Question

If I have a collection with thousands of elements, is there a way I can easily find which elements are taking up the most space (in terms of MB)?

Answer 1

没有这方面的内在询问,你必须重新收集资料,收集每份文件的篇幅,然后进行分类。如何开展工作:

var cursor = db.coll.find(); 
var doc_size = {}; 
cursor.forEach(function (x) { 
    var size = Object.bsonsize(x); 
    doc_size[x._id] = size;
});

At this point you ll have a hashmap with document ids as keys and their sizes as values. Note that with this approach you will be fetching the entire collection over the wire. An alternative is to use MapReduce and do this server-side (inside mongo):

> function mapper() {emit(this._id, Object.bsonsize(this));}
> function reducer(obj, size_in_b) { return { id : obj, size : size_in_b}; }
>
> var results = db.coll.mapReduce(mapper, reducer, {out : {inline : 1 }}).results
> results.sort(function(r1, r2) { return r2.value - r1.value; })

在线:1 告诉mongo不要为成果设立临时收集,所有工作都将留在援助团。

我收集的样本:

[
    {
        "_id" : ObjectId("4ce9339942a812be22560634"),
        "value" : 1156115
    },
    {
        "_id" : ObjectId("4ce9340442a812be24560634"),
        "value" : 913413
    },
    {
        "_id" : ObjectId("4ce9340642a812be26560634"),
        "value" : 866833
    },
    {
        "_id" : ObjectId("4ce9340842a812be28560634"),
        "value" : 483614
    },
       ...
    {
        "_id" : ObjectId("4ce9340742a812be27560634"),
        "value" : 61268
    }
]
>

Answer 2

Figured this out! I did this in two steps using Object.bsonsize():

db.myCollection.find().forEach(function(myObject) {
    db.objectSizes.save({object_id: object._id, size: Object.bsonsize(chain)});
});

db.objectSizes.find().sort({size: -1}).limit(5).pretty();

友情链接