How best to write this if-else?

[Mick]

you could try bound methods

The following is probably not the nicest of ways but it shows you how to
keep the abstraction in the loop.

import re

def printRes1(m):
    print m.group(1)

def printRes2(m):
    print m.group(1)
    print m.group(2)

re1 = (re.compile("demo.*"),printRes1)
re2 = (re.compile("testing(.*)for(.*)"),printRes2)

all = [re1,re2]

str = "testing things for now"

for i in all:
    ma = i[0].match(str)
    if ma:
        i[1](ma)


Mick


-----Original Message-----
From: python-list-admin at python.org
[mailto:python-list-admin at python.org]On Behalf Of Roy Smith
Sent: Sunday, April 22, 2001 9:02 AM
To: python-list at python.org
Subject: Re: How best to write this if-else?


"Ingo Wilken" <Ingo.Wilken at Informatik.Uni-Oldenburg.DE> wrote:
> for e in [e1, e2, e3]:
>     m = e.match(line)
>     if m:
>         text = m.group(1)
>         break
> else:
>     no match found

OK, that's pretty neat, but I realize now that my example was
unintentionally misleading.  The problem is not quite as regular as I made
it out to be.

What if I want to execute different code depending on which expression I
matched?  Something along the lines of (pseudocode):

if (m = e1.match(line)):
   text1 = m.group(1)
   do_complicated_processing (text1)
elif (m = e2.match(line)):
   text1 = m.group(1)
   text2 = m.group(2)
   print text1, text2
elif (m = e3.match(line)):
   return
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