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.

stackoverflow.com/questions/40950791/…

Rashik
– Rashik

2020-05-11 23:07:49 +00:00
Commented May 11, 2020 at 23:07 — Rashik
– Rashik, Commented May 11, 2020 at 23:07
p = re.findall("(?<=name:[\s]{1})[a-zA-Z]+",string)

Zhd Zilin
– Zhd Zilin

2020-05-11 23:08:43 +00:00
Commented May 11, 2020 at 23:08 — Zhd Zilin
– Zhd Zilin, Commented May 11, 2020 at 23:08

Mustafa Aydın · Accepted Answer · 2020-05-11 23:22:21Z

1

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']

answered May 11, 2020 at 23:22

Mustafa Aydın

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Comments

Georgina Skibinski · Accepted Answer · 2020-05-11 23:08:33Z

1

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']

answered May 11, 2020 at 23:08

Georgina Skibinski

13.5k2 gold badges16 silver badges44 bronze badges

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