Regular expression: match start or whitespace
PythonRegexPython Problem Overview
Can a regular expression match whitespace or the start of a string?
I'm trying to replace currency the abbreviation GBP with a £ symbol. I could just match anything starting GBP, but I'd like to be a bit more conservative, and look for certain delimiters around it.
>>> import re
>>> text = u'GBP 5 Off when you spend GBP75.00'
>>> re.sub(ur'GBP([\W\d])', ur'£\g<1>', text) # matches GBP with any prefix
u'\xa3 5 Off when you spend \xa375.00'
>>> re.sub(ur'^GBP([\W\d])', ur'£\g<1>', text) # matches at start only
u'\xa3 5 Off when you spend GBP75.00'
>>> re.sub(ur'(\W)GBP([\W\d])', ur'\g<1>£\g<2>', text) # matches whitespace prefix only
u'GBP 5 Off when you spend \xa375.00'
Can I do both of the latter examples at the same time?
Python Solutions
Solution 1 - Python
Use the OR "|
" operator:
>>> re.sub(r'(^|\W)GBP([\W\d])', u'\g<1>£\g<2>', text)
u'\xa3 5 Off when you spend \xa375.00'
Solution 2 - Python
\b
is word boundary, which can be a white space, the beginning of a line or a non-alphanumeric symbol (\bGBP\b
).
Solution 3 - Python
This replaces GBP if it's preceded by the start of a string or a word boundary (which the start of a string already is), and after GBP comes a numeric value or a word boundary:
re.sub(u'\bGBP(?=\b|\d)', u'£', text)
This removes the need for any unnecessary backreferencing by using a lookahead. Inclusive enough?
Solution 4 - Python
A left-hand whitespace boundary - a position in the string that is either a string start or right after a whitespace character - can be expressed with
(?<!\S) # A negative lookbehind requiring no non-whitespace char immediately to the left of the current position
(?<=\s|^) # A positive lookbehind requiring a whitespace or start of string immediately to the left of the current position
(?:\s|^) # A non-capturing group matching either a whitespace or start of string
(\s|^) # A capturing group matching either a whitespace or start of string
See a regex demo. Python 3 demo:
import re
rx = r'(?<!\S)GBP([\W\d])'
text = 'GBP 5 Off when you spend GBP75.00'
print( re.sub(rx, r'£\1', text) )
# => £ 5 Off when you spend £75.00
Note you may use \1
instead of \g<1>
in the replacement pattern since there is no need in an unambiguous backreference when it is not followed with a digit.
BONUS: A right-hand whitespace boundary can be expressed with the following patterns:
(?!\S) # A negative lookahead requiring no non-whitespace char immediately to the right of the current position
(?=\s|$) # A positive lookahead requiring a whitespace or end of string immediately to the right of the current position
(?:\s|$) # A non-capturing group matching either a whitespace or end of string
(\s|$) # A capturing group matching either a whitespace or end of string
Solution 5 - Python
I think you're looking for '(^|\W)GBP([\W\d])'
Solution 6 - Python
Yes, why not?
re.sub(u'^\W*GBP...
matches the start of the string, 0 or more whitespaces, then GBP...
edit: Oh, I think you want alternation, use the |
:
re.sub(u'(^|\W)GBP...
Solution 7 - Python
You can always trim leading and trailing whitespace from the token before you search if it's not a matching/grouping situation that requires the full line.
Solution 8 - Python
It works in Perl:
$text = 'GBP 5 off when you spend GBP75';
$text =~ s/(\W|^)GBP([\W\d])/$1\$$2/g;
printf "$text\n";
The output is:
$ 5 off when you spend $75
Note that I stipulated that the match should be global, to get all occurrences.