from itertools import *

def recursive_product(s,r=None,i=0):
    if r is None:
        r = []
    if i>len(s):
        return r
    for c in product(s, repeat=i):
        r.append("".join(c))
    return recursive_product(s,r,i+1)

print(recursive_product('ab'))
print(recursive_product('abc'))

输出：
第一个月
['', 'a', 'b', 'c', 'aa', 'ab', 'ac', 'ba', 'bb', 'bc', 'ca', 'cb', 'cc', 'aaa', 'aab', 'aac', 'aba', 'abb', 'abc', 'aca', 'acb', 'acc', 'baa', 'bab', 'bac', 'bba', 'bbb', 'bbc', 'bca', 'bcb', 'bcc', 'caa', 'cab', 'cac', 'cba', 'cbb', 'cbc', 'cca', 'ccb', 'ccc']
老实说，在这种情况下使用递归感觉真的很勉强，一个简单得多的版本也有相同的结果：

nonrecursive_product = lambda s: [''.join(c)for i in range(len(s)+1) for c in product(s,repeat=i)]

这是字符串中字符集的powerset。

from itertools import chain, combinations

s = set('ab') #split string into a set of characters

# combinations gives the elements of the  powerset of a given length r
# from_iterable puts all these into an 'iterable'
# which is converted here to a list

list(chain.from_iterable(combinations(s, r) for r in range(len(s)+1)))

import itertools as it

s='aab'
subsets = sorted(list(map("".join, it.chain.from_iterable(it.permutations(s,r) for r in range(len(s) + 1)))))

print(subsets)
# ['', 'a', 'a', 'aa', 'aa', 'aab', 'aab', 'ab', 'ab', 'aba', 'aba', 'b', 'ba', 'ba', 'baa', 'baa']