MongoDB count distinct value?

问题

Here below show my code. I have to calculate the how many times distinct value repeated. Here i have store distinct value in "results".I used collection.count() to calculate but it's not work. please any one tell me where i have to mistake. Thank you very much .

var DistinctIntoSingleDB = function(Collection,arr,opt,distVal,callback){
 Collection.find({}).distinct(distVal, function(err, results) {
      if(!err && results){
            console.log("Distinct Row Length :", results.length);
            var a,arr1 = [];
            for(var j=0; j<results.length; j++){
                collection.count({'V6': results[j]}, function(err, count) {
                      console.log(count)
                });

                arr1.push(results[j]+ " : " +a);
            }
            callback(results,arr1);
      }else{
           console.log(err, results);
           callback(results);
      }
 });

回答1:

To get occurrences of distinct values of a field 'field1' on a collection 'col1' and write to a separate collection 'distinctCount'. Also allow to use disk space in case the collection is huge.

db.col1.aggregate(
          [{$group: {
              _id: "$field1",
              count: { $sum : 1 }
            }}, {
            $group: {
              _id: "$_id",
              count: { $sum : "$count" }
            }},{
              $out: "distinctCount"
            }],
         {allowDiskUse:true}
)

回答2:

While .distinct() works well for just obtaining the distinct values for a field, in order to actually get the counts of occurrences, this is better suited to the aggregation framework:

Collection.aggregate([
    { "$group": {
        "_id": "$field",
        "count": { "$sum": 1 }
    }}
],function(err,result) {

});

Also the .distinct() method does "abstract" from where the specified "distinct" field is actually within an array. In this case you need to call $unwind first to process the array elements here:

Collection.aggregate([
    { "$unwind": "$array" },
    { "$group": {
        "_id": "$array.field",
        "count": { "$sum": 1 }
    }}
],function(err,result) {

});

So the main work is basically done in the $group by "grouping" on the field values, which means the same thing as "distinct". The $sum is a grouping operator which in this case just adds up 1 for each occurrence of that value in the field for that collection.

来源：https://stackoverflow.com/questions/25888128/mongodb-count-distinct-value