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Date: Tue, 19 Jul 2011 23:56:51 -0700
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Subject: Re: Pythonic way with more than one max possible
From: Chris Rebert <clp2@rebertia.com>
To: CM <cmpython@gmail.com>
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On Tue, Jul 19, 2011 at 10:10 PM, CM <cmpython@gmail.com> wrote:
> On Jul 19, 11:17=C2=A0pm, CM <cmpyt...@gmail.com> wrote:
>> I have three items in a dict, like this:
>>
>> the_dict =3D {'a':1, 'b':2, 'c':3}
>>
>> but the vals could be anything. =C2=A0I want to configure something else
>> based on the "winner" of such a dict, with these rules:
<snip>
> I realize, now though, (and Chris asked about this) that I was
> imprecise in my
> rules. =C2=A0They really should be stated as:
>
> 1. In this dict, if there is a UNIQUE max value, then its *key* is the
> winner.
> 2. If there are any TIES for max value, then the *key* 'b' is the
> winner by default.
>
> The point is, I am trying to determine the name of the winning
> category, either
> 'a', 'b', or 'c', not the value of its winning score.
>
> So in your solutions there is sorting by values, which makes sense.
> But how
> can I go back to keys from there? =C2=A0Sorry for the mistake (but even s=
o,
> I learned
> something already).

# still presumes at least 2 items
from heapq import nlargest
winner, runner_up =3D nlargest(2, the_dict, lambda k: the_dict[k])
if the_dict[winner] =3D=3D the_dict[runner_up]:
   winner =3D 'b'

Cheers,
Chris