CODEWITHSUNDEEP
dategroup-bymongodb
Mark Gill
Mark Gill

Reputation: 3781

group by dates in mongodb

I am working on a project in which I am tracking number of clicks on a topic.

I am using mongodb and I have to group number of click by date( i want to group data for 15 days).

I am having data store in following format in mongodb

{ 
   "_id" : ObjectId("4d663451d1e7242c4b68e000"), 
  "date" : "Mon Dec 27 2010 18:51:22 GMT+0000 (UTC)", 
  "topic" : "abc", 
  "time" : "18:51:22"
}
{ 
    "_id" : ObjectId("4d6634514cb5cb2c4b69e000"), 
    "date" : "Mon Dec 27 2010 18:51:23 GMT+0000 (UTC)", 
    "topic" : "bce", 
    "time" : "18:51:23"
}

i want to group number of clicks on topic:abc by days(for 15 days)..i know how to group that but how can I group by date which are stored in my database

I am looking for result in following format

[
  {
    "date" : "date in log",
    "click" : 9 
  },  
  {
    "date" : "date in log",
    "click" : 19
  },  
]

I have written code but it will work only if date are in string (code is here http://pastebin.com/2wm1n1ix) ...please guide me how do I group it

Upvotes: 93

Views: 152745

Answers (10)

anshuman burmman
anshuman burmman

Reputation: 629

To fetch data group by date in mongodb

db.getCollection('supportIssuesChat').aggregate([
{
        $group : {
           _id :{ $dateToString: { format: "%Y-%m-%d", date: "$createdAt"} },
           list: { $push: "$$ROOT" },
           count: { $sum: 1 }
        }
}
])

Upvotes: 46

mhalttu
mhalttu

Reputation: 1347

There are already many answers to this question, but I wasn't happy with any of them. MongoDB has improved over the years, and there are now easier ways to do it. The answer by Jonas Tomanga gets it right, but is a bit too complex.

If you are using MongoDB 3.0 or later, here's how you can group by date. I start with the $match aggregation because the author also asked how to limit the results.

db.yourCollection.aggregate([
  { $match: { date: { $gte: ISODate("2019-05-01") } } },
  { $group: { _id: { $dateToString: { format: "%Y-%m-%d", date: "$date"} }, count: { $sum: 1 } } },
  { $sort: { _id: 1} }
])

Upvotes: 73

Old Pro
Old Pro

Reputation: 25567

New answer using Mongo aggregation framework

After this question was asked and answered, 10gen released Mongodb version 2.2 with an aggregation framework, which is now the better way to do this sort of query. This query is a little challenging because you want to group by date and the values stored are timestamps, so you have to do something to convert the timestamps to dates that match. For the purposes of example I will just write a query that gets the right counts. 

db.col.aggregate(
   { $group: { _id: { $dayOfYear: "$date"},
               click: { $sum: 1 } } }
   )

This will return something like:

[
    {
        "_id" : 144,
        "click" : 165
    },
    {
        "_id" : 275,
        "click" : 12
    }
]

You need to use $match to limit the query to the date range you are interested in and $project to rename _id to date. How you convert the day of year back to a date is left as an exercise for the reader. :-)

10gen has a handy SQL to Mongo Aggregation conversion chart worth bookmarking. There is also a specific article on date aggregation operators.

Getting a little fancier, you can use:

db.col.aggregate([
  { $group: {
      _id: {
        $add: [
         { $dayOfYear: "$date"}, 
         { $multiply: 
           [400, {$year: "$date"}]
         }
      ]},   
      click: { $sum: 1 },
      first: {$min: "$date"}
    }
  },
  { $sort: {_id: -1} },
  { $limit: 15 },
  { $project: { date: "$first", click: 1, _id: 0} }
])

which will get you the latest 15 days and return some datetime within each day in the date field. For example:

[
    {
        "click" : 431,
        "date" : ISODate("2013-05-11T02:33:45.526Z")
    },
    {
        "click" : 702,
        "date" : ISODate("2013-05-08T02:11:00.503Z")
    },
            ...
    {
        "click" : 814,
        "date" : ISODate("2013-04-25T00:41:45.046Z")
    }
]

