Generate Sequence of Characters with a Certain Amount of Each Character

Question

So I'm currently generating a sequence of characters as a string using:

def nth(lists, num):
    res = []
    for a in lists:
        res.insert(0, a[num % len(a)])
        num //= len(a)
    return res

def generate_page(chars, length, iteration):
    result = nth([chars] * length, iteration)
    return "".join(result)

length = 15
seed = 16
number = generate_page('0123456789ABC', length, seed)

But what I can't wrap my head around is how would I generate a sequence of characters if I know how much of each character I want.

For example, let's say I want to generate a sequence that had 1 "A", 3 "B"'s, and 1 "C" how could I get to any arrangement of "ABBBC"?

I'm thinking I would just concatenate a list of the amounts I know I want and then scramble them up like:

A = 1
B = 3
C = 1
listOfCharacters = ["A"]*A + ["B"]*B + ["C"]*C
>>> ['A', 'B', 'B', 'B', 'C']
random.seed(1)
listOfCharacters = random.shuffle(listOfCharacters)
>>> None
listOfCharacters = ''.join(listOfCharacters)
>>> TypeError

But am obviously getting a TypeError at the moment, and am not even sure if this is the best route to retrieve all of the permutations without repeats with a seed.

So what exactly is your desired output? All permutations of "ABBBC" without duplicates? — user2390182, Dec 12 '18 at 02:21
You should iterate over shuffle. Iterating over seed will destroy any guarantees the PRNG can offer. — Richard, Dec 12 '18 at 05:09
You should post your working code as an answer to your own question, rather than editing it into the question itself. — o11c, Dec 12 '18 at 06:04

Eb946207 · Accepted Answer · 2018-12-12T02:29:33.353

1

Just do this:

import itertools
match = 'ABBBC' #string of letter combinations to match.
permutations = itertools.permutations(match) #get all
all_combos = list({''.join(per) for per in permutations}) #use sets to remove duplicates

The values of all_combos is a list of all possible combinations.

edited Dec 12 '18 at 02:29

answered Dec 12 '18 at 02:28

Eb946207

748
8
26

1

That will contain duplicates. – user2390182 Dec 12 '18 at 02:28
Oh, sorry. Let me fix that. – Eb946207 Dec 12 '18 at 02:29
@schwobaseggl fixed! – Eb946207 Dec 12 '18 at 02:29
2

Better, but for longer sequences with lots of repetitions, this will overproduce dramatically. – user2390182 Dec 12 '18 at 02:31
...however, the OP’s use case does not have long sequences. – Eb946207 Dec 12 '18 at 02:33
1

Ideally, it would be scalable. Your solution does work for smaller sequences though!! – Jacob W. Dallas Dec 12 '18 at 02:49
After looking around, any solution that removed duplicates would still have its own speed problems because of checks of "is it a duplicate index". Speed should not be that bad because of how fast computers are, so unless you have lots of repeating letters of the same type you should be okay. :) It could be made quicker by not turning it into a list, if you are okay working with `set`s – Eb946207 Dec 12 '18 at 02:59

score 0 · Answer 2 · answered Dec 12 '18 at 06:38

UPDATE: So here is what I ended up doing:

A = 1
B = 3
C = 1
listOfCharacters = ["A"]*A + ["B"]*B + ["C"]*C
>>> ['A', 'B', 'B', 'B', 'C']
random.seed(1)
random.shuffle(listOfCharacters)
listOfCharacters = ''.join(listOfCharacters)
>>>'BBCAB'

And then I just iterated over the shuffle with a for loop.

Joseph Wood · Answer 3 · 2018-12-13T18:20:57.227

A better approach in order to avoid duplicates is to use the right tool as pointed out by @BillBell here : permutations with unique values. Essentially, we are dealing with Multisets, and using sympy we can get the desired outcome:

from sympy.utilities.iterables import multiset_permutations

for item in multiset_permutations(listOfCharacters):
    print(''.join(item))

ABBBC
ABBCB
ABCBB
ACBBB
BABBC
BABCB
BACBB
BBABC
BBACB
BBBAC
BBBCA
BBCAB
BBCBA
BCABB
BCBAB
BCBBA
CABBB
CBABB
CBBAB
CBBBA

Generate Sequence of Characters with a Certain Amount of Each Character

3 Answers3