Permutations of n elements into r groups Python-CodePudding

I don't know how to create an algorithm / which to use for the following problem:

If I have a set of 20 elements (e.g. A -> T) and have a string of length 8, what are the permutations I could make? (I know there's around 390 million)

e.g. we could have ABCDEFGH or EFHGATRI.

Does anyone have any ideas to help me?

CodePudding user response：

If you want all permutations, given the number of possibilities, better keep it as a generator. For this, use itertools.permutations:

from itertools import permutations
from string import ascii_uppercase

letters = ascii_uppercase[:20]

perms = map(''.join, permutations(letters, r=8))  # this doesn't generate
                                                  # anything yet

# then use it lazily

for p in perms:
    # do something


# or get elements one by one
next(perm)

If you only want random possibilities, use random.sample:

from random import sample
from string import ascii_uppercase

''.join(sample(ascii_uppercase[:20], 8))

example output: 'IKEDQRTS'

CodePudding user response：

Python has specific function for your issue itertools.permutations

Return successive r length permutations of elements in the iterable.

So, you can use it for your task

["".join(x) for x in itertools.permutations("abcd", 3)]

> ['abc', 'abd', 'acb', 'acd', 'adb', 'adc', 'bac', 'bad', 'bca', 'bcd', 'bda', 'bdc', 'cab', 'cad', 'cba', 'cbd', 'cda', 'cdb', 'dab', 'dac', 'dba', 'dbc', 'dca', 'dcb']

Also, you can use generators to reduce RAM usage

def perms():
    for perm in itertools.permutations("abcdefghij", 5):
        yield "".join(perm)

for p in perms():
    print(p)