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，任何帮助和解释都表示赞赏。
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。由于该算法的复杂度为O（N ^ 2），因此需要很长时间才能为大字符串生成正确的输出。
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")
产量
apple pie
apples
apples
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
首先是一个辅助功能，改编自itertools pairwise recipe以产生子串。
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
>>>
这是称为“最长序列发现者”的课堂问题。我已经给出了一些对我有用的简单代码，我的输入也是一个序列的列表，也可以是一个字符串，可能会帮助你：
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
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))
为了完整起见，标准库中的difflib提供了大量的序列比较实用程序。例如find_longest_match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
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))
哪个可能更具可读性
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))
人们也可以考虑使用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'
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())
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")
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。
这不是最有效的方法，但这是我能想到的，它的工作原理。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。此外，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
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