On Wed, Feb 4, 2009 at 3:47 PM, Erik Hatcher wrote:
> What about using the luke request handler to get the distinct values count?
That wouldn't restrict results by the base query and filters.
-Yonik
What about using the luke request handler to get the distinct values
count? Although it is pretty seriously heavy on a big index, so
probably not quite workable in your case.
Erik
On Feb 4, 2009, at 12:54 PM, Yonik Seeley wrote:
On Wed, Feb 4, 2009 at 5:42 AM, Bruno Aranda
wrote
On Wed, Feb 4, 2009 at 5:42 AM, Bruno Aranda wrote:
> Unfortunately, after some tests listing all the distinct surnames or other
> fields is too slow and too memory consuming with our current infrastructure.
> Could someone confirm that if I wanted to add this functionality (just count
> the total
Unfortunately, after some tests listing all the distinct surnames or other
fields is too slow and too memory consuming with our current infrastructure.
Could someone confirm that if I wanted to add this functionality (just count
the total of different facets) what I should do is to subclass the
Sim
Thanks, I will try that though I am talking in my case about 100,000+
distinct surnames/towns maximum per query and I just needed the count and
not the whole list. In any case, this brute-force approach is still
something I can try but I wonder how this will behave speed and memory wise
when there
On Wed, Feb 4, 2009 at 2:53 PM, Bruno Aranda wrote:
> Mmh, thanks for your answer but with that I get the count of names starting
> with A*, but I would like to get the count of distinct surnames (or town
> names, or any other field that is not the name...) for the people with name
> starting wit
Mmh, thanks for your answer but with that I get the count of names starting
with A*, but I would like to get the count of distinct surnames (or town
names, or any other field that is not the name...) for the people with name
starting with A*. Is that possible?
Thanks!
Bruno
2009/2/4 Shalin Shekh
On Wed, Feb 4, 2009 at 2:14 PM, Bruno Aranda wrote:
> Maybe I am not clear, but I am not able to find anything on the net.
> Basically, if I had in my index millions of names starting with A* I would
> like to know how many distinct surnames are present in the resultset
> (similar to a distinct S
Maybe I am not clear, but I am not able to find anything on the net.
Basically, if I had in my index millions of names starting with A* I would
like to know how many distinct surnames are present in the resultset
(similar to a distinct SQL query).
I will attempt to have a look at the SOLR sources t
But as far as I understand the total number of constraints is limited (there
is a default value), so I cannot know the total if I don't set the
facet.limit to a really big number and then the request takes a long time. I
was wondering if there was a way to get the total number (e.g. 100.000
constra
Hello,
Searching for ?q=*:* with facetting turned on gives me the total number
of available constraints, if that is what you mean.
Cheers,
On Tue, 2009-02-03 at 16:03 +, Bruno Aranda wrote:
> Hi,
>
> I would like to know if is there a way to get the total number of different
> facets r
Hi,
I would like to know if is there a way to get the total number of different
facets returned by a faceted search? I see already that I can paginate
through the facets with the facet.offset and facet.limit, but there is a way
to know how many facets are found in total?
For instance,
NameSu
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