Regular expression: match start or whitespace

Question

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?

Answer 1

\b is word boundary, which can be a white space, the beginning of a line or a non-alphanumeric symbol ("\"GBP\"").

Answer 2

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...

Answer 3

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.

Answer 4

this replaces GBP if its preceded by start of string or a word boundary (which start of 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 ?

Answer 5

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'

Answer 6

I think you're looking for '(^|\W)GBP([\W\d])'

Answer 7

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.