python - 拆分为 n 个字符串时返回字符串的所有可能组合

Question

我为此搜索了stackoverflow，但找不到解决方法。它可能涉及迭代工具。

我想找到拆分字符串的所有可能结果，将字符串说thisisateststring成n（相等或不等长度，没关系，都应该包含）字符串。

例如n让3：

[["thisisat", "eststrin", "g"], ["th", "isisates", "tstring"], ............]

Answer 1

你可以itertools.combinations在这里使用。您只需选择两个分割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']

Answer 2

在结果中包含空字符串会很尴尬itertools.combinations()。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

这将适用于任何字符串的任意数量k的所需分区s。

Answer 3

这是根据Raymond Hettinger的代码将序列划分为 n 个组的方法：

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 small 的递归解决方案慢n，

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，它对于大型n（随着递归深度的增加）更快：

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