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，感谢任何帮助和解释。
问问题
139283 次
20 回答
181
为了完整起见，difflib标准库中提供了大量的序列比较实用程序。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。示例使用：
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
于 2016-09-09T06:00:18.787 回答
40
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'
或者稍微奇怪的方式：
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))
这可能更具可读性
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))
于 2013-09-10T09:59:58.167 回答
40
人们也可能认为os.path.commonprefix它适用于字符，因此可用于任何字符串。
import os
common = os.path.commonprefix(['apple pie available', 'apple pies'])
assert common == 'apple pie'
正如函数名所示，这只考虑了两个字符串的公共前缀。
于 2018-11-07T14:07:25.160 回答
17
它被称为最长公共子串问题。在这里，我提出了一个简单、易于理解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")
输出
apple pie
apples
apples
于 2013-09-10T11:28:39.030 回答
8
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")
于 2017-03-19T03:33:29.010 回答
6
与Evo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())
于 2016-07-15T21:42:49.023 回答
2
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。
于 2013-09-10T10:10:14.347 回答
2
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
于 2017-10-15T17:25:33.487 回答
2
此脚本要求您提供最小公共子字符串长度，并在两个字符串中提供所有公共子字符串。此外，它还消除了较长子字符串已经包含的较短子字符串。
def common_substrings(str1,str2):
len1,len2=len(str1),len(str2)
if len1 > len2:
str1,str2=str2,str1
len1,len2=len2,len1
#short string=str1 and long string=str2
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')
于 2020-08-10T09:28:49.183 回答
2
我发现最快的方法是使用suffix_trees包：
from suffix_trees import STree
a = ["xxxabcxxx", "adsaabc"]
st = STree.STree(a)
print(st.lcs()) # "abc"
于 2021-05-14T18:14:39.967 回答
1
Trie 数据结构效果最好，比 DP 更好。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。
于 2019-11-28T09:05:06.640 回答
1
如果我们有一个单词列表，我们需要找到所有常见的子字符串，我会检查上面的一些代码，最好的是https://stackoverflow.com/a/42882629/8520109但它有一些错误，例如“histhome”和'homehist'。在这种情况下，我们应该有'hist'和'home'作为结果。此外，如果参数的顺序改变，它会有所不同。因此，我更改了代码以查找每个子字符串块，并产生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'}
于 2020-02-25T06:10:20.887 回答
1
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]
于 2020-10-22T22:29:38.517 回答
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
于 2015-12-22T15:07:10.417 回答
0
这是名为“最长序列查找器”的课堂问题。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，也可以是一个字符串：
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
于 2018-12-05T11:55:06.133 回答
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
于 2019-10-13T15:36:28.133 回答
0
好像这个问题没有足够的答案，这里有另一种选择：
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]
一些示例案例：
LongestCommonSubstring("whose car?", "this is my car")
> ' car'
LongestCommonSubstring("apple pies", "apple? forget apple pie!")
> 'apple pie'
于 2020-08-21T21:56:46.333 回答
-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。它的作用是创建一个矩阵并将 1 放在字符匹配的位置。然后它扫描矩阵以找到最长的 1 对角线，并跟踪它的开始和结束位置。然后它以开始和结束位置作为参数返回输入字符串的子字符串。
注意：这只会找到一个最长的公共子字符串。如果有多个，您可以创建一个数组来存储结果并返回它此外，它区分大小写，因此 (Apple pie, apple pie) 将返回 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
于 2016-01-09T12:43:12.780 回答
-1
首先是一个从itertools 成对配方改编的辅助函数来生成子字符串。
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)
然后是一个函数，它迭代子字符串，最长的优先，并测试成员资格。（不考虑效率）
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
>>>
于 2016-08-12T07:02:59.483 回答
-1
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))
于 2018-10-27T21:10:57.147 回答