Regular Expression For Printing Integers Within Brackets

First time ever using regular expressions and can't get it working although there's quite a few examples in stackoverflow already. How can I extract integers which are in a string

Solution 1:

If you need to fail the match in [ +-34 ] (i.e. if you needn't extract a negative number if there is a + before it) you will need to use

\[\s*(?:\+|(-))?(\d+)\s*]

and when getting a match, concat the Group 1 and Group 2 values. See this regex demo.

Details

• \[ - a [ char
• \s* - 0+ whitespaces
• \+? - an optional + char
• (-?\d+) - Capturing group 1 (the actual output of re.findall): an optional - and 1+ digits
• \s* - 0+ whitespaces
• ] - a ] char.

In Python,

import re
text = "dijdi[d43]     d5[55++][ 43]  [+32]dm dij [    -99]x"
numbers_text = [f"{x}{y}"for x, y in re.findall(r'\[\s*(?:\+|(-))?(\d+)\s*]', text)]
numbers = list(map(int, numbers_text))

# => [43, 32, -99] for both

Solution 2:

If you've not done so I suggest you switch to the PyPI regex module. Using it here with regex.findall and the following regular expression allows you to extract just what you need.

r'\[ *\+?\K-?\d+(?= *\])'

regex engine_¯\(ツ)/¯Python code

At the regex tester pass your cursor across the regex for details about individual tokens.

The regex engine performs the following operations.

\[       : match'['
\ *      : match0+ spaces
\+?      : optionally match'+'
\K       : forget everything matched so far and reset
           startofmatchtocurrent position
-?       : optionally match'-'
\d+      : match1+ digits
(?=*\]) : use positive lookahead to assert the last digit
         : matched is followed by0+ spaces then']'