查找两个子字符串之间的字符串[重复]

2020/10/31 23:32 · python ·  · 0评论

如何找到两个子字符串（'123STRINGabc' -> 'STRING'）之间的字符串？

我当前的方法是这样的：

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

但是，这似乎效率很低而且不合Python。什么是做这样的更好的方法？

忘了提：该字符串可能无法启动，并最终start和end。他们前后可能有更多字符。

import re

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

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" )

给出：

123STRING
STRINGabc

我认为应该注意-根据您需要的行为，可以混用index或rindex调用或使用上述版本之一（等效于regex(.*)和(.*?)groups）。

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

给 

iwantthis

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

字符串格式化为Nikolaus Gradwohl建议的内容增加了一些灵活性。start并且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)

只需将OP自身的解决方案转换为答案即可：

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

如果您不想导入任何内容，请尝试使用字符串方法.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)])

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法

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

要么

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

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：here0，here1，here2

regex更好，但需要额外的lib，您可能只想使用python

要提取STRING，请尝试：

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

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

这些解决方案假定起始字符串和最终字符串不同。这是当初始和最终指示符相同时我用于整个文件的一种解决方案，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：

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)

给出：

A NEWT?
I GOT BETTER!

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)]

如果按以下方式运行该功能。

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

您可能会留下输出：

(5*2

而不是

5*2

如果要在输出末尾包含子字符串，则代码必须如下所示。

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

但是，如果您不希望子字符串最后出现，则+1必须位于第一个值上。

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

这是我执行的函数，用于返回一个列表，其中包含在搜索的string1和string2之间的字符串。

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

这基本上是cji的答案-10年7月30日在5:58。我更改了tryexcept结构，以更清楚地了解引起异常的原因。

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]

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)

我之前在Daniweb中将其作为代码段发布：

# 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')
"""

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

re_find比index_find本示例慢了近20倍。

使用来自不同电子邮件平台的定界符来解析文本构成了此问题的更大版本。他们通常有一个开始和一个停止。通配符的分隔符使正则表达式令人窒息。拆分问题在这里和其他地方都提到过-哎呀，分隔符消失了。我想到了使用replace（）给split（）消费其他东西。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)

另外从尼古拉斯Gradwohl答案，我需要得到版本号（即0.0.2（“UI：”和“ - ”）之间），从下面的文件内容（文件名：泊坞窗-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

对我来说，这似乎更直接了：

import re

s = 'asdf=5;iwantthis123jasd'
x= re.search('iwantthis',s)
print(s[x.start():x.end()])
本文地址：http://python.askforanswer.com/chazhaolianggezizifuchuanzhijiandezifuchuanzhongfu.html
文章标签： ,   ,  
版权声明：本文为原创文章，版权归 admin 所有，欢迎分享本文，转载请保留出处！

文件下载

老薛主机终身7折优惠码boke112

上一篇：
下一篇：

评论已关闭！