我有一个字符串
s = "mouse"
以及一个字符串列表
sub_strings = ["m", "o", "se", "e"]
我需要找到与字符串s最匹配的子字符串列表中最好且最短的匹配子集。 如何做到这一点呢? 理想的结果应该是["m", "o", "se"],因为它们可以拼成mose。
我有一个字符串
s = "mouse"
以及一个字符串列表
sub_strings = ["m", "o", "se", "e"]
我需要找到与字符串s最匹配的子字符串列表中最好且最短的匹配子集。 如何做到这一点呢? 理想的结果应该是["m", "o", "se"],因为它们可以拼成mose。
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)
这种方法虽然简单快捷,但并不一定能找到最佳匹配;它太贪心了。例如,
matches("bears", ["bea", "be", "ars"])
当应该返回["be", "ars"]
时,它返回了["bea"]
。
代码的解释:
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。这确保正则表达式优先选择较长的匹配项而不是较短的匹配项。
第二行创建一个正则表达式模式,其中包含所有子字符串,用|
符号分隔,表示“或”。
第三行只使用re.findall
函数在给定的字符串s
中查找模式的所有匹配项。
acora
包来高效地使用Aho–Corasick算法找到目标字符串中所有子字符串的匹配,然后使用动态规划来求解答案。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]
import difflib
print difflib.get_close_matches(target_word,list_of_possibles)
但不幸的是,它对于你上面的例子无效。你可以使用Levenstein距离代替...
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]
这将始终为您的目标提供“最接近”的单词(对于某些定义而言)。
从http://www.korokithakis.net/posts/finding-the-levenshtein-distance-in-python/窃取的Levenshtein函数。
itertools.combinations
(或者也许是排列...我经常混淆这两个)...而且它可能会很慢...非常慢... - Joran Beasley