What I want to do is obtain all combinations and all unique permutations of each combination. The combinations with replacement function only gets me so far:
from itertools import combinations_with_replacement as cwr
foo = list(cwr('ACGT', n)) ## n is an integer
My intuition on how to move forward is to do something like this:
import numpy as np
from itertools import permutations as perm
bar = []
for x in foo:
carp = list(perm(x))
for i in range(len(carp)):
for j in range(i+1,len(carp)):
if carp[i] == carp[j]:
carp[j] = ''
carp = carp[list(np.where(np.array(carp) != '')[0])]
for y in carp:
bar.append(y)
for i in range(len(bar)):
for j in range(i+1,len(bar)):
if bar[i] == bar[j]:
bar[j] = ''
bar = [bar[x2] for x2 in list(np.where(np.array(bar) != '')[0])]
Is there a more efficient algorithm?
It sounds like you're thinking of a sort of "permutations with replacement", where an input of 'AB'
with a permutation size of 2 would give the outputs
AA
AB
BA
BB
If so, that's the Cartesian product of the input with itself n
times. You want itertools.product
:
>>> import itertools
>>> list(itertools.product('AB', repeat=2))
[('A', 'A'), ('A', 'B'), ('B', 'A'), ('B', 'B')]
来源:https://stackoverflow.com/questions/17912298/itertools-to-generate-scrambled-combinations