Difference between revisions of "AGGREGATE world"
| Line 84: | Line 84: | ||
<div class=ans> | <div class=ans> | ||
pp.pprint(list(db.world.aggregate([{"$match":{"continent":{"$eq":"South America"}}},{"$project":{"_id":0,"name":1,"population":{"$divide":["$population",1000000]}}}]))) | pp.pprint(list(db.world.aggregate([{"$match":{"continent":{"$eq":"South America"}}},{"$project":{"_id":0,"name":1,"population":{"$divide":["$population",1000000]}}}]))) | ||
| + | </div> | ||
| + | </div> | ||
| + | |||
| + | |||
| + | <div class=q data-lang="py3"> | ||
| + | Show the <code>name</code> and <code>population density</code> for <b>France</b>, <b>Germany</b>, and <b>Italy</b> | ||
| + | <pre class=def> | ||
| + | pp.pprint(list( | ||
| + | db.world.aggregate([ | ||
| + | {"$match":{ | ||
| + | "name": {"$nin":['United Kingdom','United States','Brazil']}, | ||
| + | "population": {"$ne": None}, | ||
| + | "area": {"$ne": 0} | ||
| + | }}, | ||
| + | {"$project":{ | ||
| + | "_id":0, | ||
| + | "name":1, | ||
| + | "popualtion density": None, | ||
| + | }} | ||
| + | ]) | ||
| + | )) | ||
| + | </pre> | ||
| + | <div class=ans> | ||
| + | pp.pprint(list( | ||
| + | db.world.aggregate([ | ||
| + | {"$match":{ | ||
| + | "name": {"$in":['France','Germany','Brazil']}, | ||
| + | "population": {"$ne": None}, | ||
| + | "area": {"$ne": 0} | ||
| + | }}, | ||
| + | {"$project":{ | ||
| + | "_id":0, | ||
| + | "name":1, | ||
| + | "population density": {"$divide":["$population","$area"]} | ||
| + | }} | ||
| + | ]) | ||
| + | )) | ||
</div> | </div> | ||
</div> | </div> | ||
Revision as of 12:45, 17 July 2015
#ENCODING
import io
import sys
sys.stdout = io.TextIOWrapper(sys.stdout.buffer, encoding='utf-16')
#MONGO
from pymongo import MongoClient
client = MongoClient()
client.progzoo.authenticate('scott','tiger')
db = client['progzoo']
#PRETTY
import pprint
pp = pprint.PrettyPrinter(indent=4)
Country Profile
For these questions you should use aggregate([]) on the collection world
Give the name and the per capita GDP for those countries with a population of at least 200 million.
per capita GDP is the GDP divided by the population.
pp.pprint(list(
db.world.aggregate([
{"$match":{
"population":{"$gte":250000000}
}},
{"$project":{
"_id":0,
"name":1,
"per capita GDP": {"$divide": ["$gdp",1000000]}
}}
])
))
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([
{"$match":{
"name": {"$nin":['United Kingdom','United States','Brazil']},
"population": {"$ne": None},
"area": {"$ne": 0}
}},
{"$project":{
"_id":0,
"name":1,
"popualtion density": None,
}}
])
))
pp.pprint(list(
db.world.aggregate([
{"$match":{
"name": {"$in":['France','Germany','Brazil']},
"population": {"$ne": None},
"area": {"$ne": 0}
}},
{"$project":{
"_id":0,
"name":1,
"population density": {"$divide":["$population","$area"]}
}}
])
))