Replace regex in python

Question

I have a text like this:

@110605!~~!Abhay_f!~~!Abhay%20Raj%20Fac!>>! @138547!~~!testvarun!~~!Varun%20Test%20User!>>! @136588!~~!jitendra_pathak!~~!Jitendra%20Pathak!>>! #gffj #varun okjjbbbd

and in this string I want to replace !~~! with "".

I am using a template tag and I used this method but it didn't work for me:

@register.filter("metioned_user_text_encode")
def metioned_user_text_encode(string, args):
    search = args.split(args[0])[1]
    replace = args.split(args[0])[2]

    return re.sub(search, replace, string)

In my template:-

result_data_for_editing.newsText | metioned_user_text_encode:"/l(u+)pin/m\1gen"

Did you consider template strings are not [raw strings](https://stackoverflow.com/a/7262918/1600649)? — , Jun 23 '17 at 10:15

John Moutafis · Accepted Answer · 2017-07-06T09:00:09.037

Unless I completely misunderstood your intentions, you are asking one thing but your code shows a different expected behavior than the one you ask for.
I will try to answer both cases:

Case 1:

You expect to get a multitude of separators like !~~! or !>>! etc., and you need a regex to replace all of them with an empty string ('').

Try the following:

import re

SEPARATORS = [
    '!~~!', 
    '!>>!', 
    other separators...
]

@register.filter("metioned_user_text_encode")
def metioned_user_text_encode(string):       
    return re.sub('|'.join(SEPARATORS), '', string)

Explanation:

The | regex operator ensures that our pattern will try to match every separator given in SEPARATORS with the given string (s).
The re.sub() method will return the string with every pattern matching any of our SEPARATORS, replaced by the empty string.

Case 2:

You will receive as arguments a series of separators and an equally sized series of replacements for those separators. In that case, try:

@register.filter("metioned_user_text_encode")
def metioned_user_text_encode(string, args):
    returned_string = string
    search = args.split(args[0])[1]
    replace = args.split(args[0])[2]

    for i in range(len(search)):
        returned_string = re.sub(search[i], replace[i], returned_string)

    return returned_string

Explanation:

The for loop will traverse the search and replace lists for every separator and the corresponding replacement.
The re.sub() method will return in each iteration, the returned_string with the search[i] separator replaced by the replace[i] substitute.

Good luck :)

Akshay Apte · Answer 2 · 2017-06-23T07:05:40.387

0

you can use the fileinput module which makes find and replace easy

import fileinput

with fileinput.FileInput(inputFile, inplace=True, backup='.back') as f:
    for line in f:
        print(line.replace(textToSearch, textToReplace), end='')

Or you can use the replace method

str.replace("!~~!", "")

edited Jun 23 '17 at 07:05

answered Jun 23 '17 at 07:03

Akshay Apte

1,539
9
24

`str.replace("!~~!", "")` should work. where `str` is the name of the string. – Akshay Apte Jun 23 '17 at 07:07

score 0 · Answer 3 · answered Jun 23 '17 at 07:36

0

try this:

a = '@110605!~~!Abhay_f!~~!Abhay%20Raj%20Fac!>>! @138547!~~!testvarun!~~!'
b = a.replace(!~~!,'""')
print(b)

answered Jun 23 '17 at 07:36

Drako

773
10
22

Replace regex in python

3 Answers3