', 'F')
('A', 'A', 'A', '0')
('A', 'A', 'A', '1')
('A', 'A', 'A', '2')
('A', 'A', 'A', '3')
('A', 'A', 'A', '4')
('A', 'A', 'A', '5')
('A', 'A', 'A', '6')
('A', 'A', 'A', '7')
('A', 'A', 'A', '8')
('A', 'A', 'A', '9')
('A', 'A', 'B', 'A')
('A', 'A', 'B', 'B')
('A', 'A', 'B', 'C')
('A', 'A', 'B', 'D')
('A', 'A', 'B', 'E')
('A', 'A', 'B', 'F')
('A', 'A', 'B', '0')
('A', 'A', 'B', '1')
('A', 'A', 'B', '2')
.
.
.

Combine these together and you would get all the combinations of what you want.

You can probably do something like:

>>> for indices in combinations(range(len(s)), 2):
...     for letters in product('ABCDEF0123456789', repeat = 4):
...          letter_iter = iter(letters)
...          print ''.join([s[x] if x in indices else letter_iter.next() for x in range(len(s))])

NOTE 1: you can change the 2 in the call of combinations to change the amount of indices you want to stay the same. Likewise, you can change the repeat parameter in the call to product to reflect those changes (repeat = n where n = len(s) - number_in_combinations)

NOTE 2: These are stupidly large amounts of values. You know this. Please be careful that you don't destroy your memory. When I did the product call, I added an index counter and broke the loop after the index counter got greater than 20 to avoid hell breaking loose.

answered Aug 3, 2016 at 15:26

R Nar

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2 Comments

user2061944 Over a year ago

Thanks I will give it a go later

R Nar Over a year ago

@user2061944 something to keep in mind: the inner for loop is going to be MUCH larger than the outer loop considering how many more permutations it will contain (for you, n = 14 which would give 14^14 combinations). It may be worth switching the inner and outer loops to decrease time spent. Mind you, the output order would be completely different, but if you're only concerned with the final list, it would have the same effect.

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