Grouping by values in an array

Question

I have a set of records with the following structure:

{u'_id': ObjectId('4e60fc677fdfb50fc3000000'),
 u'columns': [
  {u'geodata_type': None,
   u'has_geodata': False,
   u'id': 1,
   u'is_available': True,
   u'is_key': False,
   u'name': u'NOMBRE',
   u'value': u'Martin'},
  {u'geodata_type': None,
   u'has_geodata': False,
   u'id': 2,
   u'is_available': True,
   u'is_key': False,
   u'name': u'EDAD',
   u'value': 12},
  {u'geodata_type': u'punto',
   u'has_geodata': True,
   u'id': 4,
   u'is_available': True,
   u'is_key': None,
   u'name': u'DIRECCION',
   u'value': u'humberto primero 2345'},
  {u'geodata_type': None,
   u'has_geodata': False,
   u'id': 5,
   u'is_available': True,
   u'is_key': False,
   u'name': u'BARRIO',
   u'value': u'centro'}],
 u'datasource_id': 1,
 u'map_empty': True
}

I pretend to group all the documents by some of those columns and to get counts by the given column name. The thing is, I don't find a proper way to set the key argument of the group operation to let mongo group the results properly.

Any suggestion?

Scott Willeke · Accepted Answer · 2011-09-11 22:31:17Z

1

You are trying to group on a value in an array and AFAIK group can only use a field. You can easily produce a set of column counts using map/reduce though:

Your mapper is where you'll do the grouping. Essentially for each column name, create a "group" (emit):

var mapper = function() {
    for (var k in this.columns) {
        emit(this.columns[k].name, {count:1} );
    }
}

In your reducer, aggregate the results for each group:

var reducer = function(key, values) {
    var sum = 0;
    values.forEach(function (item) {
        sum+=item.count;
    });
    return {count:sum};
}

Finally run the mapReduce operation:

var res = db.things.mapReduce(mapper, reducer, {out:"colCounts"});

answered Sep 11, 2011 at 22:31

Scott Willeke

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2 Comments

Scott Willeke Over a year ago

@tutuca: If this the correct answer, I would love it if you mark this as the "accepted" answer (check the green checkbox). - meta.stackexchange.com/questions/5234/…

Scott Willeke Over a year ago

sure! If there is something inconvenient about my suggestion let me know and I'll see if I can help. Or if you come up with a better way to do it, please let me know that too.

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