Although it works in my interactive environment.. it does not work in
my actual script.. ??
On Mar 9, 7:02 pm, Calaganne <hasan.karaha...@gmail.com> wrote:
> 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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