How to create a regex that matches numbers that have 3 or less than 3 digits?

Question

I want to delete any numbers that have 3 or less than 3 digits. Can someone please help me with a regex that does this?

Currently, my code removes all the numbers it finds.

     # Cleans Numbers
     def cleanNumbers(stringToClean):
       stringToClean = re.sub(r'[0-9]*', r'', stringToClean)

       print 'String after cleaning : %s' %stringToClean

       return stringToClean

Numbers will be surrounded by space. Example string I pass into the function :

connection breaks on Win8 client after a while. [persistence] 123 1 22 333 4444 554665 645fdgf45 ds3434 457870978934787843 345342kl

I call the above function as follows :

# Main function, calls other functions          
def main():

   # Parsing the input query
   searchQuery = open('input.txt', 'r').read()
   print 'Input query : %s' %searchQuery

   # Cleaning the input query
   string = CleanUpText.cleanNumbers(searchQuery)

Answer 1

I have corrected the question, '3 or less than 3'

Given that, it should be as simple as: \b\d{1,3}\b

Answer 2

You could use a regex like this

r'\b[0-9]{1,2}\b'

Edit: Sorry wrote my answer to fast without really thinking. You have to use boundries so you don't capture 3456 for example

Answer 3

\b[0-9]{1,3}\b finds blocks of digits that have up to three digits.

Answer 4

re.sub(r'[0-9]{,3}',r'',stringToClean)