查找两个字符串之间的公共子字符串

Question

I'd like to compare 2 strings and keep the matched, splitting off where the comparison fails.

So if I have 2 strings -

string1 = apples
string2 = appleses

answer = apples

Another example, as the string could have more than one word.

string1 = apple pie available
string2 = apple pies

answer = apple pie

I'm sure there is a simple Python way of doing this but I can't work it out, any help and explanation appreciated.

Answer 1

For completeness, difflib in the standard-library provides loads of sequence-comparison utilities. For instance find_longest_match which finds the longest common substring when used on strings. Example use:

from difflib import SequenceMatcher

string1 = "apple pie available"
string2 = "come have some apple pies"

match = SequenceMatcher(None, string1, string2).find_longest_match(0, len(string1), 0, len(string2))

print(match)  # -> Match(a=0, b=15, size=9)
print(string1[match.a: match.a + match.size])  # -> apple pie
print(string2[match.b: match.b + match.size])  # -> apple pie

Answer 2

def common_start(sa, sb):
    """ returns the longest common substring from the beginning of sa and sb """
    def _iter():
        for a, b in zip(sa, sb):
            if a == b:
                yield a
            else:
                return

    return ''.join(_iter())

>>> common_start("apple pie available", "apple pies")
'apple pie'

---

Or a slightly stranger way:

def stop_iter():
    """An easy way to break out of a generator"""
    raise StopIteration

def common_start(sa, sb):
    return ''.join(a if a == b else stop_iter() for a, b in zip(sa, sb))

Which might be more readable as

def terminating(cond):
    """An easy way to break out of a generator"""
    if cond:
        return True
    raise StopIteration

def common_start(sa, sb):
    return ''.join(a for a, b in zip(sa, sb) if terminating(a == b))

Answer 3

One might also consider os.path.commonprefix that works on characters and thus can be used for any strings.

import os
common = os.path.commonprefix(['apple pie available', 'apple pies'])
assert common == 'apple pie'

As the function name indicates, this only considers the common prefix of two strings.

Answer 4

Its called Longest Common Substring problem. Here I present a simple, easy to understand but inefficient solution. It will take a long time to produce correct output for large strings, as the complexity of this algorithm is O(N^2).

def longestSubstringFinder(string1, string2):
    answer = ""
    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

print longestSubstringFinder("apple pie available", "apple pies")
print longestSubstringFinder("apples", "appleses")
print longestSubstringFinder("bapples", "cappleses")

Output

apple pie
apples
apples

Answer 5

Fix bugs with the first's answer:

def longestSubstringFinder(string1, string2):
    answer = ""
    len1, len2 = len(string1), len(string2)
    for i in range(len1):
        for j in range(len2):
            lcs_temp=0
            match=''
            while ((i+lcs_temp < len1) and (j+lcs_temp<len2) and string1[i+lcs_temp] == string2[j+lcs_temp]):
                match += string2[j+lcs_temp]
                lcs_temp+=1
            if (len(match) > len(answer)):
                answer = match
    return answer

print longestSubstringFinder("dd apple pie available", "apple pies")
print longestSubstringFinder("cov_basic_as_cov_x_gt_y_rna_genes_w1000000", "cov_rna15pcs_as_cov_x_gt_y_rna_genes_w1000000")
print longestSubstringFinder("bapples", "cappleses")
print longestSubstringFinder("apples", "apples")

Answer 6

The same as Evo's , but with arbitrary number of strings to compare:

def common_start(*strings):
    """ Returns the longest common substring
        from the beginning of the `strings`
    """
    def _iter():
        for z in zip(*strings):
            if z.count(z[0]) == len(z):  # check all elements in `z` are the same
                yield z[0]
            else:
                return

    return ''.join(_iter())

Answer 7

Try:

import itertools as it
''.join(el[0] for el in it.takewhile(lambda t: t[0] == t[1], zip(string1, string2)))

It does the comparison from the beginning of both strings.

Answer 8

def matchingString(x,y):
    match=''
    for i in range(0,len(x)):
        for j in range(0,len(y)):
            k=1
            # now applying while condition untill we find a substring match and length of substring is less than length of x and y
            while (i+k <= len(x) and j+k <= len(y) and x[i:i+k]==y[j:j+k]):
                if len(match) <= len(x[i:i+k]):
                   match = x[i:i+k]
                k=k+1
    return match  

print matchingString('apple','ale') #le
print matchingString('apple pie available','apple pies') #apple pie     

Answer 9

This script requests you the minimum common substring length and gives all common substrings in two strings. Also, it eliminates shorter substrings that longer substrings include already.

def common_substrings(str1,str2):
    len1,len2=len(str1),len(str2)

    if len1 > len2:
        str1,str2=str2,str1
        len1,len2=len2,len1

    min_com = int(input('Please enter the minumum common substring length:'))
    
    cs_array=[]
    for i in range(len1,min_com-1,-1):
        for k in range(len1-i+1):
            if (str1[k:i+k] in str2):
                flag=1
                for m in range(len(cs_array)):
                    if str1[k:i+k] in cs_array[m]:
                    #print(str1[k:i+k])
                        flag=0
                        break
                if flag==1:
                    cs_array.append(str1[k:i+k])
    if len(cs_array):
        print(cs_array)
    else:
        print('There is no any common substring according to the parametres given')

common_substrings('ciguliuana','ciguana')
common_substrings('apples','appleses')
common_substrings('apple pie available','apple pies')

Answer 10

The fastest way I've found is to use suffix_trees package:

from suffix_trees import STree

a = ["xxxabcxxx", "adsaabc"]
st = STree.STree(a)
print(st.lcs()) # "abc"

Answer 11

A Trie data structure would work the best, better than DP. Here is the code.

class TrieNode:
    def __init__(self):
        self.child = [None]*26
        self.endWord = False

class Trie:

    def __init__(self):
        self.root = self.getNewNode()

    def getNewNode(self):
        return TrieNode()

    def insert(self,value):
        root = self.root


        for i,character in enumerate(value):
            index = ord(character) - ord('a')
            if not root.child[index]:
                root.child[index] = self.getNewNode()
            root = root.child[index]

        root.endWord = True


    def search(self,value):
        root = self.root

        for i,character in enumerate(value):
            index = ord(character) - ord('a')
            if not root.child[index]:
                return False
            root = root.child[index]
        return root.endWord

def main(): 

    # Input keys (use only 'a' through 'z' and lower case) 
    keys = ["the","anaswe"] 
    output = ["Not present in trie", 
            "Present in trie"] 

    # Trie object 
    t = Trie() 

    # Construct trie 
    for key in keys: 
        t.insert(key) 

    # Search for different keys 
    print("{} ---- {}".format("the",output[t.search("the")])) 
    print("{} ---- {}".format("these",output[t.search("these")])) 
    print("{} ---- {}".format("their",output[t.search("their")])) 
    print("{} ---- {}".format("thaw",output[t.search("thaw")])) 

if __name__ == '__main__': 
    main() 

Let me know in case of doubts.

Answer 12

In case we have a list of words that we need to find all common substrings I check some of the codes above and the best was https://stackoverflow.com/a/42882629/8520109 but it has some bugs for example 'histhome' and 'homehist' . In this case, we should have 'hist' and 'home' as a result. Furthermore, it differs if the order of arguments is changed. So I change the code to find every block of substring and it results a set of common substrings:

main = input().split(" ")    #a string of words separated by space
def longestSubstringFinder(string1, string2):
    '''Find the longest matching word'''
    answer = ""
    len1, len2 = len(string1), len(string2)
    for i in range(len1):
        for j in range(len2):
            lcs_temp=0
            match=''
            while ((i+lcs_temp < len1) and (j+lcs_temp<len2) and string1[i+lcs_temp] == string2[j+lcs_temp]):
                match += string2[j+lcs_temp]
                lcs_temp+=1         
            if (len(match) > len(answer)):
                answer = match              
    return answer

def listCheck(main):
    '''control the input for finding substring in a list of words'''
    string1 = main[0]
    result = []
    for i in range(1, len(main)):
        string2 = main[i]
        res1 = longestSubstringFinder(string1, string2)
        res2 = longestSubstringFinder(string2, string1)
        result.append(res1)
        result.append(res2)
    result.sort()
    return result

first_answer = listCheck(main)

final_answer  = []


for item1 in first_answer:    #to remove some incorrect match
    string1 = item1
    double_check = True
    for item2 in main:
        string2 = item2
        if longestSubstringFinder(string1, string2) != string1:
            double_check = False
    if double_check:
        final_answer.append(string1)

print(set(final_answer))

---

main = 'ABACDAQ BACDAQA ACDAQAW XYZCDAQ' #>>> {'CDAQ'}
main = 'homehist histhome' #>>> {'hist', 'home'}

Answer 13

Returns the first longest common substring:

def compareTwoStrings(string1, string2):
    list1 = list(string1)
    list2 = list(string2)

    match = []
    output = ""
    length = 0

    for i in range(0, len(list1)):

        if list1[i] in list2:
            match.append(list1[i])

            for j in range(i + 1, len(list1)):

                if ''.join(list1[i:j]) in string2:
                    match.append(''.join(list1[i:j]))

                else:
                    continue
        else:
            continue

    for string in match:

        if length < len(list(string)):
            length = len(list(string))
            output = string

        else:
            continue

    return output

Answer 14

This is the classroom problem called 'Longest sequence finder'. I have given some simple code that worked for me, also my inputs are lists of a sequence which can also be a string:

def longest_substring(list1,list2):
    both=[]
    if len(list1)>len(list2):
        small=list2
        big=list1
    else:
        small=list1
        big=list2
    removes=0
    stop=0
    for i in small:
        for j in big:
            if i!=j:
                removes+=1
                if stop==1:
                    break
            elif i==j:
                both.append(i)
                for q in range(removes+1):
                    big.pop(0)
                stop=1
                break
        removes=0
    return both

Answer 15

**Return the comman longest substring** 
def longestSubString(str1, str2):
    longestString = ""
    maxLength = 0
    for i in range(0, len(str1)):
        if str1[i] in str2:
            for j in range(i + 1, len(str1)):
                if str1[i:j] in str2:
                    if(len(str1[i:j]) > maxLength):
                        maxLength = len(str1[i:j])
                        longestString =  str1[i:j]
return longestString

Answer 16

As if this question doesn't have enough answers, here's another option:

from collections import defaultdict
def LongestCommonSubstring(string1, string2):
    match = ""
    matches = defaultdict(list)
    str1, str2 = sorted([string1, string2], key=lambda x: len(x))

    for i in range(len(str1)):
        for k in range(i, len(str1)):
            cur = match + str1[k]
            if cur in str2:
                match = cur
            else:
                match = ""
            
            if match:
                matches[len(match)].append(match)
        
    if not matches:
        return ""

    longest_match = max(matches.keys())
        
    return matches[longest_match][0]

Some example cases:

LongestCommonSubstring("whose car?", "this is my car")
> ' car'
LongestCommonSubstring("apple pies", "apple? forget apple pie!")
> 'apple pie'

Answer 17

def LongestSubString(s1,s2):
    if len(s1)<len(s2) :
        s1,s2 = s2,s1  
    
    maxsub =''
    for i in range(len(s2)):
        for j in range(len(s2),i,-1):
            if s2[i:j] in s1 and j-i>len(maxsub):                
                return  s2[i:j]

Answer 18

This isn't the most efficient way to do it but it's what I could come up with and it works. If anyone can improve it, please do. What it does is it makes a matrix and puts 1 where the characters match. Then it scans the matrix to find the longest diagonal of 1s, keeping track of where it starts and ends. Then it returns the substring of the input string with the start and end positions as arguments.

Note: This only finds one longest common substring. If there's more than one, you could make an array to store the results in and return that Also, it's case sensitive so (Apple pie, apple pie) will return pple pie.

def longestSubstringFinder(str1, str2):
answer = ""

if len(str1) == len(str2):
    if str1==str2:
        return str1
    else:
        longer=str1
        shorter=str2
elif (len(str1) == 0 or len(str2) == 0):
    return ""
elif len(str1)>len(str2):
    longer=str1
    shorter=str2
else:
    longer=str2
    shorter=str1

matrix = numpy.zeros((len(shorter), len(longer)))

for i in range(len(shorter)):
    for j in range(len(longer)):               
        if shorter[i]== longer[j]:
            matrix[i][j]=1

longest=0

start=[-1,-1]
end=[-1,-1]    
for i in range(len(shorter)-1, -1, -1):
    for j in range(len(longer)):
        count=0
        begin = [i,j]
        while matrix[i][j]==1:

            finish=[i,j]
            count=count+1 
            if j==len(longer)-1 or i==len(shorter)-1:
                break
            else:
                j=j+1
                i=i+1

        i = i-count
        if count>longest:
            longest=count
            start=begin
            end=finish
            break

answer=shorter[int(start[0]): int(end[0])+1]
return answer

Answer 19

First a helper function adapted from the itertools pairwise recipe to produce substrings.

import itertools
def n_wise(iterable, n = 2):
    '''n = 2 -> (s0,s1), (s1,s2), (s2, s3), ...

    n = 3 -> (s0,s1, s2), (s1,s2, s3), (s2, s3, s4), ...'''
    a = itertools.tee(iterable, n)
    for x, thing in enumerate(a[1:]):
        for _ in range(x+1):
            next(thing, None)
    return zip(*a)

Then a function the iterates over substrings, longest first, and tests for membership. (efficiency not considered)

def foo(s1, s2):
    '''Finds the longest matching substring
    '''
    # the longest matching substring can only be as long as the shortest string
    #which string is shortest?
    shortest, longest = sorted([s1, s2], key = len)
    #iterate over substrings, longest substrings first
    for n in range(len(shortest)+1, 2, -1):
        for sub in n_wise(shortest, n):
            sub = ''.join(sub)
            if sub in longest:
                #return the first one found, it should be the longest
                return sub

s = "fdomainster"
t = "exdomainid"
print(foo(s,t))

---

>>> 
domain
>>> 

Answer 20

def LongestSubString(s1,s2):
    left = 0
    right =len(s2)
    while(left<right):
        if(s2[left] not in s1):
            left = left+1
        else:
            if(s2[left:right] not in s1):
                right = right - 1
            else:
                return(s2[left:right])

s1 = "pineapple"
s2 = "applc"
print(LongestSubString(s1,s2))