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Difference between revisions of "MAPREDUCE Elite"

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(Rewording.)
Line 52: Line 52:
 
<p class="strong">What are the populations of the three main factions?</p>
 
<p class="strong">What are the populations of the three main factions?</p>
 
<div class="hint" title="Three main factions">["Alliance","Federation","Empire"]</div>
 
<div class="hint" title="Three main factions">["Alliance","Federation","Empire"]</div>
<div class="hint" title="NaN?">Some systems have no populations, make sure to exclude them using <code>!isNaN()</code>.</div>
+
<div class="hint" title="NaN?">Some systems are not populated and will have '''null''' population fields, make sure to exclude them using <code>!isNaN()</code>.</div>
 
<pre class="def">
 
<pre class="def">
 
db.systems.mapReduce(
 
db.systems.mapReduce(
Line 64: Line 64:
 
)
 
)
 
</pre>
 
</pre>
<pre class="ans">db.systems.mapReduce(function(){if (!isNaN(this.population) && this.allegiance!=null && this.allegiance!="Independent" && this.allegiance!="Anarchy"){emit(this.allegiance, this.population);}},function(k,v){return Array.sum(v);},{out:{inline:1}})</pre>
+
<pre class="ans">db.systems.mapReduce(function(){if(!isNaN(this.population)&&this.allegiance!=null&&this.allegiance!="Independent"&&this.allegiance!="Anarchy"){emit(this.allegiance,this.population);}},function(k,v){return Array.sum(v);},{out:{inline:1}})</pre>
 
</div>
 
</div>
 +
 
==Harder Questions==
 
==Harder Questions==
 
<div class="q" data-lang="mongo" data-switches='elite'>How much Hydrogen Fuel is owned by each faction?  Limit your query to the first 5000 stations.
 
<div class="q" data-lang="mongo" data-switches='elite'>How much Hydrogen Fuel is owned by each faction?  Limit your query to the first 5000 stations.

Revision as of 00:18, 23 March 2018

Introducing the elite database

These questions will introduce the "elite" database, which contains data about the video game Elite Dangerous


There are two collections, commodities and systems.
Inside systems there are nested documents called stations.
A system has many stations, and a station has many trade listings.

Keys used in this database.

    commodities:
        _id, average_price, category, name
    systems:
        _id, allegiance, faction, government, name, population, primary_economy, security, state, stations, updated_at, x, y, z
    systems.stations: 
        allegiance, distance_to_star, economies, export_commodities,has_blackmarket, has_commodities, has_rearm, has_repair,
        has_shipyard, has_outfitting, faction, government, listings, max_landing_pad, name, state, type, updated_at
    systems.stations.listings: 
        buy_price, collected_at, demand, commodity, sell_price, supply, update_count

Read more about the structure here: Elite Document Structure

Questions

The commodities collection contains the name and average_price of each commodity.

There are 99 unique commodities and 15 categories.

Find the average price of each category, round to the nearest whole number.

db.commodities.mapReduce(
  function(){
    emit(1,1);
  },
  function(k,v){
    return Array.sum(v);
  },
  {out:{inline:1}}
)
db.commodities.mapReduce(function(){emit(this.category,this.average_price);},function(k,v){return Math.round(Array.sum(v)/v.length);},{out:{inline:1}})
Each system has an allegiance. There are three main factions: The Federation, The Empire, and The Alliance.

Non-populated systems without stations do not have an allegiance, and should be ignored.

Show the amount of systems following each type of allegiance.

db.systems.mapReduce(
  function(){
    emit(1,1);
  },
  function(k,v){
    return Array.sum(v);
  },
  {out:{inline:1}}
)
db.systems.mapReduce(function(){if (this.allegiance!=null){emit(this.allegiance,1);}},function(k,v){return Array.sum(v);},{out:{inline:1}})

What are the populations of the three main factions?

["Alliance","Federation","Empire"]
Some systems are not populated and will have null population fields, make sure to exclude them using !isNaN().
db.systems.mapReduce(
  function(){
    emit(1,1);
  },
  function(k,v){
    return Array.sum(v);
  },
  {out:{inline:1}}
)
db.systems.mapReduce(function(){if(!isNaN(this.population)&&this.allegiance!=null&&this.allegiance!="Independent"&&this.allegiance!="Anarchy"){emit(this.allegiance,this.population);}},function(k,v){return Array.sum(v);},{out:{inline:1}})

Harder Questions

How much Hydrogen Fuel is owned by each faction? Limit your query to the first 5000 stations.

