從字符串列表中找到最佳子集以匹配給定的字符串

Question

我有一個弦

s = "mouse"

和一個字符串列表

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

我需要找出匹配s的列表中sub_strings的最佳和最短匹配子集是什么。 做這個的最好方式是什么？ 理想的結果是[“ m”，“ o”，“ se”]，因為它們一起拼寫了mose

Answer 1

import difflib
print difflib.get_close_matches(target_word,list_of_possibles)

但不幸的是，它對於上面的示例不起作用，您可以改用Levenstein距離...

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]

這將始終為您提供與目標“最接近”的詞（用於一些最接近的定義）

levenshtein函數從以下位置被盜：http：//www.korokithakis.net/posts/finding-the-levenshtein-distance-in-python/

Answer 2

您可以使用正則表達式：

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)

這至少是簡短而快速的，但不一定能找到最佳的匹配項。 太貪心了 例如，

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

返回["bea"] ，應在何時返回["be", "ars"] 。

---

代碼說明：

• 第一行按長度對子字符串進行排序，以使最長的字符串出現在列表的開頭。 這確保了正則表達式將首選長匹配而不是短匹配。

• 第二行創建由所有子字符串組成的正則表達式模式，用|分隔| 符號，表示“或”。

• 第三行僅使用re.findall函數查找給定字符串s模式的所有匹配項。

Answer 3

該解決方案基於用戶Running Wild的 答案 。 它使用Stefan Behnel提供的acora軟件包，通過Aho–Corasick算法有效地找到目標中子字符串的所有匹配項，然后使用動態編程來找到答案。

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]