Wrong Aggregation when nulls (Mongodb)

Question

So I have data which has null or no document for certain data points. Here is some set of commands demonstrating data

db.perfR.find({platform:"mac", "filename" : "f1.json"},{filename:1, platform:1, int_secs:1, version:1, _id:0})

{ "filename" : "f1.json", "platform" : "mac", "version" : "1.4.14", "int_secs" : 15 }

Indicates only one entry for "Mac"

db.perfR.find({platform:"win", "filename" : "f1.json"},{filename:1, platform:1, int_secs:1, version:1, _id:0}).count()

0

Indicates no entry for Win

Here is the aggregation command

db.perfR.aggregate([ {$sort:{version: -1 }},{ $group:{ _id: {platform:"win", "filename" : "f1.json", version:"$version"}, avgSecs: {$avg:"$int_secs"} } }])

And output

{ "_id" : { "platform" : "win", "filename" : "f1.json", "version" : "1.4.13" }, "avgSecs" : 759 }
{ "_id" : { "platform" : "win", "filename" : "f1.json", "version" : "1.4.14" }, "avgSecs" : 415.46153846153845 }
{ "_id" : { "platform" : "win", "filename" : "f1.json", "version" : "3.0.0_dev" }, "avgSecs" : 563.8333333333334 }

I am quiet confused where does Mongodb get all that data from and how to restrict it to look for values that are present.

Note: There are other entries for other versions, which I have not shown here for brevity.

Any pointers to improve the aggregation command would be useful

Sergey Berezovskiy · Accepted Answer · 2017-02-08 12:43:43Z

1

You are providing string literals in your group _id specification:

_id: {platform:"win", "filename" : "f1.json", version:"$version"}

"win" and "f1.json" are just string literals. Only one field from documents is used in _id - it's "$version". So you got three groups for three different version values which you have in your documents.

Correct _id definition would be

_id: {platform:"$platform", filename : "$filename", version:"$version"}

That will group documents by three fields.

NOTE: if you want to filter collection by some field, you should use $match stage for that.

answered Feb 8, 2017 at 12:43

Sergey Berezovskiy

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