On Tue, Feb 11, 2014 at 1:31 PM, pete suchsland
<visiondoctor2...@gmail.com> wrote:
> How about make it simple by using sorted(a.values()) ...
>
>>>> a = {}
>>>> a['first'] = datetime.datetime.now()
>>>> a['second'] = datetime.datetime.now()
>>>> a['third'] = datetime.datetime.now()
>>>> a['forth'] = datetime.datetime.now()
>>>> a['fifth'] = datetime.datetime.now()
>
>>>> sorted(a.values())
> [datetime.datetime(2014, 2, 10, 19, 24, 35, 163585), datetime.datetime(2014,
> 2, 10, 19, 24, 53, 244532), datetime.datetime(2014, 2, 10, 19, 25, 9,
> 483683), datetime.datetime(2014, 2, 10, 19, 25, 25, 581743),
> datetime.datetime(2014, 2, 10, 19, 25, 37, 789907)]
Works nicely if you don't care about the keys. Otherwise, you need to
somehow look them back up, which is at best going to add an extra O(n)
step (creating a reverse dictionary), and at worst will require an
O(n*n) search (if the objects aren't hashable).
ChrisA
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