In my case, I am dealing with two MongoDB collections: A and B. Each of them contains the field 'user'. I am interested in finding the number of unique users in A, B, as well as the combined collection (A + B). Here is a pseudo code example:
A.user.distinct().count()
B.user.distinct().count()
(A.user union B.user)..distinct().count()
Any recommendations or suggestions on how to achieve this?
If you're looking to find the number of unique users in your collection, using the count() method with distinct won't work. This is because distinct returns an array, not a count. Instead, you can utilize the Array.length property.
To find the number of unique users in either A or B, you can use the following:
db.A.distinct('user').length
db.B.distinct('user').length
If you want to find the number of unique users in the union of A and B, you can combine arrays using Array.prototype.concat() and filter out duplicates using Array.prototype.filter().
var users = db.A.distinct('user');
users = users.concat(db.B.distinct('user'));
var uniqueUsers = users.filter(function(u, pos) {return users.indexOf(u) == pos; });
uniqueUsers.length;
If you want to find the total number of unique users in each collection, you can execute the following commands in the mongo shell:
db.A.distinct("user").length;
db.B.distinct("user").length;
To determine the number of distinct users in the combination of collections A and B, I suggest retrieving distinct arrays for each collection and then merging the arrays to calculate the total count. For JavaScript, utilizing Underscore.js' union() function is recommended. You can find more information on how to use it here. Remember, you can load Underscore.js (or any JavaScript file) into the mongo shell by running this command within the shell:
load("path/to/underscore.js");
Subsequently, you can easily perform the following actions:
var a = db.A.distinct("user");
var b = db.B.distinct("user");
_.union(a, b).length;
Alternatively, if you prefer, you can create your own JavaScript function following the guidelines provided here, or develop a solution in your application's respective language.
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