How to aggregate by floor in MongoDB

Question

MongoDB Aggregation Framework doesn't have floor function. It only has simple arithmetic operators. So, how to compose floor function using them?

Possible duplicate of [Is there a floor function in Mongodb aggregation framework?](http://stackoverflow.com/questions/13617526/is-there-a-floor-function-in-mongodb-aggregation-framework) — Salvador Dali, Nov 13 '15 at 07:02

score 7 · Accepted Answer · answered Sep 13 '15 at 20:18

According to definition floor(number) = number - (number % step) we can compose our aggregation formula:

{$subtract: ["$number", {$mod: ["$number", <step>]}]}

where step is order of magnitude. First, it computes remainder with $mod and then it computes difference with $subtract.

So, grouping Users by age floored to whole numbers will be

db.users.aggregate(
    {$group: {_id: {$subtract: ["$age", {$mod: ["$age", 1]}] }}} )

If you want to floor to 10s or 1000s use 10 or 1000 instead of 1.

score 0 · Answer 2 · answered Jul 23 '22 at 18:37

0

Since mongoDB version 3.2 there is a $floor function, so you can simply use:

db.users.aggregate([
    {$group: {_id: {$floor: "$age"}}}
])

answered Jul 23 '22 at 18:37

nimrod serok

2 Answers2