在两个子字符串之间查找字符串[重复]

Question

如何在两个子字符串之间找到一个字符串 ('123STRINGabc' -> 'STRING')？

我现在的方法是这样的：

>>> start = 'asdf=5;'
>>> end = '123jasd'
>>> s = 'asdf=5;iwantthis123jasd'
>>> print((s.split(start))[1].split(end)[0])
iwantthis

但是，这似乎非常低效且不符合 Python 风格。有什么更好的方法来做这样的事情？

忘了说： 字符串可能不以start 和end 开头和结尾。它们前后可能有更多字符。

Answer 1

import re

s = 'asdf=5;iwantthis123jasd'
result = re.search('asdf=5;(.*)123jasd', s)
print(result.group(1))

Answer 2

s = "123123STRINGabcabc"

def find_between( s, first, last ):
    try:
        start = s.index( first ) + len( first )
        end = s.index( last, start )
        return s[start:end]
    except ValueError:
        return ""

def find_between_r( s, first, last ):
    try:
        start = s.rindex( first ) + len( first )
        end = s.rindex( last, start )
        return s[start:end]
    except ValueError:
        return ""


print find_between( s, "123", "abc" )
print find_between_r( s, "123", "abc" )

给予：

123STRING
STRINGabc

我认为应该注意 - 根据您需要的行为，您可以混合使用 index 和 rindex 调用或使用上述版本之一（它相当于正则表达式 (.*) 和 (.*?) 组） .

Answer 3

start = 'asdf=5;'
end = '123jasd'
s = 'asdf=5;iwantthis123jasd'
print s[s.find(start)+len(start):s.rfind(end)]

给予

iwantthis

Answer 4

s[len(start):-len(end)]

Answer 5

字符串格式为 Nikolaus Gradwohl 的建议增加了一些灵活性。 start 和 end 现在可以根据需要进行修改。

import re

s = 'asdf=5;iwantthis123jasd'
start = 'asdf=5;'
end = '123jasd'

result = re.search('%s(.*)%s' % (start, end), s).group(1)
print(result)

Answer 6

如果你不想导入任何东西，试试字符串方法.index()：

text = 'I want to find a string between two substrings'
left = 'find a '
right = 'between two'

# Output: 'string'
print(text[text.index(left)+len(left):text.index(right)])

Answer 7

只需将 OP 自己的解决方案转换为答案：

def find_between(s, start, end):
    return (s.split(start))[1].split(end)[0]

Answer 8

source='your token _here0@df and maybe _here1@df or maybe _here2@df'
start_sep='_'
end_sep='@df'
result=[]
tmp=source.split(start_sep)
for par in tmp:
  if end_sep in par:
    result.append(par.split(end_sep)[0])

print result

必须显示： 这里0，这里1，这里2

正则表达式更好，但它需要额外的库，您可能只想使用 python

Answer 9

这是一种方法

_,_,rest = s.partition(start)
result,_,_ = rest.partition(end)
print result

使用正则表达式的另一种方式

import re
print re.findall(re.escape(start)+"(.*)"+re.escape(end),s)[0]

或

print re.search(re.escape(start)+"(.*)"+re.escape(end),s).group(1)

Answer 10

要提取STRING，请尝试：

myString = '123STRINGabc'
startString = '123'
endString = 'abc'

mySubString=myString[myString.find(startString)+len(startString):myString.find(endString)]

Answer 11

您可以简单地使用此代码或复制下面的函数。整齐地排成一行。

def substring(whole, sub1, sub2):
    return whole[whole.index(sub1) : whole.index(sub2)]

如果按如下方式运行函数。

print(substring("5+(5*2)+2", "(", "("))

你可能会得到输出：

(5*2

而不是

5*2

如果您想在输出末尾添加子字符串，代码必须如下所示。

return whole[whole.index(sub1) : whole.index(sub2) + 1]

但如果您不希望最后的子字符串，+1 必须在第一个值上。

return whole[whole.index(sub1) + 1 : whole.index(sub2)]

Answer 12

这是我用来返回一个列表的函数，该列表在 string1 和 string2 之间搜索了一个字符串。

def GetListOfSubstrings(stringSubject,string1,string2):
    MyList = []
    intstart=0
    strlength=len(stringSubject)
    continueloop = 1

    while(intstart < strlength and continueloop == 1):
        intindex1=stringSubject.find(string1,intstart)
        if(intindex1 != -1): #The substring was found, lets proceed
            intindex1 = intindex1+len(string1)
            intindex2 = stringSubject.find(string2,intindex1)
            if(intindex2 != -1):
                subsequence=stringSubject[intindex1:intindex2]
                MyList.append(subsequence)
                intstart=intindex2+len(string2)
            else:
                continueloop=0
        else:
            continueloop=0
    return MyList


#Usage Example
mystring="s123y123o123pp123y6"
List = GetListOfSubstrings(mystring,"1","y68")
for x in range(0, len(List)):
               print(List[x])
output:


mystring="s123y123o123pp123y6"
List = GetListOfSubstrings(mystring,"1","3")
for x in range(0, len(List)):
              print(List[x])
output:
    2
    2
    2
    2

mystring="s123y123o123pp123y6"
List = GetListOfSubstrings(mystring,"1","y")
for x in range(0, len(List)):
               print(List[x])
output:
23
23o123pp123

Answer 13

这些解决方案假定起始字符串和最终字符串不同。这是我在初始指标和最终指标相同时用于整个文件的解决方案，假设使用 readlines() 读取整个文件：

def extractstring(line,flag='$'):
    if flag in line: # $ is the flag
        dex1=line.index(flag)
        subline=line[dex1+1:-1] #leave out flag (+1) to end of line
        dex2=subline.index(flag)
        string=subline[0:dex2].strip() #does not include last flag, strip whitespace
    return(string)

例子：

lines=['asdf 1qr3 qtqay 45q at $A NEWT?$ asdfa afeasd',
    'afafoaltat $I GOT BETTER!$ derpity derp derp']
for line in lines:
    string=extractstring(line,flag='$')
    print(string)

给予：

A NEWT?
I GOT BETTER!