Upvotes: 102

Jonas Tomanga
Jonas Tomanga

Reputation: 1108

This can help

return new Promise(function(resolve, reject) {
db.doc.aggregate(
            [
                { $match: {} },
                { $group: { _id: { $dateToString: { format: "%Y-%m-%d", date: "$date" } }, count: { $sum: 1 } } },
                { $sort: { _id: 1 } }
            ]
        ).then(doc => {
            /* if you need a date object */
            doc.forEach(function(value, index) {
                  doc[index]._id = new Date(value._id);
              }, this);
            resolve(doc);
        }).catch(reject);
}

Upvotes: 29

egvo
egvo

Reputation: 1895

Of course, that is a good solution. Aside from that you can group dates by days as strings (as that answer propose) or you can get the beginning of dates by projecting date field (in aggregation) like that:

{'$project': {
    'start_of_day': {'$subtract': [
        '$date',
        {'$add': [
            {'$multiply': [{'$hour': '$date'}, 3600000]},
            {'$multiply': [{'$minute': '$date'}, 60000]},
            {'$multiply': [{'$second': '$date'}, 1000]},
            {'$millisecond': '$date'}
        ]}
    ]},
}}

It gives you this:

{
    "start_of_day" : ISODate("2015-12-03T00:00:00.000Z")
},
{
    "start_of_day" : ISODate("2015-12-04T00:00:00.000Z")
}

It has some pluses: you can manipulate with your days in date type (not number or string), it allows you to use all of the date aggregation operators in following aggregation operations and gives you date type on the output.

Upvotes: 0

Blakes Seven
Blakes Seven

Reputation: 50436

If You want a Date oject returned directly

Then instead of applying the Date Aggregation Operators, instead apply "Date Math" to round the date object. This can often be desirable as all drivers represent a BSON Date in a form that is commonly used for Date manipulation for all languages where that is possible:

db.datetest.aggregate([
    { "$group": {
        "_id": {
            "$add": [
                { "$subtract": [
                    { "$subtract": [ "$date", new Date(0) ] },
                    { "$mod": [
                        { "$subtract": [ "$date", new Date(0) ] },
                        1000 * 60 * 60 * 24
                    ]}
                ]},
                new Date(0)
            ]
        },
        "click": { "$sum": 1 }
    }}
])

Or if as is implied in the question that the grouping interval required is "buckets" of 15 days, then simply apply that to the numeric value in $mod:

db.datetest.aggregate([
    { "$group": {
        "_id": {
            "$add": [
                { "$subtract": [
                    { "$subtract": [ "$date", new Date(0) ] },
                    { "$mod": [
                        { "$subtract": [ "$date", new Date(0) ] },
                        1000 * 60 * 60 * 24 * 15
                    ]}
                ]},
                new Date(0)
            ]
        },
        "click": { "$sum": 1 }
    }}
])

The basic math applied is that when you $subtract two Date objects the result returned will be the milliseconds of differnce numerically. So epoch is represented by Date(0) as the base for conversion in whatever language constructor you have.

With a numeric value, the "modulo" ( $mod ) is applied to round the date ( subtract the remainder from the division ) to the required interval. Being either:

1000 milliseconds x 60 seconds * 60 minutes * 24 hours = 1 day

Or

1000 milliseconds x 60 seconds * 60 minutes * 24 hours * 15 days = 15 days

So it's flexible to whatever interval you require.

By the same token from above an $add operation between a "numeric" value and a Date object will return a Date object equivalent to the millseconds value of both objects combined ( epoch is 0, therefore 0 plus difference is the converted date ).

