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Date: Tue, 17 May 2011 11:31:19 -0700
Subject: Re: if statement on lenght of a list
From: Chris Rebert <clp2@rebertia.com>
To: Joe Leonardo <joe.leonardo@datalogix.com>
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On Tue, May 17, 2011 at 11:02 AM, Joe Leonardo
<joe.leonardo@datalogix.com> wrote:
>
> Hey all,
>
> Totally baffled by this=E2=80=A6maybe I need a nap. Writing a small funct=
ion to reject input that is not a list of 19 fields.
>
> def breakLine(value):
> =C2=A0=C2=A0=C2=A0 if value.__class__() !=3D [] and value.__len__() !=3D =
19:
> =C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0 print 'You must pass a list th=
at contains 19 fields.'
> =C2=A0=C2=A0=C2=A0 else:
> =C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0 print 'YAY!'
>
> If I pass:
> breakLine([])
>
> I get:
> YAY!
>
> I expect:
> You must pass a list that contains 19 fields.

Your test should use `or` as opposed to `and`. Since you're indeed
passing in a list, the first part of your condition is False; since
you used `and`, this makes the entire condition False. Therefore, the
else clause ("YAY!") gets executed.

Also, your test is written quite strangely. One would more normally
and straightforwardly write it as:
    if not isinstance(value, list) or len(value) !=3D 19:

Cheers,
Chris
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