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Hadoop, mail # user - Re: best way to join?

Ted Dunning 2012-08-28, 15:32
dexter morgan 2012-08-28, 16:04
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Re: best way to join?
Ted Dunning 2012-08-31, 15:41
Yes.  I think you mis-understood.

My suggestion is that the few clusters near a point are very likely to
contain the nearest points.  If you scan these clusters computing the
distance to your original point, you should be able to find a list of
points that overlaps with your desired result.  Keep in mind that I am
suggesting that you use a LOT of clusters here, much more than is typical.
 Typically, I recommend sqrt(N)/m clusters for this kind of work where m is
a small constant 1 <= m <= 30.

The basic idea for the method is that there are many points that are
obviously far from your query.  You don't have to compute the distance to
these data points to your query since you can eliminate them from
consideration.  Many of them can be absolutely eliminated by appeal to the
triangle inequality, but many more can be eliminated safely with reasonably
high probability.  It is a reasonable heuristic that the farthest clusters
contain points that can be eliminated this way.  Another way to state this
is to say that you should only search through the points that are in
clusters that are near your original cluster.

On Fri, Aug 31, 2012 at 9:03 AM, dexter morgan <[EMAIL PROTECTED]>wrote:

> Hi Ted,
> First of all, i'd like to know how to make a map/reduce job that does join
> on the input-file it self.
> Second, maybe your clustering approach be usefull, i still think it's not
> correct.
> Reason:
> Lets say i want to find the 10 closest points for a given point. Point:
> [120,90] for example.
> Clustering approach: which cluster has [120,90] as a node? answer: the
> cluster at [300,200]
> Now, if i understood you, i should get the 10 nearest neighbors of
> [300,200] (again, you didn't elaborate much on this or i didn't understand
> it)
> But i require the 10 nearest to [120,90] , not to [300,200]. Even if i
> know the distances from [120,90] to [300,200] and to the 10 nearest points
> to [300,200] it won't help me, because maybe the 10 nearest points to
> [120,90] are actually starting from the 5000th nearest points to [300,200].
> In the end my goal is to pre-process (as i wrote at the begining) this
> list of N nearest points for every point in the file. Where N is a
> parameter given to the job. Let's say 10 points. That's it.
> No calculation after-wards, only querying that list.
> Thank you
> On Thu, Aug 30, 2012 at 11:05 PM, Ted Dunning <[EMAIL PROTECTED]>wrote:
>> I don't know off-hand.  I don't understand the importance of your
>> constraint either.
>> On Thu, Aug 30, 2012 at 5:21 AM, dexter morgan <[EMAIL PROTECTED]>wrote:
>>> Ok, but as i said before, how do i achieve the same result with out
>>> clustering , just linear. Join on the same data-set basically?
>>> and calculating the distance as i go
>>> On Tue, Aug 28, 2012 at 11:07 PM, Ted Dunning <[EMAIL PROTECTED]>wrote:
>>>> I don't mean that.
>>>> I mean that a k-means clustering with pretty large clusters is a useful
>>>> auxiliary data structure for finding nearest neighbors.  The basic outline
>>>> is that you find the nearest clusters and search those for near neighbors.
>>>>  The first riff is that you use a clever data structure for finding the
>>>> nearest clusters so that you can do that faster than linear search.  The
>>>> second riff is when you use another clever data structure to search each
>>>> cluster quickly.
>>>> There are fancier data structures available as well.
>>>> On Tue, Aug 28, 2012 at 12:04 PM, dexter morgan <
>>>> [EMAIL PROTECTED]> wrote:
>>>>> Right, but if i understood your sugesstion, you look at the end goal ,
>>>>> which is:
>>>>> 1[40.123,-50.432]\t[[41.431,-43.32],[...,...],...,[...]]
>>>>> for example, and you say: here we see a cluster basically, that
>>>>> cluster is represented by the point:  [40.123,-50.432]
>>>>> which points does this cluster contains?  [[41.431,-
>>>>> 43.32],[...,...],...,[...]]
>>>>> meaning: that for every point i have in the dataset, you create a
dexter morgan 2012-09-09, 09:22