Hi Ryan,

Digging up a rather old announcement from you, because this is exactly what
I was looking for. But I am not sure if a special parameter is required to
enable trend filtering.

I just now added a "trends" feature in my app, and thus noticed that the
search results for trends still shows some spam.


   - Is there a special parameter to indicate that the search is for a trend
   and not just another keyword search?
   - Does it make sense to have a "follower_count" parameter to the search
   API? When specified, only tweets from people having followers > than
   specified parameter will be shown. This could be useful for non-trending
   keywords as well.

cheers,
Harshad

On Fri, Nov 6, 2009 at 6:02 AM, Ryan Sarver <rsar...@twitter.com> wrote:

>
> Today on the Twitter Blog we announced that we will be changing search
> results for trending topics to improve the quality. It used to be that
> trends were a great way to quickly see what was going on on Twitter,
> but they have begun to get fairly noisy due to the sheer volume of
> tweets. We wanted to improve that experience and will start returning
> what we think are the best results. This doesn't mean that tweets are
> getting dropped from the index, instead we are just making intelligent
> decisions on which ones to return for searches on popular topics,
> beginning with trending topics. If you make a more specific search,
> you can still get to all the tweets.
>
> So for you, the developer, this means that if you will see better
> quality results over the search API for trends, but you won't be
> getting every tweet that matched the search term. If you still do want
> to get every tweet matching a trend, we recommend you check out the
> Streaming API (http://apiwiki.twitter.com/Streaming-API-Documentation).
> In the end, this is a change that is good for developers and
> end-users, but we wanted to notify you as you might be seeing some
> slightly different behavior via the API.
>
> Please let us know if you have any questions -- we are happy to answer.
>
> Ryan
>
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-- 
Harshad RJ
http://hrj.wikidot.com

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