newbie question: if var1 == var2:

John Machin sjmachin at lexicon.net
Fri Dec 12 02:59:38 CET 2008


On Dec 12, 11:39 am, MRAB <goo... at mrabarnett.plus.com> wrote:
> Jason Scheirer wrote:
> > On Dec 11, 3:49 pm, John Machin <sjmac... at lexicon.net> wrote:
> >> On Dec 12, 10:31 am, "Rhodri James" <rho... at wildebst.demon.co.uk>
> >> wrote:
>
> >>> On Thu, 11 Dec 2008 19:49:23 -0000, Steve Holden <st... at holdenweb.com>  
> >>> wrote:
> >>>> Kirk Strauser wrote:
> >>>>> At 2008-11-29T04:02:11Z, Mel <mwil... at the-wire.com> writes:
> >>>>>> You could try
> >>>>>> for item in fname:
> >>>>>>     item = item.strip()
> >>>>> This is one case where I really miss Perl's "chomp" function.  It  
> >>>>> removes a
> >>>>> trailing newline and nothing else, so you don't have to worry about  
> >>>>> losing
> >>>>> leading or trailing spaces if those are important to you.
> >>>> ... and it's so hard to write
> >>>>      item = item[:-1]
> >>> Tsk.  That would be "chop".  "chomp" would be
> >>>      if item[-1] == '\n':
> >>>          item = item[:-1]
> >> Better:
> >> if item and item[-1] == '\n':
> >>     return item[:-1]
> >> return item
>
> > Best:
>
> > return item \
> >        if not (item and item.endswith('\n')) \
> >        else item[:-1]
>
> > Though really you should be using item.rstrip()
>
> Why not just:
>
>      item[:-1] if item.endswith('\n') else item

Some possible reasons:
* because you might be supporting old versions of Python (my offering
runs on 1.5)
* because the "<true_value> if <condition> else <false_value>" syntax
gives you the screaming dry Edgar Britts
* because you'd prefer not to have the overhead of a method lookup and
method call



More information about the Python-list mailing list