Python Get Text in Quotes After Pattern

Question

I want to get the text in quotations after the occurrence of the pattern "name".

A sample string is: info = "name: joe", "name: jerry", "name: kate"

Here is what I'm doing:

import re

string = 'info = "name: joe", "name: jerry", "name: kate"'
array = re.findall(r'"(.*?)"', string)
for x in array:
       x = x.replace(" ","") #remove spaces because there might be space before colon occasionally 
       print(x.split("name:")[1])

The output prints:

joe
jerry
kate

I am just wondering if there is any easier way to do this and return the same output than what I did.

https://stackoverflow.com/questions/40950791/remove-quotes-from-string-in-python — Rashik, May 11 '20 at 23:07

score 1 · Answer 1 · answered May 11 '20 at 23:08

Try:

res=list(re.findall(r'\"name: ([^\"]+)"', string))

print(res)

2 important things - you extract only a group, hence the square brackets, and you define a name as set of one or more characters, that aren't double quotes ", since regex is on default greedy (i.e. matches the longest match only).

Outputs:

['joe', 'jerry', 'kate']

score 1 · Accepted Answer · answered May 11 '20 at 23:22

You can embed the name: pattern in the regex, account for the occasional space before the colon via \s* and also match one or more spaces after colon and the actual name (where you are splitting from):

re.findall(r'"name\s*:\s+(.*?)"', string)
# ['joe', 'jerry', 'kate']

Python Get Text in Quotes After Pattern

2 Answers2