I have a list that I need to sort by the most popular elements. Is there a method to accomplish that?
After I re-sort it, I also need to get rid of duplicates. I have an idea of a function in my mind for this but it seems inefficient, so are there built-in methods to help with this?
[1,5,4,6,4,1,4,5].group_by {|x| x}.sort_by {|x,list| [-list.size,x]}.map(&:first)
=> [4,1,5,6]
Like that?
The Array#sort method takes an optional predicate to compare two elements, so...
list.sort { |a, b| a.popularity <=> b.popularity }
To eliminate duplicates, use Array#uniq.
list.uniq
To glue them together,
list = list.sort { |a, b| a.popularity <=> b.popularity }.unique
Or simply
list.sort! { |a, b| a.popularity <=> b.popularity }.uniq!
Iterating through the list to build a hash that maps item -> number of times just need one visit of all elements of the list, then operations with the hash would be constant time, so O(n), that doesn't seem so expensive.
The uniq method takes a block, so you can specify which 'property' of your object has to be uniq.
new_list = list.sort_by{|el| el.popularity}.uniq{|el| el.popularity}
Most of these answers didn't work for me, except for Glenn Mcdonalds (up until i posted this answer) I found an answer to my own question somewhere else like this
list = [2,1,4,4,4,1] #for example
count = Hash.new(0)
list.each {|element| count[element] += 1} #or some other parameter than element
list = list.uniq.sort {|x,y| count[y] <=> count[x]}