60

使用像 leveinstein（leveinstein 或 difflib）这样的算法，很容易找到近似匹配。例如。

>>> import difflib
>>> difflib.SequenceMatcher(None,"amazing","amaging").ratio()
0.8571428571428571

可以通过根据需要确定阈值来检测模糊匹配。

当前要求：根据较大字符串中的阈值查找模糊子字符串。

例如。

large_string = "thelargemanhatanproject is a great project in themanhattincity"
query_string = "manhattan"
#result = "manhatan","manhattin" and their indexes in large_string

一种蛮力解决方案是生成所有长度为 N-1 到 N+1（或其他匹配长度）的子字符串，其中 N 是 query_string 的长度，并在它们上一一使用 levenstein 并查看阈值。

python 中是否有更好的解决方案，最好是 python 2.7 中包含的模块，或者外部可用的模块。

---------------------更新和解决方案----------------

Python regex 模块工作得很好，尽管它比用于模糊子字符串情况的内置模块慢一点re，这是由于额外操作而产生的明显结果。期望的输出是好的，并且可以很容易地定义对模糊程度的控制。

>>> import regex
>>> input = "Monalisa was painted by Leonrdo da Vinchi"
>>> regex.search(r'\b(leonardo){e<3}\s+(da)\s+(vinci){e<2}\b',input,flags=regex.IGNORECASE)
<regex.Match object; span=(23, 41), match=' Leonrdo da Vinchi', fuzzy_counts=(0, 2, 1)>
4

5 回答 5

21

怎么用difflib.SequenceMatcher.get_matching_blocks？

>>> import difflib
>>> large_string = "thelargemanhatanproject"
>>> query_string = "manhattan"
>>> s = difflib.SequenceMatcher(None, large_string, query_string)
>>> sum(n for i,j,n in s.get_matching_blocks()) / float(len(query_string))
0.8888888888888888

>>> query_string = "banana"
>>> s = difflib.SequenceMatcher(None, large_string, query_string)
>>> sum(n for i,j,n in s.get_matching_blocks()) / float(len(query_string))
0.6666666666666666

更新

import difflib

def matches(large_string, query_string, threshold):
    words = large_string.split()
    for word in words:
        s = difflib.SequenceMatcher(None, word, query_string)
        match = ''.join(word[i:i+n] for i, j, n in s.get_matching_blocks() if n)
        if len(match) / float(len(query_string)) >= threshold:
            yield match

large_string = "thelargemanhatanproject is a great project in themanhattincity"
query_string = "manhattan"
print list(matches(large_string, query_string, 0.8))

以上代码打印：['manhatan', 'manhattn']

于 2013-07-19T08:11:13.717 回答
21

即将取代 re 的新正则表达式库包括模糊匹配。

https://pypi.python.org/pypi/regex/

模糊匹配语法看起来相当有表现力，但这会给你一个匹配的插入/添加/删除。

import regex
regex.match('(amazing){e<=1}', 'amaging')
于 2013-10-30T23:59:37.883 回答
19

我使用fuzzywuzzy基于阈值进行模糊匹配，并使用fuzzysearch从匹配中模糊提取单词。

process.extractBests接受查询、单词列表和截止分数，并返回匹配元组的列表以及高于截止分数的分数。

find_near_matches获取结果process.extractBests并返回单词的开始和结束索引。我使用索引来构建单词并使用构建的单词在大字符串中查找索引。max_l_distoffind_near_matches是“Levenshtein 距离”，必须根据需要进行调整。

from fuzzysearch import find_near_matches
from fuzzywuzzy import process

large_string = "thelargemanhatanproject is a great project in themanhattincity"
query_string = "manhattan"

def fuzzy_extract(qs, ls, threshold):
    '''fuzzy matches 'qs' in 'ls' and returns list of 
    tuples of (word,index)
    '''
    for word, _ in process.extractBests(qs, (ls,), score_cutoff=threshold):
        print('word {}'.format(word))
        for match in find_near_matches(qs, word, max_l_dist=1):
            match = word[match.start:match.end]
            print('match {}'.format(match))
            index = ls.find(match)
            yield (match, index)

