

Hi Dexter, I am no sure if I understood your requirements right. So I repet it to define a starting point. 1.) You have a (static) list of points (the points.txt file) 2.) Now you want to calculate the nearest points to a set of given points. Are the points which have to be considered in a different data set or do you look for closest points "within" your big list (in the points.txt) file? Lets assume the last is what you want: I suggest Mahout to do this. This could be helpfull: " https://cwiki.apache.org/MAHOUT/algorithms.html""Vector Similarity Mahout contains implementations that allow one to compare one or more vectors with another set of vectors. This can be useful if one is, for instance, trying to calculate the pairwise similarity between all documents (or a subset of docs) in a corpus.  RowSimilarityJob – Builds an inverted index and then computes distances between items that have cooccurrences. This is a fully distributed calculation.  VectorDistanceJob – Does a map side join between a set of "seed" vectors and all of the input vectors. " If you look for pairs within just one data set you use this as seed vectors as well as input vectors, otherwise you use different files or portions of it. I hope that helps. Best wishes Mirko 2012/9/9 dexter morgan <[EMAIL PROTECTED]> > Elmar, > > Right, thanks a lot for your help, If you'll read what Ted suggested its > basically this. I'm interesting in knowing how to do this using JOIN > (mapjoin + reducerjoin i suppose) as well... though i'll go with the > mahout approach > > Best, > Dex > > On Tue, Sep 4, 2012 at 11:17 AM, BjörnElmar Macek <[EMAIL PROTECTED] > > wrote: > >> Hi Dexter, >> >> i think, what you want is a clustering of points based on the euclidian >> distance or density based clustering ( http://en.wikipedia.org/wiki/**>> Cluster_analysis < http://en.wikipedia.org/wiki/Cluster_analysis> ). I >> bet there are some implemented quite well in Mahout already: afaik this is >> the datamining framework based on Hadoop. >> >> Best luck! >> Elmar >> >> >> Am 27.08.2012 22:15, schrieb dexter morgan: >> >> Dear list, >>> >>> Lets say i have a file, like this: >>> id \t at,tlng < structure >>> >>> 1\t40.123,50.432 >>> 2\t41.431,43.32 >>> ... >>> ... >>> lets call it: 'points.txt' >>> I'm trying to build a mapreduce job that runs over this BIG points file >>> and it should output >>> a file, that will look like: >>> id[lat,lng] \t [list of points in JSON standard] < structure >>> >>> 1[40.123,50.432]\t[[41.431,**43.32],[...,...],...,[...]] >>> 2[41.431,43.32]\t[[40.123,**50.432],...[,]] >>> ... >>> >>> Basically it should run on ITSELF, and grab for each point the N (it >>> will be an argument for the job) CLOSEST points (the mappers should >>> calculate the distance).. >>> >>> Distributed cache is not an option, what else? not sure if to classify >>> it as a mapjoin , reducejoin or both? >>> Would you do this in HIVE some how? >>> Is it feasible in a single job? >>> >>> Would LOVE to hear your suggestions, code (if you think its not that >>> hard) or what not. >>> BTW using CDH3  rev 3 (20.23) >>> >>> Thanks! >>> >> >> >
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Mirko Kämpf 20120909, 09:55
I don't mean that.
