我如何找到两个子字符串('123STRINGabc' -> '字符串')之间的字符串?
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
然而，这似乎非常低效且不符合python规则。有什么更好的方法来做这样的事情吗?
忘了说: 字符串可能不是以start和end开始和结束的。他们可能会有更多的字符之前和之后。
当前回答
从Nikolaus Gradwohl的答案进一步，我需要从下面的文件内容(文件名:docker- composition .yml)中获得版本号(即0.0.2)之间('ui:'和'-'):
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
这是它如何为我工作(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
其他回答
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
在这个例子中，Re_find几乎比index_find慢20倍。
字符串格式为Nikolaus Gradwohl的建议增加了一些灵活性。开始和结束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)
这些解决方案假设起始字符串和最终字符串是不同的。下面是当初始和最终指示符相同时，我用于整个文件的解决方案，假设整个文件是使用readlines()读取的:
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:
A NEWT?
I GOT BETTER!
只是把OP自己的解决方案转换成一个答案:
def find_between(s, start, end):
return (s.split(start))[1].split(end)[0]
这本质上是cji的答案——7月30日10日5:58。 我更改了try except结构，以便更清楚地说明导致异常的原因。
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]
