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Difference between revisions of "AGGREGATE world"

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Line 100: Line 100:
 
         {"$project":{
 
         {"$project":{
 
             "_id":0,
 
             "_id":0,
             "name":1,
+
             "name":1
 
         }}
 
         }}
 
     ])
 
     ])
Line 106: Line 106:
 
</pre>
 
</pre>
 
<div class=ans>
 
<div class=ans>
pp.pprint(list(
+
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"]}}}])))
    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"]}
 
        }}
 
    ])
 
))
 
 
</div>
 
</div>
 
</div>
 
</div>

Revision as of 13:49, 17 July 2015

#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"]}}}])))