我想比较2个字符串并保持匹配，在比较失败的地方分开。
所以，如果我有2个字符串 -
string1 = apples
string2 = appleses
answer = apples
另一个例子，因为字符串可能有多个单词。
string1 = apple pie available
string2 = apple pies
answer = apple pie
我确信有一种简单的Python方法可以做到这一点，但我无法解决，任何帮助和解释都会受到赞赏。
答案 0 :(得分：111)
为了完整性，标准库中的difflib提供了大量序列比较实用程序。例如find_longest_match，它在字符串上使用时找到最长的公共子字符串。使用示例：
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
答案 1 :(得分：32)
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'
或者有点奇怪的方式：
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))
可能更具可读性
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))
答案 2 :(得分：12)
它被称为最长公共子串问题。在这里，我提出了一个简单易懂但效率低下的解决方案。为大字符串生成正确的输出需要很长时间，因为该算法的复杂度为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")
<强>输出
apple pie
apples
apples
答案 3 :(得分：8)
还可以考虑使用os.path.commonprefix来处理字符，因此可以用于任何字符串。
import os
common = os.path.commonprefix(['apple pie available', 'apple pies'])
assert common == 'apple pie'
答案 4 :(得分：5)
与Evo's相同，但要比较任意数量的字符串：
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())
答案 5 :(得分：3)
使用第一个答案修复错误：
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")
答案 6 :(得分：1)
尝试：
import itertools as it
''.join(el[0] for el in it.takewhile(lambda t: t[0] == t[1], zip(string1, string2)))
它从两个字符串的开头进行比较。
答案 7 :(得分：1)
我发现的最快方法是使用 suffix_trees 包：
from suffix_trees import STree
a = ["xxxabcxxx", "adsaabc"]
st = STree.STree(a)
print(st.lcs()) # "abc"
答案 8 :(得分：1)
如果我们有一个需要查找所有常见子字符串的单词列表，我检查了上面的一些代码，最好的是https://stackoverflow.com/a/42882629/8520109，但其中有一些错误，例如'histhome'和'homehist'。在这种情况下，结果应该是'hist'和'home'。此外，如果更改参数顺序，则有所不同。因此，我更改代码以查找子字符串的每个块，并得到一组常见的子字符串：
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'}
答案 9 :(得分：1)
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
答案 10 :(得分：1)
这不是最有效的方法，但这是我能想到的并且它有效。如果有人可以改进它，请做。它的作用是制作一个矩阵并将1放在字符匹配的位置。然后它扫描矩阵以找到1s的最长对角线，跟踪它的开始和结束位置。然后它返回输入字符串的子字符串，其中起始位置和结束位置作为参数。
注意：这只找到一个最长的公共子字符串。如果有多个，你可以创建一个数组来存储结果并返回它。另外，它区分大小写（苹果派，苹果派）将返回pple馅饼。
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
答案 11 :(得分：0)
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))
答案 12 :(得分：0)
首先从itertools pairwise recipe改编一个帮助函数来生成子串。
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)
然后，一个函数迭代子字符串，最长的第一个，并测试成员资格。 （效率未考虑）
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
>>>
答案 13 :(得分：0)
这是教室问题，称为“最长序列查找器”。我给出了一些对我有用的简单代码，我的输入也是一个序列列表，该序列也可以是字符串，可能对您有帮助：
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
答案 14 :(得分：0)
**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
答案 15 :(得分：0)
Trie数据结构将比DP更好地工作。 这是代码。
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()
如有疑问，请告诉我。
答案 16 :(得分：0)
此脚本要求您最小公共子字符串长度，并在两个字符串中给出所有公共子字符串。而且，它消除了较长的子字符串已经包含的较短的子字符串。
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')
答案 17 :(得分：0)
好像这个问题没有足够的答案，这是另一个选择：
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]
一些示例情况：
LongestCommonSubstring("whose car?", "this is my car")
> ' car'
LongestCommonSubstring("apple pies", "apple? forget apple pie!")
> 'apple pie'
答案 18 :(得分：0)
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]
答案 19 :(得分：0)
返回第一个最长的公共子字符串：
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