Sorting algorithm

Question

I have two objects, Artists and Songs.

Artists={
A0: [ "a0", "a1", ...],
A1: [ "a21", "a69", ...],
...
//255 artists
}

Songs={
s0:{
...
album: "a0",
...
},
s1:{
...
album: "a182",
...
},
...
//26,500 songs
}

I need to find ten artists with most songs.
I used this naive approach:

let artists=Object.keys(Artists)
.map(m=>[
m,
Artists[m].map(n=>
Object.keys(Songs)
.filter(f=>Songs[f].album==n).length)
.reduce((mm,n)=>mm+n)
])
.sort((m,n)=>n[1]-m[1])
.slice(0,10);

But it takes 70 seconds to execute.
Could you come up with a much faster approach?
There are 805 albums.

Artists object is here http://remi.rf.gd/JS/artists.js
Songs object is here http://remi.rf.gd/JS/songs.js
https://code.sololearn.com/WK4UfhwmJhrB/?ref=app

Answer 1

ODLNT thank you very much for the insight. I have included the Artists and Songs objects on my post. Could you please try executing it again with your machine? Much appreciated.

Answer 2

Fernando Moceces based on your post I wrote the code below. I ran the algorithm provided, on my laptop got an average time of 3.9 seconds. I also ran it on bluestack virtual android phone and got an average time of 5.7second.
I tracked the time using window.performance(), not for precision but for general knowledge.

Take a look and let me know if the objects are formatted correctly.

https://code.sololearn.com/WoM17HXrVrND/?ref=app

Answer 3

Fernando Moceces

Great now were on the same page.

When I run your algorithm on my laptop with the objects you provided I see an average time of 3.6 seconds.

I was able to come up with something faster using same object and timer on my laptop I get an average time of 0.015 seconds.

I included screenshots and algorithm below

https://imgur.com/Km2uHq3

https://imgur.com/yjeRc8c

https://code.sololearn.com/WI34X9d66qF0

I'm very interested what results you get on your device.

Answer 4

ODLNT excuse me but apparently you have your code set private so I can only see snippet of it. Also your google drive links seem to be inaccessible to public as well, they require access request.

Answer 5

My bad, I updated my previous post with public links to the screen shots and the code is now public