Question

bigString = "AGAHKGHKHASNHADKRGHFKXXX_I_AM_THERE_XXXXXMHHGRFSAHGSKHASGKHGKHSKGHAK"
smallString = "I_AM_HERE"

我应该使用哪种有效算法来查找＆＃34; bigString＆＃34;的子字符串。与＆＃34; smallString＆＃34;

紧密匹配

output = "I_AM_THERE"

与小字符串相比，输出可能没有多少插入和删除。

编辑：找到了一个很好的例子，非常接近我的问题：How to add variable error to regex fuzzy search. Python

Answer 1

您可以将almost-ready-to-be-everyones-regex包与模糊匹配结合使用：

>>> import regex
>>> bigString = "AGAHKGHKHASNHADKRGHFKXXX_I_AM_THERE_XXXXXMHHGRFSAHGSKHASGKHGKHSKGHAK"
>>> regex.search('(?:I_AM_HERE){e<=1}',bigString).group(0)
'I_AM_THERE'

或者：

>>> bigString = "AGAH_I_AM_HERE_RGHFKXXX_I_AM_THERE_XXX_I_AM_NOWHERE_EREXXMHHGRFS"
>>> print(regex.findall('I_AM_(?:HERE){e<=3}',bigString))
['I_AM_HERE', 'I_AM_THERE', 'I_AM_NOWHERE']

新的正则表达式模块将（hopefully）成为Python3.4的一部分

如果你有pip，只需输入pip install regex或pip3 install regex，直到Python 3.4出来（正则表达式部分...）

回答评论Is there a way to know the best out of the three in your second example? How to use BESTMATCH flag here?

使用最佳匹配标志(?b)来获得单个最佳匹配：

print(regex.search(r'(?b)I_AM_(?:ERE){e<=3}', bigString).group(0))
# I_AM_THE

或者与difflib结合或使用levenshtein距离以及所有可接受的匹配列表到第一个文字：

import regex

def levenshtein(s1,s2):
    if len(s1) > len(s2):
        s1,s2 = s2,s1
    distances = range(len(s1) + 1)
    for index2,char2 in enumerate(s2):
        newDistances = [index2+1]
        for index1,char1 in enumerate(s1):
            if char1 == char2:
                newDistances.append(distances[index1])
            else:
                newDistances.append(1 + min((distances[index1],
                                             distances[index1+1],
                                             newDistances[-1])))
        distances = newDistances
    return distances[-1]

bigString = "AGAH_I_AM_NOWHERE_HERE_RGHFKXXX_I_AM_THERE_XXX_I_AM_HERE_EREXXMHHGRFS"
cl=[(levenshtein(s,'I_AM_HERE'),s) for s in regex.findall('I_AM_(?:HERE){e<=3}',bigString)]

print(cl)
print([t[1] for t in sorted(cl, key=lambda t: t[0])])

print(regex.search(r'(?e)I_AM_(?:ERE){e<=3}', bigString).group(0))

打印：

[(3, 'I_AM_NOWHERE'), (1, 'I_AM_THERE'), (0, 'I_AM_HERE')]
['I_AM_HERE', 'I_AM_THERE', 'I_AM_NOWHERE']

Answer 2

使用difflib：

执行此操作时，有点愚蠢

from difflib import *

window = len(smallString) + 1  # allow for longer matches
chunks = [bigString[i:i+window] for i in range(len(bigString)-window)]
get_close_matches(smallString,chunks,1)

输出：

['_I_AM_THERE']

Answer 3

也许动态编程问题Longest Common Substring在这里会有一些用处。根据您的需要和匹配标准，您可以使用最长公共子区域

如何使用部分匹配查找子字符串

3 个答案: