将两个字符串与一个公共子字符串连接在一起？

Question

Say I have strings,

string1 = 'Hello how are you'
string2 = 'are you doing now?'

The result should be something like

Hello how are you doing now?

I was thinking different ways using re and string search. ( Longest common substring problem )

But is there any simple way (or library) that does this in python?

To make things clear i'll add one more set of test strings!

string1 = 'This is a nice ACADEMY'
string2 = 'DEMY you know!'

the result would be!,

'This is a nice ACADEMY you know!'

Answer 1

This should do:

string1 = 'Hello how are you'
string2 = 'are you doing now?'
i = 0
while not string2.startswith(string1[i:]):
    i += 1

sFinal = string1[:i] + string2

OUTPUT :

>>> sFinal
'Hello how are you doing now?'

or, make it a function so that you can use it again without rewriting:

def merge(s1, s2):
    i = 0
    while not s2.startswith(s1[i:]):
        i += 1
    return s1[:i] + s2

OUTPUT :

>>> merge('Hello how are you', 'are you doing now?')
'Hello how are you doing now?'
>>> merge("This is a nice ACADEMY", "DEMY you know!")
'This is a nice ACADEMY you know!'

Answer 2

This should do what you want:

def overlap_concat(s1, s2):
    l = min(len(s1), len(s2))
    for i in range(l, 0, -1):
        if s1.endswith(s2[:i]):
            return s1 + s2[i:]
    return s1 + s2

Examples:

>>> overlap_concat("Hello how are you", "are you doing now?")
'Hello how are you doing now?'
>>> 

>>> overlap_concat("This is a nice ACADEMY", "DEMY you know!")
'This is a nice ACADEMY you know!'
>>> 

Answer 3

Using str.endswith and enumerate :

def overlap(string1, string2):
    for i, s in enumerate(string2, 1):
         if string1.endswith(string2[:i]):
            break

    return string1 + string2[i:]

>>> overlap("Hello how are you", "are you doing now?")
'Hello how are you doing now?'

>>> overlap("This is a nice ACADEMY", "DEMY you know!")
'This is a nice ACADEMY you know!'

If you were to account for trailing special characters, you'd be wanting to employ some re based substitution.

import re
string1 = re.sub('[^\w\s]', '', string1)

Although note that this would remove all special characters in the first string.

---

A modification to the above function which will find the longest matching substring (instead of the shortest) involves traversing string2 in reverse.

def overlap(string1, string2):
   for i in range(len(s)):
      if string1.endswith(string2[:len(string2) - i]):
          break

   return string1 + string2[len(string2) - i:]

>>> overlap('Where did', 'did you go?') 
'Where did you go?'

Answer 4

Other answers were great guys but it did fail for this input.

string1 = 'THE ACADEMY has'
string2= '.CADEMY has taken'

output:

>>> merge(string1,string2)
'THE ACADEMY has.CADEMY has taken'
>>> overlap(string1,string2)
'THE ACADEMY has'

However there's this standard library difflib which proved to be effective in my case!

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.size]+string2[match.b + match.size:]) 

output:

Match(a=5, b=1, size=10)
THE ACADEMY has taken

Answer 5

which words you want to replace are appearing in the second string so you can try something like :

new_string=[string2.split()]
new=[]
new1=[j for item in new_string for j in item if j not in string1]
new1.insert(0,string1)
print(" ".join(new1))

with the first test case:

string1 = 'Hello how are you'
string2 = 'are you doing now?'

output:

Hello how are you doing now?

second test case:

string1 = 'This is a nice ACADEMY'
string2 = 'DEMY you know!'

output:

This is a nice ACADEMY you know!

Explanation :

first, we are splitting the second string so we can find which words we have to remove or replace :

new_string=[string2.split()]

second step we will check each word of this splitter string with string1 , if any word is in that string than choose only first string word , leave that word in second string :

new1=[j for item in new_string for j in item if j not in string1]

This list comprehension is same as :

new1=[]
for item in new_string:
    for j in item:
        if j not in string1:
            new1.append(j)

last step combines both string and join the list:

new1.insert(0,string1)
print(" ".join(new1))