如何从python列表中的字符中获取length=n的所有字符串

Question

I would like to know a simple and pythonic way to get all strings of length n that are composed of the characters contained in a list called L .

For example:

L = ['0','1']
n = 3

How do I get:

['000','001','010','011','100','101','110','111']

A possible solution is by using itertools.product . It works, but it's not very elegant, so I'm looking for a more pythonic solution. Here's how it would be:

L = ['0','1']
n = 3
a = [L for i in range(0,n)]
x = list(itertools.product(*a))
x = ["".join(i) for i in x]

The resulting list looks like this:

>>> x
['000', '001', '010', '011', '100', '101', '110', '111']

Is there something built-in like all_possible_strings(L, n) , which takes the elements in L and obtains all possible combinations of length n with those elements?

Answer 1

The following code with Python 2.6 and above ONLY

First, import itertools:

import itertools

Permutation (order matters):

print list(itertools.permutations([1,2,3,4], 2))
[(1, 2), (1, 3), (1, 4),
(2, 1), (2, 3), (2, 4),
(3, 1), (3, 2), (3, 4),
(4, 1), (4, 2), (4, 3)]

Documentation:

https://docs.python.org/2/library/itertools.html#itertools.permutations

Happy Coding

Answer 2

Using the repeat option of product seems to give you what you want more directly:

L = ['0','1']
n = 3
x = list(itertools.product(L, repeat=n))
x = ["".join(i) for i in x]
print(x)

Output:

['000', '001', '010', '011', '100', '101', '110', '111']