Hey,
thanx for the reply! No, the bid is *not* ordered randomly, but you
know what, your solution works anyway! ;) Test this and you'll notice
that it works.. (still haven't figured out why!):
* bs = BID.objects.all ()
* idx=81; bid=bs[idx];
* bs.order_by ('?').filter (pk__lt=bid.id).count () ## == idx
Can u explain this? I've checked even for *every* bid in bs if this
condition holds for my sample, and it was true. My test routine was:
* for idx, bid in enumerate(bs):
* idx, bs.order_by ('?').filter (pk__lt=bid.id).count () - idx ==
0
.. strange.
On Mar 7, 11:50 am, greatlemer <greatle...@googlemail.com> wrote:
> if bids is ordered by index then you could just return
> BID.objects.filter(...).filter(pk__lte=id).count() and that should be
> what your after in one step (if I've understood the question
> correctly).
>
> --
> G
>
> On Mar 7, 7:10 am, Hasan Karahan <hasan.karaha...@gmail.com> wrote:
>
> > Hi,
>
> > I'm wondering if there is an efficient way to look-up the index of a given
> > object in a query-set? I'm using now the following approach:
>
> > def get_index (id):
>
> > bids = BID.objects.filter (..)
> > obj2idx = dict (zip (bids, xrange (bids.count ()))
>
> > bid = BID.objects.get (id=ID)
>
> > return obj2idx (bid)
>
> > This has a linear time processing.. is there a more efficient way to do it?
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