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/bin/env python2.7
import re
line = "STARC-1.1.1.5 ConsCase WARNING Warning"
m = re.search('^[^\S]*?',line)
if m:
print m.group(0)
It seems to be printing the whitespace instead of STARC-1.1.1.5
So far as I understand it, this regular expression is saying: At the start of the line, find a set of nonwhitespace characters, don't be greedy
I was pretty sure this would work, the documentation says I can use /S to match whitespace in [], so i'm not sure where the issue is.
Now, I know, I know this probably looks weird, why aren't I using some other function to do this? Well, there's more than one way to skin a cat and i'm still getting the hang of regular expressions in Python so I'd like to know how I can use re.search to extract this field in this fashion.
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.
\s
matches a whitespace character.
\S
matches a non-whitespace character.
[...]
matches a character in the set ...
.
[^...]
matches a character not in the set ...
.
[^\S]
matches a character that is not a non-whitespace character, i.e. it matches a whitespace character.
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