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From Jeroen van Dijk <jeroentjevand...@gmail.com>
Subject Re: Help with an advanced view to build a recommender?
Date Tue, 12 Apr 2011 08:21:54 GMT
This feature (proposal) would be of great help
http://wiki.apache.org/couchdb/Forward_document_references I believe.
Anybody aware of progress on this?


On Mon, Apr 11, 2011 at 10:10 PM, Jeroen van Dijk <
jeroentjevandijk@gmail.com> wrote:

> Hi all,
>
> The last couple of days I have been trying to build a view that would act
> as a recommender. I have tried all the stuff that I could find/think of, but
> I can't find a solution. I hope someone can tell me how I can do it or maybe
> tell me that it is just not possible with one map reduce. Below is the
> problem description, I hope it is clear enough.
>
> The basic idea is to use co-occurrences of apps attached to a user to
> calculate the similarity between apps. This is how the two types of
> documents; users and apps, look like:
>
> { _id: 'user-1', _type: 'user', app_ids: 'app-1', 'app-2' }
>
> { _id: 'app-1', _type: 'app', user_ids: 'user-1', 'user-2' }
>
> I was hoping the map reduce approach below would work when adding the
> include_docs=true option. Unfortunately this doesn't work with a reduce
> function. So the remaining problem so far seems to be to obtain the total
> app count together with the co-occurrence counts.
>
>   //map
>   function(doc) {
>     if(doc.type == "user") {
>       var app_count = doc.app_ids.length;
>       for(var i = 0; i < app_count; i++) {
>         for(var j = i + 1; j < app_count; j++) {
>           emit([doc.app_ids[j], doc.app_ids[i]], [0, 1, 0, {_id:
> doc.app_ids[i]}]);
>           emit([doc.app_ids[i], doc.app_ids[j]], [0, 1, 0, {_id:
> doc.app_ids[j]}]);
>         }
>       }
>     }
>   }
>
>   //reduce
>   function(keys, values, rereduce) {
>     //output is [similarity, number of co-occurrences, total number, doc]
>     var output = [0, 0, 0, null];
>
>     values.forEach(function(pair) {
>       output[1] += pair[1];
>       output[2] = pair[2].user_ids.length;
>       output[0] = output[1] / output[2];
>       output[3] = pair[3];
>     });
>
>     return output;
>   }
>
>
> Hopefully someone has new insights that can help me a bit further. Thanks.
>
> Jeroen
>

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