Regular expression for matching non-whitespace in Python

Question

I want to use re.search to extract the first set of non-whitespace characters. I have the following pseudoscript that recreates my problem:

#!/usr/b         


        

Answer 1

The [^\S] is a negated character class that is equal to \s (whitespace pattern). The *? is a lazy quantifier that matches zero or more characters, but as few as possible, and when used at the end of the pattern never actually matches any characters.

Replace you m = re.search('^[^\S]*?',line) line with

m = re.match(r'\S+',line)

or - if you want to also allow an empty string match:

m = re.match(r'\S*',line)

The re.match method anchors the pattern at the start of the string. With re.search, you need to keep the ^ anchor at the start of the pattern:

m = re.search(r'^\S+',line)

See the Python demo:

import re
line = "STARC-1.1.1.5             ConsCase    WARNING    Warning"
m = re.search('^\S+',line)
if m:
    print m.group(0)
# => STARC-1.1.1.5

However, here, in this case, you may just use a mere split():

res = line.split() 
print(res[0])

See another Python demo.