在python中查找两个字符串的所有常见连续子字符串[重复]

Question

我有两个字符串，我想找到所有常用词。例如，

s1 = 'Today is a good day, it is a good idea to have a walk.'

s2 = 'Yesterday was not a good day, but today is good, shall we have a walk?'

考虑 s1 匹配 s2

'Today is' 匹配 'today is' 但 'Today is a' 不匹配 s2 中的任何字符。因此，“今天是”是常见的连续字符之一。同样，我们有“美好的一天”、“是”、“美好的”、“散步”。所以常用词是

common = ['today is', 'a good day', 'is', 'a good', 'have a walk']

我们可以使用正则表达式来做到这一点吗？

非常感谢。

Answer 1

import string
s1 = 'Today is a good day, it is a good idea to have a walk.'
s2 = 'Yesterday was not a good day, but today is good, shall we have a walk?'
z=[]
s1=s1.translate(None, string.punctuation) #remove punctuation
s2=s2.translate(None, string.punctuation)
print s1
print s2
sw1=s1.lower().split()                   #split it into words
sw2=s2.lower().split()
print sw1,sw2
i=0
while i<len(sw1):          #two loops to detect common strings. used while so as to change value of i in the loop itself
    x=0
    r=""
    d=i
    #print r
    for j in range(len(sw2)):
        #print r
        if sw1[i]==sw2[j]:
            r=r+' '+sw2[j]                       #if string same keep adding to a variable
            x+=1
            i+=1
        else:
            if x>0:     # if not same check if there is already one in buffer and add it to result (here z)
                z.append(r)
                i=d
                r=""
                x=0
    if x>0:                                            #end case of above loop
        z.append(r)
        r=""
        i=d
        x=0
    i+=1 
    #print i
print list(set(z)) 

#O(n^3)

Answer 2

引用自Find common substring between two strings

修改了几行，增加了几行 如果未找到任何子字符串，则默认返回 answer = "NULL" 。

已添加 继续搜索，直到你得到 NULL 并存储到 List

def longestSubstringFinder(string1, string2):
    answer = "NULL"
    len1, len2 = len(string1), len(string2)
    for i in range(len1):
        match = ""
        for j in range(len2):
            if (i + j < len1 and string1[i + j] == string2[j]):
                match += string2[j]
            else:
                if (len(match) > len(answer)): answer = match
                match = ""
    return answer


mylist = []

def call():
    s1 = 'Today is a good day, it is a good idea to have a walk.'

    s2 = 'Yesterday was not a good day, but today is good, shall we have a walk?'
    s1 =  s1.lower()
    s2 = s2.lower()
    x = longestSubstringFinder(s2,s1)
    while(longestSubstringFinder(s2,s1) != "NULL"): 
        x = longestSubstringFinder(s2,s1)
        print(x)
        mylist.append(x)
        s2 = s2.replace(x,' ')

call()
print ('[%s]' % ','.join(map(str, mylist)))

输出

[ a good day, , have a walk,today is , good]

输出的不同

common = ['today is', 'a good day', 'is', 'a good', 'have a walk']

您对第二个 "is" 的期望是错误的，正如您在 s2 中看到的那样，只有一个“is”