如何在python的字符串中的每个字符后添加换行符，例如“。[xxx]”

Question

I have the following string :

It reported the proportion of the edits made from America was 51% for the Wikipedia, and 25% for the simple Wikipedia.[142] The Wikimedia Foundation hopes to increase the number in the Global South to 37% by 2015.[143]

I am trying to replace every characters lik this .[xxx] with .[xxx] \\n ;

x are digits here

I am taking help from different stalk overflow answers; one such is :

Python insert a line break in a string after character "X"

Regex: match fullstop and one word in python

import re
str = "It reported the proportion of the edits made from America was 51% 
for the Wikipedia, and 25% for the simple Wikipedia.[142] The Wikimedia 
Foundation hopes to increase the number in the Global South to 37% by 
2015.[143] "
x = re.sub("\.\[[0-9]{2,5}\]\s", "\.\[[0-9]{2,5}\]\s\n",str)
print(x)

I expect the following output:

It reported the proportion of the edits made from America was 51% for the Wikipedia, and 25% for the simple Wikipedia.[142]                          
The Wikimedia Foundation hopes to increase the number in the Global South to 37% by 2015.[143]”

But I am getting:

It reported the proportion of the edits made from America was 51% for the Wikipedia, and 25% for the simple Wikipedia\\.\[[0-9]{2,5}\]\s   The Wikimedia Foundation hopes to increase the number in the Global South to 37% by 2015\\.\[[0-9]{2,5}\]\s

Answer 1

You probably want to use capturing groups and back-referrences in re.sub . You also don't need to escape the replacement string ( regex101 ):

import re
s = '''It reported the proportion of the edits made from America was 51% for the Wikipedia, and 25% for the simple Wikipedia.[142] The Wikimedia Foundation hopes to increase the number in the Global South to 37% by 2015.[143] '''
x = re.sub(r'\.\[([0-9]{2,5})\]\s', r'.[\1] \n', s)
print(x)

Prints:

It reported the proportion of the edits made from America was 51% for the Wikipedia, and 25% for the simple Wikipedia.[142] 
The Wikimedia Foundation hopes to increase the number in the Global South to 37% by 2015.[143] 

Answer 2

You may use

(\.\[[^][]*\])\s*

And replace this with \\1\\n , see a demo on regex101.com .

---

This reads

 ( \\.\\[ # ".[" literally [^][]* # neither "[" nor "]" 0+ times \\] # "]" literally )\\s* # consume whitespaces, eventually 

Answer 3

Use findall() to identify list of matching patterns. Then you can replace it with original string+'\\n'