AGGREGATE Tutorial
Contents
- 1 Country Profile
- 2 $group on continent
- 3 Per Capita GDP
- 4 Population Density in South America
- 5 Population Density for "V"
- 6 Population in millions
- 7 Population density
- 8 Continents by area
- 9 Big Continents
- 10 First and last country by continent
- 11 Countries beginning with...
- 12 Harder Questions
- 13 Messing with continent names
- 14 Messing with continent names 2
- 15 Messing with continent names 3
- 16 Messing with continent names 4
Country Profile
For these questions you should use aggregate([])
on the collection world
You may find these AGGREGATE examples useful.
$group on continent
The aggregate method allows a $group
- you must specify the _id
and you can use aggregating functions such as $sum
$min
$max
$push
The sample code shows the total population of each continent.
db.world.aggregate({ $group: { _id: '$continent', res: { $sum: '$population' } } });
db.world.aggregate({ $group: { _id: '$continent', res: { $sum: 1 } } });
Per Capita GDP
Give the name
and the per capita GDP
for those countries with a population
of at least 200 million.
per capita GDP is the GDP divided by the population.
db.world.aggregate([ {$match: { population: {$gte: 250000000} }}, {$project: { _id: 0, name: 1, "per capita GDP": {$divide: ['$gdp', 1000000]} }} ]);
db.world.aggregate([{"$match":{"population":{"$gte":200000000}}},{"$project":{"_id":0,"name":1,"per capita GDP": {"$divide": ["$gdp","$population"]}}}]);
Population Density in South America
Give the name
and the population density
of all countries in South America.
population density is the population divided by the area
Use a $match
. {"area":{"$ne":0}}
db.world.aggregate([ {$match: {continent: 'Asia'}}, {$project: { _id: 0, name: 1, density: {$divide: ["$population", "$area"]} }} ]);
db.world.aggregate([{$match:{continent:'South America'}},{$project:{_id:0,name:1,density:{$divide:["$population","$area"]}}}]);
Population Density for "V"
Give the name
and the population density
of all countries with name after V in the alphabet.
Note that because Vatican City (with area 0) is in Europe you will get a divide by zero error unless you filter first.
Use a $match
.
{ $match: { area: { "$ne": 0 } } }
db.world.aggregate([ {$match: {name: {$gt: 'V'}}}, {$project: { _id: 0, name: 1, area: 1 }} ]);
db.world.aggregate([{$match:{name:{$gt:'V'}}},{$match:{area:{"$ne":0}}},{$project:{_id:0,name:1,density:{$divide:["$population","$area"]}}}]);
Population in millions
Show the name
and population
in millions for the countries of the continent South America. Divide the population by 1000000 to get population in millions.
db.world.aggregate([ {$match:{ }}, {$project:{ _id: 0, name: 1 }} ]);
db.world.aggregate([{"$match":{"continent":{"$eq":"South America"}}},{"$project":{"_id":0,"name":1,"population":{"$divide":["$population",1000000]}}}]);
Population density
Show the name
and population density
for France, Germany, and Italy
db.world.aggregate([ {$match:{ name: {$in: ['United Kingdom', 'United States', 'Brazil']}, population: {$ne: null}, area: {$ne: 0} }}, {$project:{ _id: 0, name: 1 }} ]);
db.world.aggregate([{"$match":{"name":{"$in":['France','Germany','Italy']},"population":{"$ne":null},"area":{"$ne":0}}},{"$project":{"_id":0,"name":1,"population density":{"$divide":["$population","$area"]}}}]);
Continents by area
Order the continents
by area
from most to least.
db.world.aggregate([ {$group: { _id: "$name", area: {$max: "$area"} }}, {$sort: { area: -1 }}, {$project: { _id: 1, area: 1 }} ]);
db.world.aggregate([{"$group":{"_id":"$continent","area":{"$sum":"$area"}}},{"$sort":{"area":-1}},{"$project":{"_id":1,"area":1}}]);
Big Continents
Show the only two continents with total area greater than 25000000 and then sort from largest to smallest.
