Re: regex recursive matching (regex 2015.07.19)

[MRAB]

On 2015-08-18 15:25, Neal Becker wrote:

Trying regex 2015.07.19

I'd like to match recursive parenthesized expressions, with groups such that
'(a(b)c)'

would give
group(0) -> '(a(b)c)'
group(1) -> '(b)'

but that's not what I get

import regex

#r = r'\((?>[^()]|(?R))*\)'
r = r'\(([^()]|(?R))*\)'
#r = r'\((?:[^()]|(?R))*\)'
m = regex.match (r, '(a(b)c)')

  m.groups()
Out[28]: ('c',)

You can't capture them into different groups in the general case; it
won't create capture groups dynamically.

Capture into 1 group and then use the .captures method:

import regex

r = r'(\([^()]*(?:(?R)[^()]*)*\))'
m = regex.match(r, '(a(b)c)')

print(m.captures(1))

--
https://mail.python.org/mailman/listinfo/python-list