How to count distinct date items in a timestamp field in Mongoose/NodeJS?

Question

I use Mongoose in my NodeJS, and I have a collection with documents that look like this -

var SomeSchema = new Schema({
    date: {
        type: Date,
        default: new Date()
    },
    some_item: {
        type: String
    }
});

The date field contains date with the time component (for example, 2014-05-09 09:43:02.478Z). How do I get a count of distinct items for a given date, say 2014-05-08?

EDIT: Ideally, I would like to get a count of records for each distinct date.

As per mongoose documentation . Model.distinct(field, conditions, callback); — Sumeet Kumar Yadav
– Sumeet Kumar Yadav, Commented May 9, 2014 at 11:46
Distinct is not a good idea. It is just a wrapper around mapReduce and therefore runs a lot slower than the aggregation framework alternative. — Neil Lunn
– Neil Lunn, Commented May 9, 2014 at 11:53

Neil Lunn · Accepted Answer · 2014-05-09 13:24:35Z

3

What you want is usage of the aggregation framework with the aggregate() command, at least for finding one date as you ask:

SomeSchema.aggregate(
    [
        { "$match": {
            "date": { 
                "$gte": new Date("2014-05-08"), "$lt": new Date("2014-05-09")
            }
        }},
        { "$group": {
            "_id": "$some_item",
            "count": { "$sum": 1 }
        }}
    ],
    function(err,result) {
        // do something with result

    }
);

If you are specifically looking for "counts" over several dates then you can do something like this:

SomeSchema.aggregate(
    [
        { "$match": {
            "date": { 
                "$gte": new Date("2014-05-01"), "$lt": new Date("2014-06-01")
            }
        }},
        { "$group": {
            "_id": { 
                "year":  { "$year": "$date" },
                "month": { "$month": "$date" },
                "day":   { "$dayOfMonth": "$date" }
            }
            "count": { "$sum": 1 }
        }}
    ],
    function(err,result) {
        // do something with result

    }
);

And that gives you "per day" grouping. Look for the date aggregation operators in the documentation if you wish to take this further.

edited May 9, 2014 at 13:24

answered May 9, 2014 at 11:52

Neil Lunn

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4 Comments

GPX Over a year ago

Fantastic. Will I be able to do this for all unique dates too? Or would I have to select each date in a loop and then count distinct records for that date?

JohnnyHK Over a year ago

Good answer. Neil, you forgot your _id definition in your second example.

Neil Lunn Over a year ago

@JohnnyHK Thankyou, Thankyou once again. How are you not a moderator yet?

Neil Lunn Over a year ago

@GPX I think there is an edit in there that shows how to find singular day boundaries and group by them

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