On 5/16/11 11:25 AM, Tracubik wrote:
pls help me fixing this:
import re
s = "linka la baba"
re_s = re.compile(r'(link|l)a' , re.IGNORECASE)
print re_s.findall(s)
output:
['link', 'l']
why?
i want my re_s to find linka and la, he just find link and l and forget
about the ending a.
can anyone help me? trying the regular expression in redemo.py (program
provided with python to explore the use of regular expression) i get what
i want, so i guess re_s is ok, but it still fail...
why?
The parentheses () create a capturing group, which specifies that the contents
of the group should be extracted. See the "(...)" entry here:
http://docs.python.org/library/re#regular-expression-syntax
You can use the non-capturing version of parentheses if you want to just isolate
the | from affecting the rest of the regex:
"""
(?:...) A non-capturing version of regular parentheses. Matches whatever
regular expression is inside the parentheses, but the substring matched by the
group cannot be retrieved after performing a match or referenced later in the
pattern.
"""
[~]
|1> import re
[~]
|2> s = "linka la baba"
[~]
|3> re_s = re.compile(r'(?:link|l)a' , re.IGNORECASE)
[~]
|4> print re_s.findall(s)
['linka', 'la']
--
Robert Kern
"I have come to believe that the whole world is an enigma, a harmless enigma
that is made terrible by our own mad attempt to interpret it as though it had
an underlying truth."
-- Umberto Eco
--
http://mail.python.org/mailman/listinfo/python-list
