CODEWITHSUNDEEP
pythonregexstring
BoltzmannBrain
BoltzmannBrain

Reputation: 5392

Python - remove parts of a string

I have many fill-in-the-blank sentences in strings,

e.g. "6d) We took no [pains] to hide it ."

How can I efficiently parse this string (in Python) to be 

"We took no to hide it"?

I also would like to be able to store the word in brackets (e.g. "pains") in a list for use later. I think the regex module could be better than Python string operations like split().

Upvotes: 0

Views: 1172

Answers (5)

Antoine Chkaiban
Antoine Chkaiban

Reputation: 1

import re
m = re.search(".*\) (.*)\[.*\] (.*)","6d) We took no [pains] to hide it .")
if m:
    g = m.groups()
    print g[0] + g[1]

Output :

We took no to hide it .

Upvotes: 0

tbraun89
tbraun89

Reputation: 2234

For your example you could use this regex:

(.*\))(.+)\[(.+)\](.+)

You will get four groups that you can use to create your resulting string and save the 3. group for later use:

  1. 6d)
  2. We took no
  3. pains
  4. to hide it .

I used .+ here because I don't know if your strings always look like your example. You can change the .+ to alphanumeric or sth. more special to your case.

import re

s = '6d) We took no [pains] to hide it .'
m = re.search(r"(.*\))(.+)\[(.+)\](.+)", s)

print(m.group(2) + m.group(4)) # "We took no  to hide it ."
print(m.group(3))              # pains

Upvotes: 0

Joran Beasley
Joran Beasley

Reputation: 113940

just for fun (to do the gather and substitution in one iteration)

matches = []
def subber(m):
    matches.append(m.groups()[0])
    return ""

new_text = re.sub("\[(.*?)\]",subber,s)
print new_text
print matches

Upvotes: 2

letsc
letsc

Reputation: 2567

import re

s = 'this is [test] string'
m = re.search(r"\[([A-Za-z0-9_]+)\]", s)
print m.group(1)

Output

'test'

Upvotes: 1

burhan
burhan

Reputation: 924

This will give you all the words inside the brackets.

import re
s="6d) We took no [pains] to hide it ."
matches = re.findall('\[(.*?)\]', s)

Then you can run this to remove all bracketed words.

re.sub('\[(.*?)\]', '', s)

Upvotes: 3

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