I mean that a kmeans 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 cluster. > If you don't mean that, but you do mean to create clusters based on some > randomseed points or what not, that would mean > that i'll have points (talking about the "end goal") that won't have > enough points in their list. > > one of the criterions for a clustering is that for any clusters: C_i and > C_j (where i != j), C_i intersect C_j is empty > > and again, how can i accomplish my task with out running mahout / knn > algo? just by calculating distance between points? > join of a file with it self. > > Thanks > > On Tue, Aug 28, 2012 at 6:32 PM, Ted Dunning <[EMAIL PROTECTED]>wrote: > >> >> >> On Tue, Aug 28, 2012 at 9:48 AM, dexter morgan <[EMAIL PROTECTED]>wrote: >> >>> >>> I understand your solution ( i think) , didn't think of that, in that >>> particular way. >>> I think that lets say i have 1M datapoints, and running knn , that the >>> k=1M and n=10 (each point is a cluster that requires up to 10 points) >>> is an overkill. >>> >> >> I am not sure I understand you. n = number of points. k = number of >> clusters. For searching 1 million points, I would recommend thousands of >> clusters. >> >> >>> How can i achieve the same result WITHOUT using mahout, just running on >>> the dataset , i even think it'll be in the same complexity (o(n^2)) >>> >> >> Running with a good knn package will give you roughly O(n log n) >> complexity. >> >> >
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Ted Dunning 20120828, 20:07
Ok, but as i said before, how do i achieve the same result with out clustering , just linear. Join on the same dataset 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 kmeans 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 cluster. >> If you don't mean that, but you do mean to create clusters based on some >> randomseed points or what not, that would mean >> that i'll have points (talking about the "end goal") that won't have >> enough points in their list. >> >> one of the criterions for a clustering is that for any clusters: C_i and >> C_j (where i != j), C_i intersect C_j is empty >> >> and again, how can i accomplish my task with out running mahout / knn >> algo? just by calculating distance between points? >> join of a file with it self. >> >> Thanks >> >> On Tue, Aug 28, 2012 at 6:32 PM, Ted Dunning <[EMAIL PROTECTED]>wrote: >> >>> >>> >>> On Tue, Aug 28, 2012 at 9:48 AM, dexter morgan <[EMAIL PROTECTED] >>> > wrote: >>> >>>> >>>> I understand your solution ( i think) , didn't think of that, in that >>>> particular way. >>>> I think that lets say i have 1M datapoints, and running knn , that the >>>> k=1M and n=10 (each point is a cluster that requires up to 10 points) >>>> is an overkill. >>>> >>> >>> I am not sure I understand you. n = number of points. k = number of >>> clusters. For searching 1 million points, I would recommend thousands of >>> clusters. >>> >>> >>>> How can i achieve the same result WITHOUT using mahout, just running on >>>> the dataset , i even think it'll be in the same complexity (o(n^2)) >>>> >>> >>> Running with a good knn package will give you roughly O(n log n) >>> complexity. >>> >>> >> >
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dexter morgan 20120830, 09:21
I don't know offhand. 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 dataset 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 kmeans 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 >>> cluster. >>> If you don't mean that, but you do mean to create clusters based on some >>> randomseed points or what not, that would mean >>> that i'll have points (talking about the "end goal") that won't have >>> enough points in their list. >>> >>> one of the criterions for a clustering is that for any clusters: C_i and >>> C_j (where i != j), C_i intersect C_j is empty >>> >>> and again, how can i accomplish my task with out running mahout / knn >>> algo? just by calculating distance between points? >>> join of a file with it self. >>> >>> Thanks >>> >>> On Tue, Aug 28, 2012 at 6:32 PM, Ted Dunning <[EMAIL PROTECTED]>wrote: >>> >>>> >>>> >>>> On Tue, Aug 28, 2012 at 9:48 AM, dexter morgan < >>>> [EMAIL PROTECTED]> wrote: >>>> >>>>> >>>>> I understand your solution ( i think) , didn't think of that, in that >>>>> particular way. >>>>> I think that lets say i have 1M datapoints, and running knn , that >>>>> the k=1M and n=10 (each point is a cluster that requires up to 10 points) >>>>> is an overkill. >>>>> >>>> >>>> I am not sure I understand you. n = number of points. k = number of >>>> clusters. For searching 1 million points, I would recommend thousands of >>>> clusters. >>>> >>>> >>>>> How can i achieve the same result WITHOUT using mahout, just running >>>>> on the dataset , i even think it'll be in the same complexity (o(n^2)) >>>>> >>>> >>>> Running with a good knn package will give you roughly O(n log n) >>>> complexity. >>>> >>>> >>> >> >
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Ted Dunning 20120830, 20:05
Hi Ted,
First of all, i'd like to know how to make a map/reduce job that does join on the inputfile 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 preprocess (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 afterwards, only querying that list.
Thank you
On Thu, Aug 30, 2012 at 11:05 PM, Ted Dunning <[EMAIL PROTECTED]> wrote:
> I don't know offhand. 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 dataset 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 kmeans 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 >>>> cluster. >>>> If you don't mean that, but you do mean to create clusters based on >>>> some randomseed points or what not, that would mean >>>> that i'll have points (talking about the "end goal") that won't have >>>> enough points in their list. >>>> >>>> one of the criterions for a clustering is that for any clusters: C_i >>>> and C_j (where i != j), C_i intersect C_j is empty >>>> >>>> and again, how can i accomplish my task with out running mahout / knn >>>> algo? just by calculating distance between points? >>>> join of a file with it self. >>>> >>>> Thanks >>>> >>>> On Tue, Aug 28, 2012 at 6:32 PM, Ted Dunning <[EMAIL PROTECTED]>wrote: >>>> >>>>> >>>>> >>>>> On Tue, Aug 28, 2012 at 9:48 AM, dexter morgan < >>>>> [EMAIL PROTECTED]> wrote: >>>>> >>>>>> >>>>>> I understand your solution ( i think) , didn't think of that, in that >>>>>> particular way. >>>>>> I think that lets say i have 1M datapoints, and running knn , that >>>>>> the k=1M and n=10 (each point is a cluster that requires up to 10 points) >>>>>> is an overkill. >>>>>> >>>>> >>>>> I am not sure I understand you. n = number of points. k = number of >>>>> clusters. For searching 1 million points, I would recommend thousands of >>>>> clusters. >>>>> >>>>> >>>
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dexter morgan 20120831, 13:03
Thanks, for the deeper explanation. Now i understand what you ment. Either way, any clustering process requires calculating the distance of all points (not between all the points, but of all of them to some relative point). Because i'll need a clustering MR job, ill probably use it, despite as you said, it has high probability to be correct (not 100%)...
I'll post another question regarding joining file with itself... (wheter its using mapperjoin / reducer join) , without using hive/ pig On Fri, Aug 31, 2012 at 6:41 PM, Ted Dunning <[EMAIL PROTECTED]> wrote:
> Yes. I think you misunderstood. > > 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 inputfile 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 preprocess (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 afterwards, only querying that list. >> >> Thank you >> >> On Thu, Aug 30, 2012 at 11:05 PM, Ted Dunning <[EMAIL PROTECTED]>wrote: >> >>> I don't know offhand. 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 dataset 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 kmeans 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
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dexter morgan 20120902, 16:26
On Sun, Sep 2, 2012 at 12:26 PM, dexter morgan <[EMAIL PROTECTED]>wrote:
> ... Either way, any clustering process requires calculating the distance > of all points (not between all the points, but of all of them to some > relative point). Because i'll need a clustering MR job, ill probably use > it, despite as you said, it has high probability to be correct (not 100%)... >
This is probably right as stated, but I think that there is confusion here.
Many people assume that each point in the training data has to have distance computed to all centroids in the clustering. Even this is not true.
It is true that you have to compute distance to at least one something, but not necessarily to all of the clusters.
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Ted Dunning 20120903, 19:47
Dear list,
Lets say i have a file, like this: id \t at,tlng < structure
1\t40.123,50.432 2\t41.431,43.32 ... ... lets call it: 'points.txt' I'm trying to build a mapreduce job that runs over this BIG points file and it should output a file, that will look like: id[lat,lng] \t [list of points in JSON standard] < structure
1[40.123,50.432]\t[[41.431,43.32],[...,...],...,[...]] 2[41.431,43.32]\t[[40.123,50.432],...[,]] ...
Basically it should run on ITSELF, and grab for each point the N (it will be an argument for the job) CLOSEST points (the mappers should calculate the distance)..
Distributed cache is not an option, what else? not sure if to classify it as a mapjoin , reducejoin or both? Would you do this in HIVE some how? Is it feasible in a single job?
Would LOVE to hear your suggestions, code (if you think its not that hard) or what not. BTW using CDH3  rev 3 (20.23)
Thanks!
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dexter morgan 20120827, 20:15
Hi Dexter, i think, what you want is a clustering of points based on the euclidian distance or density based clustering ( http://en.wikipedia.org/wiki/Cluster_analysis ). I bet there are some implemented quite well in Mahout already: afaik this is the datamining framework based on Hadoop. Best luck! Elmar Am 27.08.2012 22:15, schrieb dexter morgan: > Dear list, > > Lets say i have a file, like this: > id \t at,tlng < structure > > 1\t40.123,50.432 > 2\t41.431,43.32 > ... > ... > lets call it: 'points.txt' > I'm trying to build a mapreduce job that runs over this BIG points > file and it should output > a file, that will look like: > id[lat,lng] \t [list of points in JSON standard] < structure > > 1[40.123,50.432]\t[[41.431,43.32],[...,...],...,[...]] > 2[41.431,43.32]\t[[40.123,50.432],...[,]] > ... > > Basically it should run on ITSELF, and grab for each point the N (it > will be an argument for the job) CLOSEST points (the mappers should > calculate the distance).. > > Distributed cache is not an option, what else? not sure if to > classify it as a mapjoin , reducejoin or both? > Would you do this in HIVE some how? > Is it feasible in a single job? > > Would LOVE to hear your suggestions, code (if you think its not that > hard) or what not. > BTW using CDH3  rev 3 (20.23) > > Thanks!
