[英]how to return the intersection of all possible combinations of n different lists
[英]Return all possible combinations of a string when splitted into n strings
我找了一个关于这个的stackoverflow但是找不到办法去做。 它可能涉及itertools。
我想找到分割字符串的所有可能结果,比如将字符串thisisateststring
写成n
(等长或不等长,无关紧要,两者都应包括在内)字符串。
例如,让n
为3
:
[["thisisat", "eststrin", "g"], ["th", "isisates", "tstring"], ............]
你可以在这里使用itertools.combinations
。 您只需要选择两个分裂点来生成每个结果字符串:
from itertools import combinations
s = "thisisateststring"
pools = range(1, len(s))
res = [[s[:p], s[p:q], s[q:]] for p, q in combinations(pools, 2)]
print res[0]
print res[-1]
输出:
['t', 'h', 'isisateststring']
['thisisateststri', 'n', 'g']
使用itertools.combinations()
在结果中包含空字符串会相当尴尬。 编写自己的递归版本可能最简单:
def partitions(s, k):
if not k:
yield [s]
return
for i in range(len(s) + 1):
for tail in partitions(s[i:], k - 1):
yield [s[:i]] + tail
这适用于任何数量的工作k
所需分区的任何字符串s
。
根据Raymond Hettinger的代码,这是一个将序列分成n组的方法:
import itertools as IT
def partition_into_n(iterable, n, chain=IT.chain, map=map):
"""
Based on http://code.activestate.com/recipes/576795/ (Raymond Hettinger)
Modified to include empty partitions, and restricted to partitions of length n
"""
s = iterable if hasattr(iterable, '__getslice__') else tuple(iterable)
m = len(s)
first, middle, last = [0], range(m + 1), [m]
getslice = s.__getslice__
return (map(getslice, chain(first, div), chain(div, last))
for div in IT.combinations_with_replacement(middle, n - 1))
In [149]: list(partition_into_n(s, 3))
Out[149]:
[['', '', 'thisisateststring'],
['', 't', 'hisisateststring'],
['', 'th', 'isisateststring'],
['', 'thi', 'sisateststring'],
...
['thisisateststrin', '', 'g'],
['thisisateststrin', 'g', ''],
['thisisateststring', '', '']]
它比小n
的递归解决方案慢,
def partitions_recursive(s, n):
if not n>1:
yield [s]
return
for i in range(len(s) + 1):
for tail in partitions_recursive(s[i:], n - 1):
yield [s[:i]] + tail
s = "thisisateststring"
In [150]: %timeit list(partition_into_n(s, 3))
1000 loops, best of 3: 354 µs per loop
In [151]: %timeit list(partitions_recursive(s, 3))
10000 loops, best of 3: 180 µs per loop
但正如您所料,对于大n
来说它更快(随着递归深度的增加):
In [152]: %timeit list(partition_into_n(s, 10))
1 loops, best of 3: 9.2 s per loop
In [153]: %timeit list(partitions_recursive(s, 10))
1 loops, best of 3: 10.2 s per loop
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