Ruby—sort_by multiple levels of sorting [duplicate]

What I eventually did:

(s1+s2).chars.sort_by{ |s| 
  [-([s1.count(s), s2.count(s)].max), 
    s1.count(s) > s2.count(s) ? -1 : 1, 
    s]
}
#=> "lloohebdgy"

The generalised answer is that giving an Array to the sort_by block allows you to specify levels of sorting.

Using String#count is relatively inefficient, as the entire string is traversed for each unique letter. It would be more efficient, at the expense of more lines of code, to create a counting hash, which can be done as follows (see Hash::new).

s1 = "hello"
s2 = "goodbye"

h1 = s1.each_char.with_object(Hash.new(0)) { |c,h| h[c] += 1 }
  #=> {"h"=>1, "e"=>1, "l"=>2, "o"=>1} 
h2 = s2.each_char.with_object(Hash.new(0)) { |c,h| h[c] += 1 }
  #=> {"g"=>1, "o"=>2, "d"=>1, "b"=>1, "y"=>1, "e"=>1}
hmx = (h1.keys | h2.keys).each_with_object({}) do |c,h| h[c] =
  case h1[c] <=> h2[c]
  when 1 then [-h1[c], -1]
  when 0 then [-h1[c], 0]
  else [-h2[c], 1]
  end
end
  #=> {"h"=>[-1, -1], "e"=>[-1, 0], "l"=>[-2, -1], "o"=>[-2, 1],
  #    "g"=>[-1, 1], "d"=>[-1, 1], "b"=>[-1, 1], "y"=>[-1, 1]} 

We can now sort using Enumerable#sort_by

(s1+s2).each_char.sort_by { |c| [*hmx[c], c] }.join
  #=> "lloooheebdgy"

See Array#<=> (third paragraph) for an explanation of how arrays are sorted.

I've made the assumption that if two characters tie for the most occurrences in one of the two strings, each character is assigned a score of -1, 0 or 1. The character with the lowest score (if they don't have the same score) precedes the other character in the sort. A character c is assigned a score of -1 if the string s1 contains more c's than does string s1; a score of 0 if both strings contain the same number of c's; and a score of 1 if s2 contains more c's than does s1.