Difference between revisions of "AGGREGATE world"
| Line 112: | Line 112: | ||
<div class=q data-lang="py3"> | <div class=q data-lang="py3"> | ||
Order the <code>continents</code> by <code>area</code> from most to least. | Order the <code>continents</code> by <code>area</code> from most to least. | ||
| + | <pre class=def> | ||
| + | pp.pprint(list( | ||
| + | db.world.aggregate([ | ||
| + | {"$group":{ | ||
| + | "_id":"$name", | ||
| + | "area":{"$max": "$area"} | ||
| + | }}, | ||
| + | {"$sort":{ | ||
| + | "area": -1 | ||
| + | }}, | ||
| + | {"$project":{ | ||
| + | "_id":1, | ||
| + | "area":1 | ||
| + | }} | ||
| + | ]) | ||
| + | )) | ||
| + | </pre> | ||
| + | <div class=ans> | ||
| + | pp.pprint(list( | ||
| + | db.world.aggregate([ | ||
| + | {"$group":{ | ||
| + | "_id":"$continent", | ||
| + | "area":{"$sum": "$area"} | ||
| + | }}, | ||
| + | {"$sort":{ | ||
| + | "area": -1 | ||
| + | }}, | ||
| + | {"$project":{ | ||
| + | "_id":1, | ||
| + | "area":1 | ||
| + | }} | ||
| + | ]) | ||
| + | )) | ||
| + | </div> | ||
| + | </div> | ||
| + | |||
| + | |||
| + | <div class=q data-lang="py3"> | ||
| + | Do the same again, but combine the Americas. | ||
| + | <div class="hint" title="Using $cond"> | ||
| + | |||
| + | </div> | ||
<pre class=def> | <pre class=def> | ||
pp.pprint(list( | pp.pprint(list( | ||
Revision as of 14:00, 17 July 2015
#ENCODING
import io
import sys
sys.stdout = io.TextIOWrapper(sys.stdout.buffer, encoding='utf-16')
#MONGO
from pymongo import MongoClient
client = MongoClient()
client.progzoo.authenticate('scott','tiger')
db = client['progzoo']
#PRETTY
import pprint
pp = pprint.PrettyPrinter(indent=4)
Country Profile
For these questions you should use aggregate([]) on the collection world
Give the name and the per capita GDP for those countries with a population of at least 200 million.
per capita GDP is the GDP divided by the population.
pp.pprint(list(
db.world.aggregate([
{"$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([
{"$match":{
"name": {"$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
}},
{"$project":{
"_id":1,
"area":1
}}
])
))
Do the same again, but combine the Americas.
pp.pprint(list(
db.world.aggregate([
{"$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
}}
])
))