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
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!
其他回答
要提取STRING，请尝试:
myString = '123STRINGabc'
startString = '123'
endString = 'abc'
mySubString=myString[myString.find(startString)+len(startString):myString.find(endString)]
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。
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]
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()])
您可以简单地使用这段代码或复制下面的函数。全都整齐地排在一条线上。
def substring(whole, sub1, sub2):
return whole[whole.index(sub1) : whole.index(sub2)]
如果按照如下方式运行该函数。
print(substring("5+(5*2)+2", "(", "("))
你可能会得到这样的输出:
(5*2
而不是
5*2
如果您希望在输出的末尾有子字符串，代码必须如下所示。
return whole[whole.index(sub1) : whole.index(sub2) + 1]
但如果不希望子字符串在末尾，则+1必须在第一个值上。
return whole[whole.index(sub1) + 1 : whole.index(sub2)]
