问题
I am performing a count of documents in a mongo (versions 2.4 and 3.2) collection. The collection is very big, 3821085 documents. I need to count all documents with a reference _id. I have tried two different queries:
db.SampleCollection.find({"field._id" : ObjectId("UUID")}).count()
db.SampleCollection.count({"field._id" : ObjectId("UUID")})
This query takes a very long time. So much time that I have not let it complete, more than 5 minutes and I get scared and kill it.
For this collection field._id is not an index. I do not have relevant info to use an index with this query.
Is there a better approach to count document in mongo.
UPDATE:
I understand that I need an index on the field field._id. If I did have an index for the field which approach would perform better on a large collection db.SampleCollection.find(...).count() or db.SampleCollection.count(...)? Or is there no difference between the two?
回答1:
In your scenario, you should have an index.
Indexes support the efficient execution of queries in MongoDB. Without indexes, MongoDB must perform a collection scan, i.e. scan every document in a collection, to select those documents that match the query statement.
https://docs.mongodb.com/manual/indexes/
UPDATE:
the question asked now is different. Is "collection.find({}).count()" more fast then "collection.count()"?
According to the MongoDB documentation:
count() is equivalent to the db.collection.find(query).count() construct. https://docs.mongodb.com/manual/reference/method/db.collection.count/
回答2:
You should add an index on field._id like this:
db.SampleCollection.createIndex( { "field._id": 1 } );
Then all the queries trying to find/count documents by that field will use this index and will perform faster. For example:
db.SampleCollection.count({"field._id" : ObjectId("UUID")});
See - https://docs.mongodb.com/manual/core/index-single/ and MongoDB 'count()' is very slow. How do we refine/work around with it?
来源：https://stackoverflow.com/questions/40061221/mongodb-count-vs-find-with-count