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Mongodb官方网站提供了一个美国人口统计数据，下载地址如下
http://media.mongodb.org/zips.json
数据示例：
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| [iyunv@localhost cluster]# head zips.json
{ "_id" : "01001", "city" : "AGAWAM", "loc" : [ -72.622739, 42.070206 ], "pop" : 15338, "state" : "MA" }
{ "_id" : "01002", "city" : "CUSHMAN", "loc" : [ -72.51564999999999, 42.377017 ], "pop" : 36963, "state" : "MA" }
{ "_id" : "01005", "city" : "BARRE", "loc" : [ -72.10835400000001, 42.409698 ], "pop" : 4546, "state" : "MA" }
{ "_id" : "01007", "city" : "BELCHERTOWN", "loc" : [ -72.41095300000001, 42.275103 ], "pop" : 10579, "state" : "MA" }
{ "_id" : "01008", "city" : "BLANDFORD", "loc" : [ -72.936114, 42.182949 ], "pop" : 1240, "state" : "MA" }
{ "_id" : "01010", "city" : "BRIMFIELD", "loc" : [ -72.188455, 42.116543 ], "pop" : 3706, "state" : "MA" }
{ "_id" : "01011", "city" : "CHESTER", "loc" : [ -72.988761, 42.279421 ], "pop" : 1688, "state" : "MA" }
{ "_id" : "01012", "city" : "CHESTERFIELD", "loc" : [ -72.833309, 42.38167 ], "pop" : 177, "state" : "MA" }
{ "_id" : "01013", "city" : "CHICOPEE", "loc" : [ -72.607962, 42.162046 ], "pop" : 23396, "state" : "MA" }
{ "_id" : "01020", "city" : "CHICOPEE", "loc" : [ -72.576142, 42.176443 ], "pop" : 31495, "state" : "MA" }
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使用mongoimport将数据导入mongodb数据库
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| [iyunv@localhost cluster]# mongoimport -d test -c "zipcodes" --file zips.json -h 192.168.199.219:27020
2016-01-16T18:31:29.424+0800 connected to: 192.168.199.219:27020
2016-01-16T18:31:32.420+0800 [################........] test.zipcodes 2.1 MB/3.0 MB (68.5%)
2016-01-16T18:31:34.471+0800 [########################] test.zipcodes 3.0 MB/3.0 MB (100.0%)
2016-01-16T18:31:34.471+0800 imported 29353 documents
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一、单一目的的聚合操作
求count，distinct等简单操作
实例1.1：求zipcodes集合的文档数
实例1.2 求MA州的文档总数
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| db.zipcodes.count({state:"MA"})
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实例1.3 求zipcodes中有哪些州
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| db.zipcodes.distinct("state")
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二、使用aggregate聚合框架，进行更复杂的聚合操作
实例2.1：统计每个州的人口总数
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| db.zipcodes.aggregate(
[
{ $group: { _id: "$state", total: { $sum: "$pop" } } }
]
)
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使用集合的aggregate方法，进行聚合查询。
$group关键字后面指定分组的字段(引用字段时，一定要用$前缀)，以及聚合函数。
_id:是关键字，代表返回结果集的主键。
该查询等价的SQL为
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| select state as _id,sum(pop) as total
from zipcodes
group by state
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实例2.2：统计每个州每个城市的人口总数
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| db.zipcodes.aggregate(
[
{ $group: { _id: {state:"$state",city:"$city"}, pop: { $sum: "$pop" } } },
]
)
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分组的字段如果多于一个，那么每个字段都要给定一个别名,如 state:"$state"
实例2.3：统计每个州人口多于10000的城市的人口总和
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| db.zipcodes.aggregate(
[
{ $match: {"pop":{$gt: 10000} }},
{ $group: { _id: {state:"$state"}, pop: { $sum: "$pop" } } },
]
)
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$match 关键字后面跟上集合的过滤条件 。该语句等价于如下SQL
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| select state,sum(pop) as pop
from zipcodes
where pop>10000
group by state
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实例2.4：查询人口总数超过1千万的州
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| db.zipcodes.aggregate(
[
{ $group: { _id: {state:"$state"}, pop: { $sum: "$pop" } } },
{ $match: {"pop":{$gt: 1000*10000} }}
]
)
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将$match放在$group后面，相当于是先执行group操作，再对结果集进行过滤。等价的sql如下
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| select state,sum(pop) as pop
from zipcodes
group by state
having sum(pop)>1000*10000
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实例5：求每个州城市的平均人口
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| db.zipcodes.aggregate(
[
{ $group: { _id: {state:"$state",city:"$city"}, pop: { $sum: "$pop" } } },
{ $group: {_id:"$_id.state",avgPop:{$avg: "$pop"}}}
]
)
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我们的aggregate函数支持多次迭代，该语句的等价sql为
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| select state,avg(pop) as avgPop
from
(select state,city,sum(pop) pop
from zipcodes
group by state,city)
group by state
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实例2.5 ：求每个州人口最多及最少的城市名及对应的人口数量
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| db.zipcodes.aggregate(
[
{ $group: { _id: {state:"$state",city:"$city"}, cityPop: { $sum: "$pop" } } },
{ $sort: { cityPop: 1 } },
{ $group: {
_id:"$_id.state",
biggestCity:{$last:"$_id.city"},
biggestPop:{$last:"$cityPop"},
smallestCity:{$first:"$_id.city"},
smallestPop:{$first:"$cityPop"}
}}
]
)
