4

我有一个字符串

s = "mouse"

和一个字符串列表

sub_strings = ["m", "o", "se", "e"]

我需要找出与 s 匹配的列表中 sub_strings 的最佳和最短匹配子集是什么。做这个的最好方式是什么？理想的结果是 ["m", "o", "se"] 因为它们一起拼写为 mose

4

3 回答 3

2 
import difflib
print difflib.get_close_matches(target_word,list_of_possibles)

但不幸的是，它不适用于您上面的示例，您可以改用 Levenstein 距离...

def levenshtein_distance(first, second):
    """Find the Levenshtein distance between two strings."""
    if len(first) > len(second):
        first, second = second, first
    if len(second) == 0:
        return len(first)
    first_length = len(first) + 1
    second_length = len(second) + 1
    distance_matrix = [[0] * second_length for x in range(first_length)]
    for i in range(first_length):
       distance_matrix[i][0] = i
    for j in range(second_length):
       distance_matrix[0][j]=j
    for i in xrange(1, first_length):
        for j in range(1, second_length):
            deletion = distance_matrix[i-1][j] + 1
            insertion = distance_matrix[i][j-1] + 1
            substitution = distance_matrix[i-1][j-1]
            if first[i-1] != second[j-1]:
                substitution += 1
            distance_matrix[i][j] = min(insertion, deletion, substitution)
    return distance_matrix[first_length-1][second_length-1]

sub_strings = ["mo", "m,", "o", "se", "e"]
s="mouse"
print sorted(sub_strings,key = lambda x:levenshtein_distance(x,s))[0]

这将始终为您提供与目标“最接近”的词（对于最接近的一些定义）

levenshtein 功能从：http://www.korokithakis.net/posts/finding-the-levenshtein-distance-in-python/ 窃取

于 2012-09-12T18:22:21.167 回答
2

您可以使用正则表达式：

import re

def matches(s, sub_strings):
    sub_strings = sorted(sub_strings, key=len, reverse=True)
    pattern = '|'.join(re.escape(substr) for substr in sub_strings)
    return re.findall(pattern, s)

这至少短而快，但不一定会找到最佳匹配集；太贪心了。例如，

matches("bears", ["bea", "be", "ars"])

返回["bea"]，什么时候应该返回["be", "ars"]。

代码说明：

  • 第一行按长度对子字符串进行排序，以便最长的字符串出现在列表的开头。这确保了正则表达式将更喜欢较长的匹配而不是较短的匹配。

  • 第二行创建一个由所有子字符串组成的正则表达式模式，由|符号分隔，表示“或”。

  • 第三行仅使用该re.findall函数查找给定字符串中模式的所有匹配项s。

于 2012-09-12T18:36:00.343 回答
2

此解决方案基于用户Running Wild的此答案。它使用Stefan Behnel的包使用Aho-Corasick 算法有效地找到目标中子串的所有匹配项，然后使用动态规划来找到答案。acora

import acora
import collections

def best_match(target, substrings):
    """
    Find the best way to cover the string `target` by non-overlapping
    matches with strings taken from `substrings`. Return the best
    match as a list of substrings in order. (The best match is one
    that covers the largest number of characters in `target`, and
    among all such matches, the one using the fewest substrings.)

    >>> best_match('mouse', ['mo', 'ou', 'us', 'se'])
    ['mo', 'us']
    >>> best_match('aaaaaaa', ['aa', 'aaa'])
    ['aaa', 'aa', 'aa']
    >>> best_match('abracadabra', ['bra', 'cad', 'dab'])
    ['bra', 'cad', 'bra']
    """
    # Find all occurrences of the substrings in target and store them
    # in a dictionary by their position.
    ac = acora.AcoraBuilder(*substrings).build()
    matches = collections.defaultdict(set)
    for match, pos in ac.finditer(target):
        matches[pos].add(match)

    n = len(target)
    # Array giving the best (score, list of matches) found so far, for
    # each initial substring of the target.
    best = [(0, []) for _ in xrange(n + 1)]
    for i in xrange(n):
        bi = best[i]
        bj = best[i + 1]
        if bi[0] > bj[0] or bi[0] == bj[0] and len(bi[1]) < bj[1]:
            best[i + 1] = bi
        for m in matches[i]:
            j = i + len(m)
            bj = best[j]
            score = bi[0] + len(m)
            if score > bj[0] or score == bj[0] and len(bi[1]) < len(bj[1]):
                best[j] = (score, bi[1] + [m])
    return best[n][1]
于 2012-09-13T10:59:49.033 回答