去测试：

query_string = "manhattan"
print('query: {}\nstring: {}'.format(query_string, large_string))
for match,index in fuzzy_extract(query_string, large_string, 70):
    print('match: {}\nindex: {}'.format(match, index))

query_string = "citi"
print('query: {}\nstring: {}'.format(query_string, large_string))
for match,index in fuzzy_extract(query_string, large_string, 30):
    print('match: {}\nindex: {}'.format(match, index))

query_string = "greet"
print('query: {}\nstring: {}'.format(query_string, large_string))
for match,index in fuzzy_extract(query_string, large_string, 30):
    print('match: {}\nindex: {}'.format(match, index))

输出：

query: manhattan  
string: thelargemanhatanproject is a great project in themanhattincity  
match: manhatan  
index: 8  
match: manhattin  
index: 49  

query: citi  
string: thelargemanhatanproject is a great project in themanhattincity  
match: city  
index: 58  

query: greet  
string: thelargemanhatanproject is a great project in themanhattincity  
match: great  
index: 29
于 2015-06-04T21:00:24.603 回答
16

最近我为 Python 写了一个对齐库：https ://github.com/eseraygun/python-alignment

使用它，您可以在任何一对序列上使用任意评分策略执行全局和局部比对。实际上，在您的情况下，您需要半本地对齐，因为您不关心query_string. 我在下面的代码中使用局部对齐和一些启发式方法模拟了半局部算法，但是很容易扩展库以获得正确的实现。

这是为您的案例修改的 README 文件中的示例代码。

from alignment.sequence import Sequence, GAP_ELEMENT
from alignment.vocabulary import Vocabulary
from alignment.sequencealigner import SimpleScoring, LocalSequenceAligner

large_string = "thelargemanhatanproject is a great project in themanhattincity"
query_string = "manhattan"

# Create sequences to be aligned.
a = Sequence(large_string)
b = Sequence(query_string)

# Create a vocabulary and encode the sequences.
v = Vocabulary()
aEncoded = v.encodeSequence(a)
bEncoded = v.encodeSequence(b)

# Create a scoring and align the sequences using local aligner.
scoring = SimpleScoring(1, -1)
aligner = LocalSequenceAligner(scoring, -1, minScore=5)
score, encodeds = aligner.align(aEncoded, bEncoded, backtrace=True)

# Iterate over optimal alignments and print them.
for encoded in encodeds:
    alignment = v.decodeSequenceAlignment(encoded)

    # Simulate a semi-local alignment.
    if len(filter(lambda e: e != GAP_ELEMENT, alignment.second)) != len(b):
        continue
    if alignment.first[0] == GAP_ELEMENT or alignment.first[-1] == GAP_ELEMENT:
        continue
    if alignment.second[0] == GAP_ELEMENT or alignment.second[-1] == GAP_ELEMENT:
        continue

    print alignment
    print 'Alignment score:', alignment.score
    print 'Percent identity:', alignment.percentIdentity()
    print

的输出minScore=5如下。

m a n h a - t a n
m a n h a t t a n
Alignment score: 7
Percent identity: 88.8888888889

m a n h a t t - i
m a n h a t t a n
Alignment score: 5
Percent identity: 77.7777777778

m a n h a t t i n
m a n h a t t a n
Alignment score: 7
Percent identity: 88.8888888889

如果您删除minScore参数，您将只获得最佳得分匹配。

m a n h a - t a n
m a n h a t t a n
Alignment score: 7
Percent identity: 88.8888888889

m a n h a t t i n
m a n h a t t a n
Alignment score: 7
Percent identity: 88.8888888889

请注意，库中的所有算法都具有O(n * m)时间复杂度，n并且m是序列的长度。

于 2013-10-30T22:09:30.943 回答
15

上面的方法很好，但我需要在很多干草中找到一根小针，最后像这样接近它：

from difflib import SequenceMatcher as SM
from nltk.util import ngrams
import codecs

needle = "this is the string we want to find"
hay    = "text text lots of text and more and more this string is the one we wanted to find and here is some more and even more still"

needle_length  = len(needle.split())
max_sim_val    = 0
max_sim_string = u""

for ngram in ngrams(hay.split(), needle_length + int(.2*needle_length)):
    hay_ngram = u" ".join(ngram)
    similarity = SM(None, hay_ngram, needle).ratio() 
    if similarity > max_sim_val:
        max_sim_val = similarity
        max_sim_string = hay_ngram

print max_sim_val, max_sim_string

产量：

0.72972972973 this string is the one we wanted to find
于 2015-07-15T14:35:54.583 回答