Descending sort by value of a Hash in Ruby

Question

My input hash: h = { \"a\" => 20, \"b\" => 30, \"c\" => 10 }

Ascending sort: h.sort {|a,b| a[1]<=>b[1]} #=> [[\"c\", 10], [\"a

Answer 1

Super simple: h.sort_by { |k, v| -v }

Answer 2

h.sort {|a,b| b[1]<=>a[1]}

Answer 3

You can have it cleaner, clearer and faster, all at once! Like this:

h.sort_by {|k,v| v}.reverse

I benchmarked timings on 3000 iterations of sorting a 1000-element hash with random values, and got these times:

h.sort {|x,y| -(x[1]<=>y[1])} -- 16.7s
h.sort {|x,y| y[1] <=> x[1]} -- 12.3s
h.sort_by {|k,v| -v} -- 5.9s
h.sort_by {|k,v| v}.reverse -- 3.7

Answer 4

<=> compares the two operands, returning -1 if the first is lower, 0 if they're equal and 1 if the first is higher. This means that you can just do -(a[1]<=>b[1]) to reverse the order.