AGGREGATE world

#ENCODING
import io
import sys
sys.stdout = io.TextIOWrapper(sys.stdout.buffer, encoding='utf-16')
#MONGO
from pymongo import MongoClient
client = MongoClient()
client.progzoo.authenticate('scott','tiger')
db = client['progzoo']
#PRETTY
import pprint
pp = pprint.PrettyPrinter(indent=4)

Country Profile

For these questions you should use aggregate([]) on the collection world

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.

pp.pprint(list(
    db.world.aggregate([
        {"$match":{
            "population":{"$gte":250000000}
        }},
        {"$project":{
            "_id":0,
            "name":1,
            "per capita GDP": {"$divide": ["$gdp",1000000]}
        }}
    ])
))

pp.pprint(list(db.world.aggregate([{"$match":{"population":{"$gte":200000000}}},{"$project":{"_id":0,"name":1,"per capita GDP": {"$divide": ["$gdp","$population"]}}}])))

Give the name and the population density of all countries. Ignore results where the density is "None".

population density is the population divided by the area

Use a $match. {"area":{"$ne":0}}


pp.pprint(list(
    db.world.aggregate([
        {"$project":{
            "_id":0,
            "name":1,
            "density": {"$divide": [10000,"$area"]}
        }},
        {"$match":{
            "density": {"$ne":None}
        }}
    ])
))


pp.pprint(list(db.world.aggregate([{"$match":{"area":{"$ne":0}}},{"$project":{"_id":0,"name":1,"density":{"$divide":["$population","$area"]}}},{"$match":{"density":{"$ne":None}}}])))


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. 

pp.pprint(list(
    db.world.aggregate([
        {"$match":{
            
        }},
        {"$project":{
            "_id":0,
            "name":1
        }}
    ])
))


pp.pprint(list(db.world.aggregate([{"$match":{"continent":{"$eq":"South America"}}},{"$project":{"_id":0,"name":1,"population":{"$divide":["$population",1000000]}}}])))


Show the name and population density for France, Germany, and Italy

pp.pprint(list(
    db.world.aggregate([
        {"$match":{
            "name": {"$in":['United Kingdom','United States','Brazil']},
            "population": {"$ne": None},
            "area": {"$ne": 0}
        }},
        {"$project":{
            "_id":0,
            "name":1
        }}
    ])
))


pp.pprint(list(db.world.aggregate([{"$match":{"name":{"$in":['France','Germany','Brazil']},"population":{"$ne":None},"area":{"$ne":0}}},{"$project":{"_id":0,"name":1,"population density":{"$divide":["$population","$area"]}}}])))


Order the continents by area from most to least.

pp.pprint(list(
    db.world.aggregate([
        {"$group":{
            "_id":"$name",
            "area":{"$max": "$area"}
        }},
        {"$sort":{
            "area": -1
        }},
        {"$project":{
            "_id":1,
            "area":1
        }}
    ])
))


pp.pprint(list(

   db.world.aggregate([
       {"$group":{
           "_id":"$continent",
           "area":{"$sum": "$area"}
       }},
       {"$sort":{
           "area": -1
       }},
       {"$project":{
           "_id":1,
           "area":1
       }}
   ])

))

Harder Questions


Using Conditions


$cond is similar to a CASE statement in other languages.

It has the form "$cond": [{<comparison> :[<field or value>,<field or value>]},<true case>,<false case>]


Using $cond, reattempt the above question but change Eurasia to Europe

pp.pprint(list(
    db.world.aggregate([
        {"$group":{
            "_id":{
                "$cond": [{"$eq":["$continent","Eurasia"]},"Europe","$continent"]
            },
            "area":{"$sum": "$area"}
        }},
        {"$sort":{
            "area": -1
        }},
        {"$project":{
            "_id":1,
            "area":1
        }}
    ])
))



pp.pprint(list(db.world.aggregate([{"$group":{"_id":{"$cond":[{"$eq":["$continent","Eurasia"]},"Europe","$continent"]},"area":{"$sum":"$area"}}},{"$sort":{"area":-1}},{"$project":{"_id":1,"area":1}}])))


Combine North America and South America to America, and then list the continents by area. Biggest first.

pp.pprint(list(
    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
        }}
    ])
))


pp.pprint(list(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}}])))


Show the name and the continent - but substitute Australasia for Oceania - for countries beginning with N.

pp.pprint(list(
    db.world.aggregate([
        {"$match":{
            "name":{"$regex":"^N"}
        }},
        {"$project":{
            "_id":0,
            "name":1
        }}
    ])
))


pp.pprint(list(db.world.aggregate([{"$match":{"name":{"$regex":"^N"}}},{"$project":{"_id":0,"name":1,"continent":{"$cond":[{"$eq":["$continent","Oceania"]},"Australasia","$continent"]}}}])))


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

You may want to experiment with $and and $or