Easily represented and reproducible in the following listing:

var now = new Date();
var bulk = db.datetest.initializeOrderedBulkOp();

for ( var x = 0; x < 60; x++ ) {
    bulk.insert({ "date": new Date( now.valueOf() + ( 1000 * 60 * 60 * 24 * x ))});
}

bulk.execute();

And running the second example with 15 day intervals:

{ "_id" : ISODate("2016-04-14T00:00:00Z"), "click" : 12 }
{ "_id" : ISODate("2016-03-30T00:00:00Z"), "click" : 15 }
{ "_id" : ISODate("2016-03-15T00:00:00Z"), "click" : 15 }
{ "_id" : ISODate("2016-02-29T00:00:00Z"), "click" : 15 }
{ "_id" : ISODate("2016-02-14T00:00:00Z"), "click" : 3 }

Or similar distribution depending on the current date when the listing is run, and of course the 15 day intervals will be consistent since the epoch date.

Using the "Math" method is a bit easier to tune, especially if you want to adjust time periods for different timezones in aggregation output where you can similarly numerically adjust by adding/subtracting the numeric difference from UTC.

Upvotes: 1

phnessu4
phnessu4

Reputation: 131

thanks for @mindthief, your answer help solve my problem today. The function below can group by day a little more easier, hope can help the others.

/**
 * group by day
 * @param query document {key1:123,key2:456}
 */
var count_by_day = function(query){
    return db.action.group(
    {
        keyf: function(doc) {
            var date = new Date(doc.time);
            var dateKey = (date.getMonth()+1)+"/"+date.getDate()+"/"+date.getFullYear();
            return {'date': dateKey};
        },
        cond:query,
        initial: {count:0},
        reduce: function(obj, prev) {
          prev.count++;
        }
    });
}

count_by_day({this:'is',the:'query'})

Upvotes: 3

golja
golja

Reputation: 1093

Another late answer, but still. So if you wanna do it in only one iteration and get the number of clicks grouped by date and topic you can use the following code:

db.coll.group(
{
   $keyf : function(doc) {
       return { "date" : doc.date.getDate()+"/"+doc.date.getMonth()+"/"+doc.date.getFullYear(),
                "topic": doc.topic };
    },
    initial: {count:0},
    reduce: function(obj, prev) { prev.count++; }
 })

Also If you would like to optimize the query as suggested you can use an integer value for date (hint: use valueOf(), for the key date instead of the String, though for my examples the speed was the same.

Furthermore it's always wise to check the MongoDB docs regularly, because they keep adding new features all the time. For example with the new Aggregation framework, which will be released in the 2.2 version you can achieve the same results much easier http://docs.mongodb.org/manual/applications/aggregation/

Upvotes: 2

mindthief
mindthief

Reputation: 13393

Late answer, but for the record (for anyone else that comes to this page): You'll need to use the 'keyf' argument instead of 'key', since your key is actually going to be a function of the date on the event (i.e. the "day" extracted from the date) and not the date itself. This should do what you're looking for:

db.coll.group(
{
    keyf: function(doc) {
        var date = new Date(doc.date);
        var dateKey = (date.getMonth()+1)+"/"+date.getDate()+"/"+date.getFullYear()+'';
        return {'day':dateKey};
    },
    cond: {topic:"abc"},
    initial: {count:0},
    reduce: function(obj, prev) {prev.count++;}
});

For more information, take a look at MongoDB's doc page on aggregation and group: http://www.mongodb.org/display/DOCS/Aggregation#Aggregation-Group

Upvotes: 34

enricog
enricog

Reputation: 4273

Haven't worked that much with MongoDB yet, so I am not completely sure. But aren't you able to use full Javascript?
So you could parse your date with Javascript Date class, create your date for the day out of it and set as key into an "out" property. And always add one if the key already exists, otherwise create it new with value = 1 (first click). Below is your code with adapted reduce function (untested code!):

db.coll.group(
{
   key:{'date':true},
   initial: {retVal: {}},
   reduce: function(doc, prev){
              var date = new Date(doc.date);
              var dateKey = date.getFullYear()+''+date.getMonth()+''+date.getDate();
              (typeof prev.retVal[dateKey] != 'undefined') ? prev.retVal[dateKey] += 1 : prev.retVal[dateKey] = 1;
            }, 
   cond: {topic:"abc"}
}
)

Upvotes: 4

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