Answer 14

这基本上是 cji 的回答 - 2010 年 7 月 30 日 5:58。 我更改了 try except 结构，以便更清楚地了解导致异常的原因。

def find_between( inputStr, firstSubstr, lastSubstr ):
'''
find between firstSubstr and lastSubstr in inputStr  STARTING FROM THE LEFT
    http://stackoverflow.com/questions/3368969/find-string-between-two-substrings
        above also has a func that does this FROM THE RIGHT   
'''
start, end = (-1,-1)
try:
    start = inputStr.index( firstSubstr ) + len( firstSubstr )
except ValueError:
    print '    ValueError: ',
    print "firstSubstr=%s  -  "%( firstSubstr ), 
    print sys.exc_info()[1]

try:
    end = inputStr.index( lastSubstr, start )       
except ValueError:
    print '    ValueError: ',
    print "lastSubstr=%s  -  "%( lastSubstr ), 
    print sys.exc_info()[1]

return inputStr[start:end]    

Answer 15

我的方法是做类似的事情，

find index of start string in s => i
find index of end string in s => j

substring = substring(i+len(start) to j-1)

Answer 16

这是我之前发布的code snippet in Daniweb：

# picking up piece of string between separators
# function using partition, like partition, but drops the separators
def between(left,right,s):
    before,_,a = s.partition(left)
    a,_,after = a.partition(right)
    return before,a,after

s = "bla bla blaa <a>data</a> lsdjfasdjöf (important notice) 'Daniweb forum' tcha tcha tchaa"
print between('<a>','</a>',s)
print between('(',')',s)
print between("'","'",s)

""" Output:
('bla bla blaa ', 'data', " lsdjfasdj\xc3\xb6f (important notice) 'Daniweb forum' tcha tcha tchaa")
('bla bla blaa <a>data</a> lsdjfasdj\xc3\xb6f ', 'important notice', " 'Daniweb forum' tcha tcha tchaa")
('bla bla blaa <a>data</a> lsdjfasdj\xc3\xb6f (important notice) ', 'Daniweb forum', ' tcha tcha tchaa')
"""

Answer 17

from timeit import timeit
from re import search, DOTALL


def partition_find(string, start, end):
    return string.partition(start)[2].rpartition(end)[0]


def re_find(string, start, end):
    # applying re.escape to start and end would be safer
    return search(start + '(.*)' + end, string, DOTALL).group(1)


def index_find(string, start, end):
    return string[string.find(start) + len(start):string.rfind(end)]


# The wikitext of "Alan Turing law" article form English Wikipeida
# https://en.wikipedia.org/w/index.php?title=Alan_Turing_law&action=edit&oldid=763725886
string = """..."""
start = '==Proposals=='
end = '==Rival bills=='

assert index_find(string, start, end) \
       == partition_find(string, start, end) \
       == re_find(string, start, end)

print('index_find', timeit(
    'index_find(string, start, end)',
    globals=globals(),
    number=100_000,
))

print('partition_find', timeit(
    'partition_find(string, start, end)',
    globals=globals(),
    number=100_000,
))

print('re_find', timeit(
    're_find(string, start, end)',
    globals=globals(),
    number=100_000,
))

结果：

index_find 0.35047444528454114
partition_find 0.5327825636197754
re_find 7.552149639286381

在此示例中，re_find 比 index_find 慢了近 20 倍。

Answer 18

使用来自不同电子邮件平台的分隔符解析文本提出了这个问题的更大版本。他们通常有一个开始和一个停止。通配符的分隔符一直阻塞正则表达式。此处和其他地方都提到了拆分的问题 - 哎呀，分隔符消失了。我想到使用 replace() 给 split() 其他东西来消费。代码块：

nuke = '~~~'
start = '|*'
stop = '*|'
julien = (textIn.replace(start,nuke + start).replace(stop,stop + nuke).split(nuke))
keep = [chunk for chunk in julien if start in chunk and stop in chunk]
logging.info('keep: %s',keep)

Answer 19

从 Nikolaus Gradwohl 的进一步回答中，我需要从下面的文件内容（文件名： docker-compose.yml）：

    version: '3.1'
services:
  ui:
    image: repo-pkg.dev.io:21/website/ui:0.0.2-QA1
    #network_mode: host
    ports:
      - 443:9999
    ulimits:
      nofile:test

这就是它对我的工作方式（python 脚本）：

import re, sys

f = open('docker-compose.yml', 'r')
lines = f.read()
result = re.search('ui:(.*)-', lines)
print result.group(1)


Result:
0.0.2

Answer 20

这对我来说似乎更直接：

import re

s = 'asdf=5;iwantthis123jasd'
x= re.search('iwantthis',s)
print(s[x.start():x.end()])