Regular expression to find and remove a sub-string or its variants from a string

Question

I have text of the type:

"Choice values selected: Option 1, or Option 2, or Option 3"
"Choice value selected: Option 1, or Option 2, or Option 3"
"Choice value selected = Option 1 , or Option 2, or Option 3"

I need to extract everything that is after the : or =

I have tried to go about it this way:

import regex as re
r = re.compile(r'Choice(.+?)selected')
r.split(str)

I don't know how to capture the : or =

No issues, I have done the same in my answer below! Please check it and accept it if it helped you :) @alannaC — Devesh Kumar Singh
– Devesh Kumar Singh, Commented May 2, 2019 at 10:38

Devesh Kumar Singh · Accepted Answer · 2019-05-02 10:32:18Z

4

You don't need to use regex, just use re.split to split both on : and =

li = ["Choice values selected: Option 1, or Option 2, or Option 3", "Choice value selected: Option 1, or Option 2, or Option 3",
      "Choice value selected = Option 1 , or Option 2, or Option 3"]

import re
for item in li:
    #Split on : and =, get the last element from list and strip it
    print(re.split(':|=',item)[1].strip())

The output will be

Option 1, or Option 2, or Option 3
Option 1, or Option 2, or Option 3
Option 1 , or Option 2, or Option 3

answered May 2, 2019 at 10:32

Devesh Kumar Singh

20.5k5 gold badges25 silver badges43 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

Pushpesh Kumar Rajwanshi · Accepted Answer · 2019-05-02 10:33:58Z

You can use this regex,

[:=]\s*(.*)

And get your value from group1

This regex starts by capturing either : or = and then optionally \s* matches optional space and then (.*) captures the remaining text in the line and captures in group1

Regex Demo

Python code,

import regex as re

arr = ['Choice values selected: Option 1, or Option 2, or Option 3','Choice value selected: Option 1, or Option 2, or Option 3','Choice value selected = Option 1 , or Option 2, or Option 3']

for s in arr:
 m = re.search(r'[:=]\s*(.*)', s)
 if m:
  print(s, '-->', m.group(1))

Output,

Choice values selected: Option 1, or Option 2, or Option 3 --> Option 1, or Option 2, or Option 3
Choice value selected: Option 1, or Option 2, or Option 3 --> Option 1, or Option 2, or Option 3
Choice value selected = Option 1 , or Option 2, or Option 3 --> Option 1 , or Option 2, or Option 3

Also, in case you want to use re.split then you can split it using [=:] regex which represents either = or :

import regex as re
arr = ['Choice values selected: Option 1, or Option 2, or Option 3','Choice value selected: Option 1, or Option 2, or Option 3','Choice value selected = Option 1 , or Option 2, or Option 3']

for s in arr:
 r = re.compile(r'[:=]')
 print(r.split(s)[1])

Output,

 Option 1, or Option 2, or Option 3
 Option 1, or Option 2, or Option 3
 Option 1 , or Option 2, or Option 3

Collectives™ on Stack Overflow

Regular expression to find and remove a sub-string or its variants from a string

2 Answers 2

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related