简体   繁体   中英

splitting on a regular expression

I have a string that I want to split using a regular expression.

Input:

S1:1- first split begins.s2:1- first split ends.s1:2-second split begins.S2:2-second split ends,S1:3-third split begins.S2:3-third split ends.

Output:should be a list containing the expression on which we split

[S1:1-first split begins,s2:1-first split ends,S1:2-second split begins,S2:2-second split ends,S1:3-third split begins.....]

I want to split on pattern [s/S][1/2]:[0-9][0-9]
This is what I have, but it gives me an extra line and removes the regex on which I split.

import re
text="""S1:1- first Split begins.continue the sentence
       s2:1- first split ends
       s1:2-second split begins
       S2:2-second split ends
       S1:3-third split begins
       S2:3-third split ends """
output=re.split("[Ss][12]:[0-9]*", text)

I'm not quite sure I understand where you want to split this input, but if you want the text that you're splitting on to be included in the output then you need to have a capturing pattern:

 re.split("([Ss][1-2]:[0-9]-)\s*", text)

result:

['',
 'S1:1-',
 'first Split begins.continue the sentence\n       ',
 's2:1-',
 'first split ends\n       ',
 's1:2-',
 'second split begins\n       ',
 'S2:2-',
 'second split ends\n       ',
 'S1:3-',
 'third split begins\n       ',
 'S2:3-',
 'third split ends ']

Try to use Positive Lookahead (?= CODE ) in Regex to accomplish this in order to preserve your regex. Your regex will be something like this:

\s(?=[sS][12]:[0-9])

Complete Code:

import re
text="""S1:1- first Split begins.continue the sentence
       s2:1- first split ends
       s1:2-second split begins
       S2:2-second split ends
       S1:3-third split begins
       S2:3-third split ends """
output=re.split("\s(?=[sS][12]:[0-9])", text)

Outcome:

['S1:1- first Split begins.continue the sentence\n ', 's2:1- first split ends\n ', 's1:2-second split begins\n ', 'S2:2-second split ends\n ', 'S1:3-third split begins\n ', 'S2:3-third split ends ']

The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM