将两个字符串与一个公共子字符串连接在一起？

Question

说我有琴弦

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

结果应该是这样的

Hello how are you doing now?

我在考虑使用re和字符串搜索的不同方式。 （ 最长的普通子串问题 ）

但是有没有简单的方法（或库）在python中做到这一点？

为了清楚起见，我将再添加一组测试字符串！

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

结果就是！，

'This is a nice ACADEMY you know!'

Answer 1

应该这样做：

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

sFinal = string1[:i] + string2

输出：

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

或者，使其成为一个函数，以便您无需重写即可再次使用它：

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

输出：

>>> 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

这应该做您想要的：

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

例子：

>>> 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

使用str.endswith和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!'

如果要考虑尾随特殊字符，则需要使用一些基于re的替换。

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

尽管注意，这会删除第一个字符串中的所有特殊字符。

---

对上述函数的修改将找到最长的匹配子字符串（而不是最短的子字符串），涉及反向遍历string2 。

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

其他答案都是好人，但对于此输入确实失败了。

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

输出：

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

但是，有一个标准库difflib在我的情况下被证明是有效的！

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:]) 

输出：

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

Answer 5

您要替换的单词出现在第二个字符串中，因此您可以尝试以下操作：

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))

与第一个测试用例：

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

输出：

Hello how are you doing now?

第二个测试用例：

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

输出：

This is a nice ACADEMY you know!

说明：

首先，我们将第二个字符串拆分，以便可以找到必须删除或替换的单词：

new_string=[string2.split()]

第二步，我们将使用string1检查此分隔符字符串的每个单词，如果该字符串中有任何单词而不是仅选择第一个字符串单词，则将该单词保留在第二个字符串中：

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

此列表理解与：

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

最后一步结合了字符串和连接列表：

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