Thanks Bill. Your reply is quite helpful.
On Nov 4, 12:01 pm, Bill Freeman <ke1g...@gmail.com> wrote:
> On Wed, Nov 4, 2009 at 1:32 PM, David <ww...@yahoo.com> wrote:
>
> > Hello,
>
> > I just wonder if anybody has handled this before. Here is a 2-D
> > dictionary.
>
> > dict[key_1] = {'a':'aa', 'b':'bb', 'c':'cc'}
> > dict[key_2] = {'a':'dd', 'b':'ee', 'c':'ff'}
> > dict[key_3] = {'a':'eef', 'b': 'ff', 'c':'ghh'}
> > ............
>
> > Assume that this dict is so long that I need to paginate. Do I need to
> > convert it to a list in order to use Paginator
> > ?
>
> > Is there a good way to paginate this dictionary?
>
> > Thanks for your ideas.
>
> You at least need to list the keys, say by using the keys() method.
> The trouble with
> thinking of paginating a dictionary is that it has no inherent order.
> When you go back
> for items n through 2n-1, for example, there is no guarantee that what
> you get won't
> share items with the first 0-n. That the order may seem stable in
> casual testing is an
> implementation detail that you shouldn't depend on.
>
> Probably you want to sort this somehow too, so that the presentation
> order makes sense.
> So something like:
>
> ks = d.keys()
> ks.sort()
>
> Then for page p of n items:
>
> for k in ks[n*(p-1):n*p]:
> v = d[k]
>
> Note that if pages are displayed across multiple requests, you need to
> put ks somewhere that its value will be preserved until the next
> request. Or if you're sure that d won't change between requests, just
> do the call to the keys method and the sort each time.
>
> [Note, too, re your sample, that "dict" is a bad name for a dictionary
> variable, because it is also the name of the type. It will work, but
> it will get confusing.]
>
> Bill
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