Thank you all very much for your help.
I did the following and it works:
imgs=v.keys()
imgs.sort(lambda a,b: cmp(
time.strptime(str(v[a][9]['date']), '%Y:%m:%d %H:%M:%S'),
time.strptime(str(v[b][9]['date']), '%Y:%m:%d %H:%M:%S'))
)
for i in imgs:
...
Regards,
Rory
On 08/03/05, Diez B. Roggisch ([EMAIL PROTECTED]) wrote:
> l = v.items()
> l.sort(lambda a, b: cmp(a[9]['date'], b[9]['date'])
On 08/03/05, Scott David Daniels ([EMAIL PROTECTED]) wrote:
> >You can't sort dicts - they don't impose an order on either key or value.
> >There are ordered dict implementations out there, but AFAIK the only keep
> >the keys sorted, or maybe the (key,values) in the insertion order.
> >
> >But maybe this helps you:
> >
> >l = v.items()
> >l.sort(lambda a, b: cmp(a[9]['date'], b[9]['date'])
>
> In 2.4, this is simple:
>
> ordered_keys = sorted(v, key=lambda name: v[name][9]['date'])
>
> In 2.3, or earlier, use "decorate-sort-undecorate":
>
> decorated = [(value[9]['date'], key)
> for key, value in v.iteritems()]
> decorated.sort()
> result = [key for key, date in decorated]
On 08/03/05, Batista, Facundo ([EMAIL PROTECTED]) wrote:
> >>> temp_list = [ (x[1][1], x[0]) for x in d.items() ]
...
> >>> temp_list.sort()
> >>> for (tmp, key) in temp_list:
--
Rory Campbell-Lange
<[EMAIL PROTECTED]>
<www.campbell-lange.net>
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