The amount of stations in a system and the amount of listings to a station aren't fixed. query can be used to ensure that they exist.

db.systems.mapReduce(
  function(){
    emit(1,1);
  },
  function(k,v){
    return Array.sum(v);
  },
  {out:{inline:1}}
)
db.systems.mapReduce(function(){if(this.stations)for(let i=0;i<this.stations.length;i++){let t=this.stations[i];if(t.listings&&t.allegiance)for(let s=0;s<t.listings.length;s++){let n=t.listings[s];"Hydrogen Fuel"===n.commodity&&emit(t.allegiance,n.supply)}}},function(i,t){return Array.sum(t)},{out:{inline:1},limit:5e3});
A power_control_faction or Power is an individual or organisation who is in control of a system.

These powers have allegiance to a faction, but the systems they control do not nescessarily have the same allegiance that they do.

At the time of writing Zemina Torval is allied with the Empire and controls 47 systems.
    {   '_id': 'Zemina Torval',
        'value': {   'alliance': 0.0,
                     'anarchy': 0.0,
                     'empire': 39.0,
                     'federation': 3.0,
                     'independent': 5.0}}]

Show the allegiance of each of the power's systems

db.systems.mapReduce(
  function(){
    emit(1,1);
  },
  function(k,v){
    return Array.sum(v);
  },
  {
    query: {"power_control_faction": {"$exists": 1}},
    out: {inline: 1}
  }
)
db.systems.mapReduce(function(){switch(this.allegiance){case "Alliance":emit(this.power_control_faction,{Alliance:1,Anarchy:0,Empire:0,Federation:0,Independent:0});break;case "Anarchy":emit(this.power_control_faction,{Alliance:0,Anarchy:1,Empire:0,Federation:0,Independent:0});break;case "Empire":emit(this.power_control_faction,{Alliance:0,Anarchy:0,Empire:1,Federation:0,Independent:0});break;case "Federation":emit(this.power_control_faction,{Alliance:0,Anarchy:0,Empire:0,Federation:1,Independent:0});break;case "Independent":emit(this.power_control_faction,{Alliance:0,Anarchy:0,Empire:0,Federation:0,Independent:1});break;}},function(k,v){let a=v[0];for(let i=1;i<v.length;i++){let b=v[i];a.Alliance+=b.Alliance;a.Anarchy+=b.Anarchy;a.Empire+=b.Empire;a.Federation+=b.Federation;a.Independent+=b.Independent;}return a;},{out:{"inline":1},query:{"power_control_faction":{"$exists":1}}})
Our dataset doesn't contain the allegiance of a power:

Using the result from the previous question, guess the power's allegiance by the faction that the majority of their systems follow.

To achieve this, you'll need to use the finalize: function(k, v){} in the third argument to find the key with the largest value.

Zemina Torval: Empire
db.systems.mapReduce(
  function(){
    emit(1,1);
  },
  function(k,v){
    return Array.sum(v);
  },
  {
    finally: function(k, v){
      return v;
    },
    query: {"power_control_faction": {"$exists": 1}},
    out: {inline: 1}
  }
)
db.systems.mapReduce(function(){switch(this.allegiance){case "Alliance":emit(this.power_control_faction,{Alliance:1,Anarchy:0,Empire:0,Federation:0,Independent:0});break;case "Anarchy":emit(this.power_control_faction,{Alliance:0,Anarchy:1,Empire:0,Federation:0,Independent:0});break;case "Empire":emit(this.power_control_faction,{Alliance:0,Anarchy:0,Empire:1,Federation:0,Independent:0});break;case "Federation":emit(this.power_control_faction,{Alliance:0,Anarchy:0,Empire:0,Federation:1,Independent:0});break;case "Independent":emit(this.power_control_faction,{Alliance:0,Anarchy:0,Empire:0,Federation:0,Independent:1});break;}},function(k,v){let a=v[0]; for(let i=1;i<v.length;i++){let b=v[i];a.Alliance+=b.Alliance;a.Anarchy+=b.Anarchy;a.Empire+=b.Empire;a.Federation+=b.Federation;a.Independent+=b.Independent;}return a;},{finalize:function(k,v){return Object.keys(v).reduce((a,b)=>v[a]>v[b]?a:b);},out:{"inline":1},query:{"power_control_faction":{"$exists":1}}})