python - 将整数字符串拆分为所有可能的数字列表

Question

我有一串整数，例如s = "1234" 我想将其拆分为整数的各个顺序组合 split = [ 1234, 1, 2, 3, 4, 12, 123, 23, 234, 34 ] 我如何用 Python 编写这个代码？

我尝试过:

for i in range(0,len(number)-1):
x =["" + number[j] for j in range(i, len(number))]
print(x)

输出:

['1', '2', '3', '4', '5']
['2', '3', '4', '5']
['3', '4', '5']
['4', '5']

Answer 1

您可以使用combinations来自itertools库结合 list comprehension :

>>> from itertools import combinations
>>> s = "1234"
>>> [int(''.join(x)) for i in range(len(s)) for x in combinations(s, i + 1)]
[1, 2, 3, 4, 12, 13, 14, 23, 24, 34, 123, 124, 134, 234, 1234]

更新 由于您只需要顺序组合，因此可以使用字符串中的所有子字符串(使用 How To Get All The Contiguous Substrings Of A String In Python? ):

>>> l = len(s)
>>> [int(s[i:j+1]) for i in range(l) for j in range(i,l)]
[1, 12, 123, 1234, 2, 23, 234, 3, 34, 4]