在Python中检查存在于较长字符串中的模糊/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.

>>> 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

一个强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,并且逐个使用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,或外部可用的模块.

更新:Python正则表达式模块工作得很好,虽然它比re模糊子字符串情况的内置模块慢一点,由于额外的操作,这是一个明显的结果.期望的输出是良好的,并且可以容易地定义对模糊度的控制.

>>> 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)>

Answer 1

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

UPDATE

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 2

我用fuzzywuzzy基于阈值和模糊匹配fuzzysearch从匹配模糊提取物的话.

process.extractBests 获取查询,单词列表和截止分数,并返回匹配和得分高于截止分数的元组列表.

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_dist的find_near_matches是"Levenshtein距离",这必须进行调整以适应需求.

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;

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;
查询:曼哈顿
字符串:thelargemanhatanproject是一个伟大的项目在themanhatcity
match:manhatan
index:8
match:manhattin
index:49

查询:citi
string:thelargemanhatanproject是一个伟大的项目在themanhatcity
匹配:城市
指数:58

查询:greet
string:thelargemanhatanproject是一个伟大的项目在themanhatcity
匹配:伟大的
索引:29

Answer 3

很快应该替换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具有表现力,但这会使您与一个或更少的插入/添加/删除匹配.

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

Answer 4

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

使用它,您可以在任何序列对上使用任意评分策略执行全局和局部对齐.实际上,在您的情况下,您需要半局部对齐,因为您不关心的子串query_string.我在下面的代码中使用局部对齐和一些启发式模拟了半局部算法,但是很容易扩展库以实现正确的实现.

以下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

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.

Answer 5

上面的方法很好,但我需要在很多干草中找到一根小针,最后接近它如下:

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