MongoDB: how to find 10 random document in a collection of 100?

问题

Is MongoDB capable of funding number of random documents without making multiple queries?

e.g. I implemented on the JS side after loading all the document in the collection, which is wasteful - hence just wanted to check if this can be done better with one db query?

The path I took on the JS side:

get all data
make an array of the IDs
shuffle array of IDs (random order)
splice the array to the number of document required
create a list of document by selecting them by ID which we have left after two previous operations, one by one from the whole collection

Two major drawback are that I am loading all data - or I make multiple queries.

Any suggestion much appreciated

回答1:

This was answered long time ago and, since then, MongoDB has greatly evolved.

As posted in another answer, MongoDB now supports sampling within the Aggregation Framework since version 3.2:

The way you could do this is:

db.products.aggregate([{$sample: {size: 5}}]); // You want to get 5 docs

Or:

db.products.aggregate([
  {$match: {category:"Electronic Devices"}}, // filter the results
  {$sample: {size: 5}} // You want to get 5 docs
]);

However, there are some warnings about the $sample operator:

(as of Nov, 6h 2017, where latest version is 3.4) => If any of this is not met:

$sample is the first stage of the pipeline
N is less than 5% of the total documents in the collection
The collection contains more than 100 documents

If any of the above conditions are NOT met, $sample performs a collection scan followed by a random sort to select N documents.

Like in the last example with the $match

OLD ANSWER

You could always run:

db.products.find({category:"Electronic Devices"}).skip(Math.random()*YOUR_COLLECTION_SIZE)

But the order won't be random and you will need two queries (one count to get YOUR_COLLECTION_SIZE) or estimate how big it is (it is about 100 records, about 1000, about 10000...)

You could also add a field to all documents with a random number and query by that number. The drawback here would be that you will get the same results every time you run the same query. To fix that you can always play with limit and skip or even with sort. you could as well update those random numbers every time you fetch a record (implies more queries).

--I don't know if you are using Mongoose, Mondoid or directly the Mongo Driver for any specific language, so I'll write all about mongo shell.

Thus your, let's say, product record would look like this:

{
 _id: ObjectId("..."),
 name: "Awesome Product",
 category: "Electronic Devices",
}

and I would suggest to use:

{
 _id: ObjectId("..."),
 name: "Awesome Product",
 category: "Electronic Devices",
 _random_sample: Math.random()
}

Then you could do:

db.products.find({category:"Electronic Devices",_random_sample:{$gte:Math.random()}})

then, you could run periodically so you update the document's _random_sample field periodically:

var your_query = {} //it would impact in your performance if there are a lot of records
your_query = {category: "Electronic Devices"} //Update 
//upsert = false, multi = true
db.products.update(your_query,{$set:{_random_sample::Math.random()}},false,true)

or just whenever you retrieve some records you could update all of them or just a few (depending on how many records you've retrieved)

for(var i = 0; i < records.length; i++){
   var query = {_id: records[i]._id};
   //upsert = false, multi = false
   db.products.update(query,{$set:{_random_sample::Math.random()}},false,false);
}

EDIT

Be aware that

db.products.update(your_query,{$set:{_random_sample::Math.random()}},false,true)

won't work very well as it will update every products that matches your query with the same random number. The last approach works better (updating some documents as you retrieve them)

回答2:

Since 3.2 there is an easier way to to get a random sample of documents from a collection:

$sample New in version 3.2.

Randomly selects the specified number of documents from its input.

The $sample stage has the following syntax:

{ $sample: { size: <positive integer> } }

Source: MongoDB Docs

In this case:

db.products.aggregate([{$sample: {size: 10}}]);

回答3:

Here is what I came up in the end:

var numberOfItems = 10;


// GET LIST OF ALL ID's
SchemaNameHere.find({}, { '_id': 1 }, function(err, data) {

    if (err) res.send(err);

    // shuffle array, as per here  https://github.com/coolaj86/knuth-shuffle
    var arr = shuffle(data.slice(0));

    // get only the first numberOfItems of the shuffled array
    arr.splice(numberOfItems, arr.length - numberOfItems);

    // new array to store all items
    var return_arr = [];

    // use async each, as per here http://justinklemm.com/node-js-async-tutorial/
    async.each(arr, function(item, callback) {

        // get items 1 by 1 and add to the return_arr
        SchemaNameHere.findById(item._id, function(err, data) {

            if (err) res.send(err);
            return_arr.push(data);

            // go to the next one item, or to the next function if done
            callback();

        });

    }, function(err) {

        // run this when looped through all items in arr
        res.json(return_arr);

    });

});

回答4:

skip didn't work out for me. Here is what I wound up with:

var randomDoc = db.getCollection("collectionName").aggregate([ {
    $match : {
// criteria to filter matches
    }
}, {
    $sample : {
        size : 1
    }
} ]).result[0];

gets a single random result, matching the criteria.

来源：https://stackoverflow.com/questions/24806721/mongodb-how-to-find-10-random-document-in-a-collection-of-100

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