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第一个$group求出按state，city分组的人口数。
$sort操作按照人口数排序
第二个$group 按照state分组，此时每个state分组的数据已经安装cityPop排序。每个组的第一行数据（$first 取得）是人口最少的city，最后一行（$last 取得）是人口最多的city。
实例2.6 利用$project重新格式化结果
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| db.zipcodes.aggregate(
[
{ $group: { _id: {state:"$state",city:"$city"}, cityPop: { $sum: "$pop" } } },
{ $sort: { cityPop: 1 } },
{
$group: {
_id:"$_id.state",
biggestCity:{$last:"$_id.city"},
biggestPop:{$last:"$cityPop"},
smallestCity:{$first:"$_id.city"},
smallestPop:{$first:"$cityPop"}
}
},
{
$project: {
_id:0,
state: "$_id",
biggestCity: { name: "$biggestCity", pop: "$biggestPop" },
smallestCity: { name: "$smallestCity", pop: "$smallestPop" }
}
}
]
)
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实例2.7 对数组中的内容做聚合统计
我们假设有一个学生选课的集合，数据示例如下
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| db.course.insert({name:"张三",age:10,grade:"四年级",course:["数学","英语","政治"]})
db.course.insert({name:"李四",age:9,grade:"三年级",course:["数学","语文","自然"]})
db.course.insert({name:"王五",age:11,grade:"四年级",course:["数学","英语","语文"]})
db.course.insert({name:"赵六",age:9,grade:"四年级",course:["数学","历史","政治"]})
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求每门课程有多少人选修
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| db.course.aggregate(
[
{ $unwind: "$course" },
{ $group: { _id: "$course", sum: { $sum: 1 } } },
{ $sort: { sum: -1 } }
]
)
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$unwind，用来将数组中的内容拆包，然后再按照拆包后的数据进行分组，另外aggregate中没有$count关键字，使用$sum:1 来计算count 。
实例2.8 求每个州有哪些city。
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| db.zipcodes.aggregate(
[
{ $group: { _id: "$state", cities: { $addToSet: "$city"} } },
]
)
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$addToSet 将每个分组的city内容，写到一个数组中。
假设我们有如下数据结构
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| db.book.insert({
_id: 1,
title: "MongoDB Documentation",
tags: [ "Mongodb", "NoSQL" ],
year: 2014,
subsections: [
{
subtitle: "Section 1: Install MongoDB",
tags: [ "NoSQL", "Document" ],
content: "Section 1: This is the content of section 1."
},
{
subtitle: "Section 2: MongoDB CRUD Operations",
tags: [ "Insert","Mongodb" ],
content: "Section 2: This is the content of section 2."
},
{
subtitle: "Section 3: Aggregation",
tags: [ "Aggregate" ],
content: {
text: "Section 3: This is the content of section3.",
tags: [ "MapReduce","Aggregate" ]
}
}
]
})
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该文档描述书的章节内容，每章节有tags字段，书本身也有tags字段。
如果客户有需要，查询带有标签Mongodb的书，以及只显示有标签Mongodb的章节。我们使用find()方法是无法满足的。
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| db.book.find(
{
$or:
[{tags:{$in: ['Mongodb']}},
{"subsections.tags":{$in: ['Mongodb']}}
]
}
)
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上面类似的查询，会显示命中文档的所有部分，把不包含Mongodb标签的章节也显示出来了。
Aggregate提供了一个$redact表达式，可以对结果进行裁剪。
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| db.book.aggregate(
[
{$redact: {
$cond: {
if: {
$gt:[ {$size: {$setIntersection: ["$tags",["Mongodb"]] }},0]
},
then:"$$DESCEND" ,
else: "$$PRUNE"
}
}}
]
)
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$$DESCEND 如果满足条件，则返回条件tags字段，对于内嵌文档，则返回父级字段。所有判断条件会作用到内嵌文档中。
$$PRUNE 如果不满足条件，则不显示该字段。
查询结果如下
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| {
"_id" : 1,
"title" : "MongoDB Documentation",
"tags" : [
"Mongodb",
"NoSQL"
],
"year" : 2014,
"subsections" : [
{
"subtitle" : "Section 2: MongoDB CRUD Operations",
"tags" : [
"Insert",
"Mongodb"
],
"content" : "Section 2: This is the content of section 2."
}
]
}
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三、使用mapReduce
实例3.1 ：统计每个州的人口总数
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| db.zipcodes.mapReduce(
function () {emit(this.state, this.pop)}, //mapFunction
(key, values)=>{return Array.sum(values)},//reduceFunction
{ out: "zipcodes_groupby_state"}
)
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使用mapReduce，最少有三个参数，map函数、reduce函数、out输出参数。
map函数中，this表示处理的当前文档。emit函数，将传入的键值对传出给reduce函数。
reduce接受map函数的输出，作为输入。reduce中的values是一个列表。对上例来说，state是键，相同state的每条记录对应的pop组成一个列表作为值。形式如下
state = "CA" values=[51841,40629,...]
reduce函数的key是默认一定会返回的，return的返回值，将values中的值相加。作为值。
out：输出结果保存的集合
实例3.2 统计每个城市的人口数，及每个城市的文档个数。
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| db.zipcodes.mapReduce(
function () {
var key = {state:this.state,city:this.city}
emit(key, {count:1,pop:this.pop})
}, //mapFunction
(key, values)=>{
var retval = {count:0,pop:0}
for (var i =0;i< values.length;i++){
retval.count += values.count
retval.pop += values.pop
}
return retval
},//reduceFunction
{ out: "zipcodes_groupby_state_city"}
)
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我们将{state，city}作为一个对象当成值，传递给map函数的key。将{count：1，pop:this.pop}对象传递给map的value 。
再reduce函数中再次计算count，pop的值。返回。
等价的sql如下
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| select state,city,count(*) as count,sum(pop) as pop
from zipcodes
group by state,city
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