MongoDB统计文档(Document)的数组(Array)中的各个元素出现的次数

Posted 2020-10-16 hapjin

一，问题描述

【使用 unwind 操作符 “解包” Document 里面的Array中的每个元素，然后使用 group 分组统计，最后使用 sort 对分组结果排序】

 

从 images.json 文件中导入数据到MongoDB服务器

mongoimport --drop -d test -c images images.json

其中Document的示例如下：

> db.images.find()
{ "_id" : 3, "height" : 480, "width" : 640, "tags" : [ "kittens", "travel" ] }
{ "_id" : 1, "height" : 480, "width" : 640, "tags" : [ "cats", "sunrises", "kittens", "travel", "vacation", "work" ] }
{ "_id" : 0, "height" : 480, "width" : 640, "tags" : [ "dogs", "work" ] }
{ "_id" : 6, "height" : 480, "width" : 640, "tags" : [ "work" ] }
{ "_id" : 4, "height" : 480, "width" : 640, "tags" : [ "dogs", "sunrises", "kittens", "travel" ] }
{ "_id" : 5, "height" : 480, "width" : 640, "tags" : [ "dogs", "cats", "sunrises", "kittens", "work" ] }
{ "_id" : 7, "height" : 480, "width" : 640, "tags" : [ "dogs", "sunrises" ] }
{ "_id" : 8, "height" : 480, "width" : 640, "tags" : [ "dogs", "cats", "sunrises", "kittens", "travel" ] }

现在要统计： 所有Document中的 tags 数组里面的每个元素 出现的次数。即："kittens"出现了多少次？"travel"出现了多少次？"dogs"出现了多少次？……

 

二，实现步骤

使用MongoDB的Aggregate操作进行实现

①使用 unwind 分解 tags 数组，得到的结果如下：

> db.images.aggregate(
... [
... {$unwind:"$tags"}
... ])

{ "_id" : 3, "height" : 480, "width" : 640, "tags" : "kittens" }
{ "_id" : 3, "height" : 480, "width" : 640, "tags" : "travel" }
{ "_id" : 1, "height" : 480, "width" : 640, "tags" : "cats" }
{ "_id" : 1, "height" : 480, "width" : 640, "tags" : "sunrises" }
{ "_id" : 1, "height" : 480, "width" : 640, "tags" : "kittens" }
{ "_id" : 1, "height" : 480, "width" : 640, "tags" : "travel" }
{ "_id" : 1, "height" : 480, "width" : 640, "tags" : "vacation" }
{ "_id" : 1, "height" : 480, "width" : 640, "tags" : "work" }
{ "_id" : 0, "height" : 480, "width" : 640, "tags" : "dogs" }
{ "_id" : 0, "height" : 480, "width" : 640, "tags" : "work" }
{ "_id" : 6, "height" : 480, "width" : 640, "tags" : "work" }
{ "_id" : 4, "height" : 480, "width" : 640, "tags" : "dogs" }
{ "_id" : 4, "height" : 480, "width" : 640, "tags" : "sunrises" }
.....
.....

 

②将分解后的每个 tag 进行 group 操作

对于group操作而言，_id 指定了 分组 的字段(对哪个字段进行 group by 操作)，分组操作之后生成的结果由 num_of_tag 字段标识

> db.images.aggregate(
... [
... {$unwind:"$tags"},
... {$group:{_id:"$tags",num_of_tag:{$sum:1}}}
... ]
... )

{ "_id" : "dogs", "num_of_tag" : 49921 }
{ "_id" : "work", "num_of_tag" : 50070 }
{ "_id" : "vacation", "num_of_tag" : 50036 }
{ "_id" : "travel", "num_of_tag" : 49977 }
{ "_id" : "kittens", "num_of_tag" : 49932 }
{ "_id" : "sunrises", "num_of_tag" : 49887 }
{ "_id" : "cats", "num_of_tag" : 49772 }

 

③使用 project 去掉不感兴趣的 _id 字段（其实这里是将 _id 字段名 替换为 tags 字段名）（这一步可忽略）

project操作，_id:0 表示去掉_id 字段；tags:"$_id"，将 _id 字段值 使用tags 字段标识；num_of_tag:1 保留 num_of_tag 字段

> db.images.aggregate( [ {$unwind:"$tags"},{$group:{_id:"$tags",num_of_tag:{$sum:1}}},{$project:{_id:0,tags:"$_id",num_of_tag:1}} ])
{ "num_of_tag" : 49921, "tags" : "dogs" }
{ "num_of_tag" : 50070, "tags" : "work" }
{ "num_of_tag" : 50036, "tags" : "vacation" }
{ "num_of_tag" : 49977, "tags" : "travel" }
{ "num_of_tag" : 49932, "tags" : "kittens" }
{ "num_of_tag" : 49887, "tags" : "sunrises" }
{ "num_of_tag" : 49772, "tags" : "cats" }

 

④使用 sort 对 num_of_tag 字段排序

> db.images.aggregate( [ {$unwind:"$tags"},{$group:{_id:"$tags",num_of_tag:{$sum:1}}},{$project:{_id:0,tags:"$_id",num_of_tag:1}},{$sort:{num_of_tag:-1}} ])
{ "num_of_tag" : 50070, "tags" : "work" }
{ "num_of_tag" : 50036, "tags" : "vacation" }
{ "num_of_tag" : 49977, "tags" : "travel" }
{ "num_of_tag" : 49932, "tags" : "kittens" }
{ "num_of_tag" : 49921, "tags" : "dogs" }
{ "num_of_tag" : 49887, "tags" : "sunrises" }
{ "num_of_tag" : 49772, "tags" : "cats" }

 

三，总结

本文是MongoDB University M101课程 For Java Developers中的一次作业。结合Google搜索和MongoDB的官方文档，很容易就能实现MongoDB的各种组合查询。

相关MongoDB文章：

MongoDB 更新数组中的元素

MongoDB 组合多个条件查询（$and、$in、$gte、$lte）

 

原文：http://www.cnblogs.com/hapjin/p/7944404.html