如何在MongoDB中按楼层汇总 [英] How to aggregate by floor in MongoDB

问题描述

MongoDB聚合框架不具有floor功能. 它只有简单的算术运算符. 那么，如何使用它们组成floor函数呢?

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

推荐答案

根据定义floor(number) = number - (number % step)，我们可以编写聚合公式:

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

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

其中，step是数量级.首先，它使用$mod计算余数，然后使用$subtract计算差异.

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

因此，将按age划分为整个个数字的用户分组为

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

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

如果您想降低到10s或1000s，请使用10或1000而不是1.

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

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