Difference between revisions of "AGGREGATE world"
| Line 233: | Line 233: | ||
<pre class=def></pre> | <pre class=def></pre> | ||
<div class=ans> | <div class=ans> | ||
| + | pp.pprint(list( | ||
| + | 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"]} | ||
| + | ]} | ||
| + | }} | ||
| + | ]) | ||
| + | )) | ||
</div> | </div> | ||
</div> | </div> | ||
Revision as of 14:12, 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
}}
])
))
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 $or
pp.pprint(list(
db.world.aggregate([
{"$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"]}
]}
}}
])
))