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BjörnElmar Macek 20120904, 08:17
Mahout is getting some very fast knn code in version 0.8.
The basic work flow is that you would first do a largescale clustering of the data. Then you would make a second pass using the clustering to facilitate fast search for nearby points.
The clustering will require two mapreduce jobs, one to find the cluster definitions and the second maponly to assign points to clusters in a form to be used by the second pass. The second pass is a maponly process as well.
In order to make this as efficient as possible, it is desirable to use a distribution of Hadoop that allows you to directly map the cluster data structures into shared memory. IF you have NFS access to your cluster, this is easy. Otherwise, it is considerably trickier.
On Mon, Aug 27, 2012 at 4:15 PM, dexter morgan <[EMAIL PROTECTED]>wrote:
> Dear list, > > Lets say i have a file, like this: > id \t at,tlng < structure > > 1\t40.123,50.432 > 2\t41.431,43.32 > ... > ... > lets call it: 'points.txt' > I'm trying to build a mapreduce job that runs over this BIG points file > and it should output > a file, that will look like: > id[lat,lng] \t [list of points in JSON standard] < structure > > 1[40.123,50.432]\t[[41.431,43.32],[...,...],...,[...]] > 2[41.431,43.32]\t[[40.123,50.432],...[,]] > ... > > Basically it should run on ITSELF, and grab for each point the N (it will > be an argument for the job) CLOSEST points (the mappers should calculate > the distance).. > > Distributed cache is not an option, what else? not sure if to classify it > as a mapjoin , reducejoin or both? > Would you do this in HIVE some how? > Is it feasible in a single job? > > Would LOVE to hear your suggestions, code (if you think its not that hard) > or what not. > BTW using CDH3  rev 3 (20.23) > > Thanks! >
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Ted Dunning 20120827, 21:52
Dear Ted,
I understand your solution ( i think) , didn't think of that, in that particular way. I think that lets say i have 1M datapoints, and running knn , that the k=1M and n=10 (each point is a cluster that requires up to 10 points) is an overkill.
How can i achieve the same result WITHOUT using mahout, just running on the dataset , i even think it'll be in the same complexity (o(n^2)) and calculating the distance between each indifferent points?
and maybe the reducer would just sort them in DESC order for each point.
Thank you!
On Tue, Aug 28, 2012 at 12:52 AM, Ted Dunning <[EMAIL PROTECTED]> wrote:
> Mahout is getting some very fast knn code in version 0.8. > > The basic work flow is that you would first do a largescale clustering of > the data. Then you would make a second pass using the clustering to > facilitate fast search for nearby points. > > The clustering will require two mapreduce jobs, one to find the cluster > definitions and the second maponly to assign points to clusters in a form > to be used by the second pass. The second pass is a maponly process as > well. > > In order to make this as efficient as possible, it is desirable to use a > distribution of Hadoop that allows you to directly map the cluster data > structures into shared memory. IF you have NFS access to your cluster, > this is easy. Otherwise, it is considerably trickier. > > > On Mon, Aug 27, 2012 at 4:15 PM, dexter morgan <[EMAIL PROTECTED]>wrote: > >> Dear list, >> >> Lets say i have a file, like this: >> id \t at,tlng < structure >> >> 1\t40.123,50.432 >> 2\t41.431,43.32 >> ... >> ... >> lets call it: 'points.txt' >> I'm trying to build a mapreduce job that runs over this BIG points file >> and it should output >> a file, that will look like: >> id[lat,lng] \t [list of points in JSON standard] < structure >> >> 1[40.123,50.432]\t[[41.431,43.32],[...,...],...,[...]] >> 2[41.431,43.32]\t[[40.123,50.432],...[,]] >> ... >> >> Basically it should run on ITSELF, and grab for each point the N (it will >> be an argument for the job) CLOSEST points (the mappers should calculate >> the distance).. >> >> Distributed cache is not an option, what else? not sure if to classify >> it as a mapjoin , reducejoin or both? >> Would you do this in HIVE some how? >> Is it feasible in a single job? >> >> Would LOVE to hear your suggestions, code (if you think its not that >> hard) or what not. >> BTW using CDH3  rev 3 (20.23) >> >> Thanks! >> > >
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dexter morgan 20120828, 13:48

