在Python中检查较长字符串中存在的模糊/近似子字符串

Question

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：在较大的字符串中根据阈值找到模糊子字符串。

例如。

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

一种暴力解决方案是生成长度为N-1到N+1（或其他匹配长度）的所有子串，其中N是query_string的长度，并逐一对其使用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正则表达式模块工作得很好，尽管它比内置的re模块在模糊子字符串情况下慢一点，这是由于额外的操作而产生的明显结果。期望的输出是好的，并且可以容易地定义对模糊性大小的控制。

>>> 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)>

Answer 1

即将取代的新regex库重新包含模糊匹配。

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

模糊匹配语法看起来相当有表现力，但这会让您通过一个或更少的插入/添加/删除进行匹配。

import regex
regex.match('(amazing){e<=1}', 'amaging')

Answer 2

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获取查询、单词列表和截止分数，并返回匹配的元组列表和高于截止分数的分数。

CCD_ 3取CCD_ 4的结果并返回单词的起始索引和结束索引。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。find_near_matches的max_l_dist是"Levenstein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)

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 

Answer 3

使用difflib.SequenceMatcher.get_matching_blocks怎么样？

>>> 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']

Answer 4

上面的方法很好，但我需要在大量的干草中找到一根小针，结果像这样接近它：

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

收益率：

0.72972972973 this string is the one we wanted to find

Answer 5

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

使用它，您可以在任意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，您需要半局部对齐，因为您不关心query_string的子字符串。在下面的代码中，我已经使用局部对齐和一些启发式方法模拟了半局部算法，但为了正确的实现，扩展库很容易。

以下是针对您的案例修改的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

minScore=5的输出如下。

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

如果去掉minScore参数，则只能得到得分最好的匹配项。

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

注意，库中的所有算法都具有O(n * m)时间复杂度，n和m是序列的长度。

Answer 6

我遇到了这个问题，我发现前两个答案都不起作用。相反，我使用以下算法来检测错误最小的模糊匹配：

def fuzzy_substring_search(cls, major: str, minor: str, errs: int = 4) -> Optional[regex.Match]:
    """Find the closest matching fuzzy substring.
    Args:
        major: the string to search in
        minor: the string to search with
        errs: the total number of errors
    Returns:
        Optional[regex.Match] object
    """
    errs_ = 0
    s = regex.search(f"({minor}){{e<={errs_}}}", major)
    while s is None and errs_ <= errs:
        errs_ += 1
        s = regex.search(f"({minor}){{e<={errs_}}}", major)
    return s

这样做的好处是，如果存在，可以返回精确的匹配，并根据需要升级模糊性。