Retrieve Exactly 1 Digit Using Regular Expression In Python

I want to print only ages that are less than 10. In this string, only the value 1 should be printed. Somehow, that is not happening. I used the following codes (using regular exp

Solution 1:

You are trying to match "any 1 number", but you want to match "any 1 number, not followed or preceded by another number".

One way to do that is to use lookarounds

re.findall(r'(?<![0-9])[0-9](?![0-9])', s5)

Possible lookarounds:

(?<!R)S   // negativelookbehind: match S thatisnotprecededby R
(?<=R)S   // positivelookbehind: match S thatisprecededby R
(?!R)S   // negativelookahead: match S thatisnotfollowedby R
(?=R)S   // positivelookahead: match S thatisfollowedby R

Maybe a simpler solution is to use a capturing group (). if regex in findall has one capturing group, it will return list of matches withing the group instead of whole matches:

re.findall(r'[^0-9]([0-9])[^0-9]', s5)

Also note that you can replace any 0-9 with \d - character group of numbers

Solution 2:

Try this :

k = re.findall('(?<!\S)\d(?!\S)', s5)
print(k)

This also works :

re.findall('(?<!\S)\d(?![^\s.,?!])', s5)

Solution 3:

import re

s = "The baby is 1 year old, Sri is 45 years old, Ann is 50 years old; their father, Sumo is 78 years old and their grandfather, Kris, is 100 years old"

m = re.findall('\d+',s)


for i in m:
    ifint(i)<10:
        print(i)