In the following text, I want to remove everything inside the parenthesis including number and string. I use the following syntax but I got result of 22701 instead of 2270. What would be a way to show 2270 only using re.sub? Thanks

import regex as re
import numpy as np
import pandas as pd

text = "2270 (1st xyz)"
text_new = re.sub(r"[a-zA-Z()\s]","",text)
text_new

CodePudding user response：

Does the text always follow the same pattern? Try:

import re
import numpy as np
import pandas as pd

text = "2270 (1st xyz)"
text_new = re.sub(r"\s\([^)]*\)","",text)
print(text_new)

Output:

2270

CodePudding user response：

Simply use the regex pattern \(.*?\):

import re

text = "2270 (1st xyz)"
text_new = re.sub("\(.*?\)", "", text)
print(text_new)

Output:

2270 

Explanation on the pattern \(.*?\):

• The \ behind each parenthesis is to tell re to treat the parenthesis as a regular character, as they are by default special characters in re.
• The . matches any character except the newline character.
• The * matches zero or more occurrences of the pattern immediately specified before the *.
• The ? tells re to match as little text as possible, thus making it non-greedy.

Note the trailing space in the output. To remove it, simply add it to the pattern:

import re

text = "2270 (1st xyz)"
text_new = re.sub(" \(.*?\)", "", text)
print(text_new)

Output:

2270