How does mongodb use an index to count documents?

Question

According to docs, db.collection.countDocuments() wraps this:

db.collection.aggregate([
   { $match: <query> },
   { $group: { _id: null, n: { $sum: 1 } } }
])

Even if there is an index, all of the matched docs will be passed into the $group to be counted, no?

If not, how is mongodb able to count the docs without processing all matching docs?

Sometimes it processes all of the matching docs. Depending on the filter, some counts can be covered by the index - https://www.mongodb.com/docs/manual/core/query-optimization/#covered-query — user20042973, Feb 14 '23 at 17:36

score 1 · Accepted Answer · answered Feb 15 '23 at 17:03

1

The MongoDB query planner can make some optimizations.

In that sample aggregation, it can see that no fields are required except for the ones referenced in <query>, so it can add an implicit $project stage to select only those fields.

If those fields and the _id are all included in a single index, there is no need to fetch the documents to execute that query, all the necessary information is available from the index.

answered Feb 15 '23 at 17:03

Joe

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But still, does it have to count the index? Is there enough information to derive the total count with some arithmetic? That's actually the meat of the question. – Bear Bile Farming is Torture Feb 15 '23 at 17:07
1

It scans the index, yes. There is no mechanism to return the number of keys in an index range without reading each index entry. – Joe Feb 15 '23 at 17:14
Log lines, profiler entries, explain output etc help confirm this – user20042973 Feb 16 '23 at 00:43

How does mongodb use an index to count documents?

1 Answers1