db.world.aggregate([ {$match: { continent: "North America" }}, {$project: { _id: 0, name: 1 }} ]);
db.world.aggregate([{$group:{_id:"$continent",area:{$sum:"$area"}}},{$sort:{area:-1}},{$match:{area:{$gt:25000000}}}]);
First and last country by continent
For each continent show the first and last country alphabetically like this:
{ "_id" : "Africa", "from" : "Algeria", "to" : "Zimbabwe" }
{ "_id" : "Asia", "from" : "Afghanistan", "to" : "Yemen" }
{ "_id" : "Caribbean", "from" : "Antigua and Barbuda", "to" : "Trinidad and Tobago" }
{ "_id" : "Eurasia", "from" : "Armenia", "to" : "Russia" }
{ "_id" : "Europe", "from" : "Albania", "to" : "Vatican City" }
{ "_id" : "North America", "from" : "Belize", "to" : "United States" }
{ "_id" : "Oceania", "from" : "Australia", "to" : "Vanuatu" }
{ "_id" : "South America", "from" : "Argentina", "to" : "Venezuela" }
db.world.aggregate([ {$group: { _id: "$continent" }}, {$sort: { _id: 1 }} ]);
db.world.aggregate([{$sort:{name:1}},{$group:{_id:'$continent',from:{$first:'$name'},to:{$last:'$name'}},},{$sort:{_id:1}}])
Countries beginning with...
Group countries according to the first letter of the name. As shown. Only give "U" through to "Z".
You will need to use the $substr function and the $push aggregate function.
{ "_id" : "U", "list" : [ "Uganda", "Ukraine", "United Arab Emirates", "United Kingdom", "United States", "Uruguay", "Uzbekistan" ] }
{ "_id" : "V", "list" : [ "Vanuatu", "Vatican City", "Venezuela", "Vietnam" ] }
{ "_id" : "Y", "list" : [ "Yemen" ] }
{ "_id" : "Z", "list" : [ "Zambia", "Zimbabwe" ] }
db.world.aggregate([ {$project: { _id: '$name', startsWith: {$substr: ['$name', 0, 1]} }}, {$match: { _id: {$gte: 'U'} }}, {$sort: {_id: 1}} ]);
db.world.aggregate([{$group:{_id:{$substr:['$name',0,1]},list:{$push:'$name'}}},{$match:{_id:{$gte:'U'}}},{$sort:{_id:1}}]);
Harder Questions
Messing with continent names
Combine North America and South America to America, and then list the continents by area. Biggest first.
db.world.aggregate([ {$group: { _id: { $cond: [ {$eq: ["$continent", "North America"]}, "America", {$cond: [ {$eq: ["$continent", "Asia"]}, "The East", "$continent" ]} ] }, area: {$sum: "$area"} }}, {$sort: {area: -1}}, {$project: { _id: 1, area: 1 }} ]);
db.world.aggregate([{"$group":{"_id":{"$cond":[{"$eq":["$continent","South America"]},"America",{"$cond":[{"$eq":["$continent","North America"]},"America","$continent"]}]},"area":{"$sum":"$area"}}},{"$sort":{"area":-1}},{"$project":{"_id":1,"area":1}}]);
Messing with continent names 2
Show the name and the continent for countries beginning with N - but replace the continent Oceania with Australasia.
db.world.aggregate([ {$match: { name: {$regex: "^N"} }}, {$project: { _id: 0, name: 1 }} ]);
db.world.aggregate([{"$match":{"name":{"$regex":"^N"}}},{"$project":{"_id":0,"name":1,"continent":{"$cond":[{"$eq":["$continent","Oceania"]},"Australasia","$continent"]}}}]);
Messing with continent names 3
Show the name and the continent but:
- substitute Eurasia for Europe and Asia.
- substitute America - for each country in North America or South America or Caribbean.
Only show countries beginning with A or B
If you're struggling you may want to experiment with $and
,$or
, etc.
db.world.aggregate([{$match:{name:{$regex:"^A|^B"}}},{$project:{_id:0,name:1,continent:{$cond:[{$or:[{$eq:["$continent","Europe"]},{$eq:["$continent","Asia"]}]},"Eurasia",{$cond:[{$or:[{$eq:["$continent","North America"]},{$eq:["$continent","South America"]},{$eq:["$continent","Caribbean"]}]},"America","$continent"]}]}}}]);
Messing with continent names 4
Put the continents right...
- Oceania becomes Australasia
- Countries in Eurasia and Turkey go to Europe/Asia
- Caribbean islands starting with 'B' go to North America, other Caribbean islands go to South America
Show the name, the original continent and the new continent of all countries.
db.world.aggregate([{"$project":{"_id":0,"name":1,"original":"$continent","new":{"$cond":[{"$or":[{"$eq":["$continent","Eurasia"]},{"$eq":["$name","Turkey"]}]},"Europe/Asia",{"$cond":[{"$eq":["$continent","Oceania"]},"Australasia",{"$cond":[{"$and":[{"$eq":["$continent","Caribbean"]},{"$eq":[{"$substr":["$name",0,1]},"B"]}]},"North America",{"$cond":[{"$and":[{"$eq":["$continent","Caribbean"]},{"$ne":[{"$substr":["$name",0,1]},"B"]}]},"South America","$continent"]}]}]}]}}}]);