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query5.js
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query5.js
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//find the oldest friend for each user who has a friend. For simplicity, use only year of birth to determine age, if there is a tie, use the one with smallest user_id
//return a javascript object : key is the user_id and the value is the oldest_friend id
//You may find query 2 and query 3 helpful. You can create selections if you want. Do not modify users collection.
//
//You should return something like this:(order does not matter)
//{user1:userx1, user2:userx2, user3:userx3,...}
function oldest_friend(dbname){
db = db.getSiblingDB(dbname)
// Create a dictionary which maps each user_id to that user's year of birth
var dict = {};
var userIdYear = db.users.find({}, {user_id: 1, YOB: 1, _id: 0});
for (var i = 0; i < userIdYear.length(); i++) {
dict[userIdYear[i]["user_id"]] = userIdYear[i]["YOB"];
}
// Create an array which stores two things for a given user:
// The YOB of the oldest friend (so far)
// The id of the oldest friend (so far)
var oldestData = [];
for (var i = 0; i < userIdYear.length(); i++) {
oldestData[userIdYear[i]["user_id"]] = [100000, 100000];
}
// Create a collection which contains each user paired with each friend with a higher id
db.users.aggregate([
{$project: {user_id: 1, friends: 1, _id: 0}},
{$unwind: "$friends"},
{$out: "flat_users"}
]);
// Create a collection which contains each user paired with each friend with a lower id
db.flat_users.find().forEach(
function(user) {
db.r_flat_users.insertOne({user_id: user.friends, friends: user.user_id});
}
);
// Iterate through each user's higher-id friends and determine which friend is the oldest
db.flat_users.find().forEach(
function(doc)
{
if (dict[doc.friends] < oldestData[doc.user_id][0] ||
(dict[doc.friends] == oldestData[doc.user_id][0] &&
doc.friends < oldestData[doc.user_id][1]))
{
oldestData[doc.user_id][0] = dict[doc.friends];
oldestData[doc.user_id][1] = doc.friends;
}
}
);
// Iterate through each user's lower-id friends and determine which friend is the oldest
// (this also compares the lower-id friends' ages with the higher-id friends' ages)
db.r_flat_users.find().forEach(
function(doc)
{
if (dict[doc.friends] < oldestData[doc.user_id][0] ||
(dict[doc.friends] == oldestData[doc.user_id][0] &&
doc.friends < oldestData[doc.user_id][1]))
{
oldestData[doc.user_id][0] = dict[doc.friends];
oldestData[doc.user_id][1] = doc.friends;
}
}
);
// Create a dictionary that maps each user to their oldest friend
var resultDict = {};
for (var i = 0; i < oldestData.length; i++) {
resultDict[i] = oldestData[i][1];
}
return resultDict;
}