查找字符串中最重复(不是最常见)序列的算法(也称为串联重复)

Question

我正在寻找能够在字符串中找到最重复序列的算法(可能在Python中实现).对于REPETITIVE,我指的是在不中断的情况下反复重复的任何字符组合(串联重复).

我寻找的算法不一样的"寻找最常见的词"之一.实际上,重复块不需要是字符串中最常见的单词(substring).

例如:

s = 'asdfewfUBAUBAUBAUBAUBAasdkBAjnfBAenBAcs'
> f(s)
'UBAUBAUBAUBAUBA' #the "most common word" algo would return 'BA'

不幸的是,我不知道如何解决这个问题.非常欢迎任何帮助.

---

UPDATE

一个额外的例子来澄清我希望返回具有最多重复次数的序列,无论其基本构建块是什么.

g = 'some noisy spacer'
s = g + 'AB'*5 + g + '_ABCDEF'*2 + g + 'AB'*3
> f(s)
'ABABABABAB' #the one with the most repetitions, not the max len

@rici的例子:

s = 'aaabcabc'
> f(s)
'abcabc'

s = 'ababcababc'
> f(s)
'ababcababc' #'abab' would also be a solution here
             # since it is repeated 2 times in a row as 'ababcababc'.
             # The proper algorithm would return both solutions.

Answer 1

结合re.findall()(使用特定的正则表达式)和max()功能:

import re

#  extended sample string
s = 'asdfewfUBAUBAUBAUBAUBAasdkjnfencsADADADAD sometext'

def find_longest_rep(s):
    result = max(re.findall(r'((\w+?)\2+)', s), key=lambda t: len(t[0]))
    return result[0]

print(find_longest_rep(s))

输出:

UBAUBAUBAUBAUBA

关键模式:

• ((\w+?)\2+):
  • (....) - 作为第一捕获组的最外侧捕获组
  • (\w+?) - 包含在第二个捕获组中的任何非空白字符序列; +? - 量词,在一次和无限次之间匹配,尽可能少,根据需要进行扩展
  • \2+ - 匹配第二个捕获组最近匹配的相同文本

Answer 2

以下是基于((\w+?)\2+)正则表达式的解决方案,但有其他改进:

import re
from itertools import chain


def repetitive(sequence, rep_min_len=1):
    """Find the most repetitive sequence in a string.

    :param str sequence: string for search
    :param int rep_min_len: minimal length of repetitive substring
    :return the most repetitive substring or None
    """
    greedy, non_greedy = re.compile(r'((\w+)\2+)'), re.compile(r'((\w+?)\2+)')

    all_rep_seach = lambda regex: \
        (regex.search(sequence[shift:]) for shift in range(len(sequence)))

    searched = list(
        res.groups()
        for res in chain(all_rep_seach(greedy), all_rep_seach(non_greedy))
        if res)

    if not sequence:
        return None

    cmp_key = lambda res: res[0].count(res[1]) if len(res[1]) >= rep_min_len else 0
    return max(searched, key=cmp_key)[0]

您可以像这样测试它:

def check(seq, expected, rep_min_len=1):
    result = repetitive(seq, rep_min_len)
    print('%s => %s' % (seq, result))
    assert result == expected, expected


check('asdfewfUBAUBAUBAUBAUBAasdkBAjnfBAenBAcs', 'UBAUBAUBAUBAUBA')
check('some noisy spacerABABABABABsome noisy spacer_ABCDEF_ABCDEFsome noisy spacerABABAB', 'ABABABABAB')
check('aaabcabc', 'aaa')
check('aaabcabc', 'abcabc', rep_min_len=2)
check('ababcababc', 'ababcababc')
check('ababcababcababc', 'ababcababcababc')

主要特点:

1. 使用贪婪((\w+)\2+)和非贪婪的((\w+)\2+?)正则表达式;
2. 在所有子串中搜索重复的子串,从头开始移位(例如'string'=> ['string','tring','ring','ing','ng','g']);
3. 选择是基于不在子序列长度上的重复次数(例如,'ABABABAB_ABCDEF_ABCDEF'结果将是'ABABABAB',而不是'_ABCDEF_ABCDEF');
4. 重复序列的最小长度是重要的(参见'aaabcabc'检查).