Descending Sort by Value of a Hash in Ruby

Descending sort by value of a Hash in Ruby

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

Sort a hash by value in descending order and then key in ascending order ruby

In a comparison, arrays get evaluated first by their first elements, then by their second, etc. You can use this fact to enumerate sequential comparisons. Comparing by [-v, k] first sorts by value (in reverse order) then by key.

>> trial_hash.sort_by{|k, v| [-v, k]}
=> [["key1", 1000], ["key3", 500], ["key4", 500], ["key5", 500], ["key6", 500], ["key2", 34]]

How to sort a hash by value in descending order and output a hash in ruby?

Try:

Hash[h.sort.reverse]

This should return what you want.

Edit:

To do it by value:

Hash[h.sort_by{|k, v| v}.reverse]

Sort array of hash by key in descending order in Ruby

You can do

h.sort_by! { |k| -k[:bookings_nd] }

or

h.sort_by! { |k| k[:bookings_nd] }.reverse!

Also i guess this question is duplicate for Sorting an array in descending order in Ruby

Sort hash by length of values (descending)

You cannot sort a hash - that might be causing your confusion. There is no "internal" ordering of elements of a hash as it appears in an array.

You can, however, iterate over a hash in a certain order, e.g.

hash.sort_by {|k,v| v.length}.reverse.each do |k, v|
  puts "k = #{k}, v = #{v}"
end

Sort array of hashes by value of hash property in descending order preserving the initial order of hash

results = [
  { "rating"=>6, "id"=>10699 },
  { "rating"=>3, "id"=>19985 },
  { "rating"=>6, "id"=>1029 }
]

results.sort_by.with_index { |h,i| [-h["rating"], i] }
  #=> [{"rating"=>6, "id"=>10699},
  #    {"rating"=>6, "id"=>1029}
  #    {"rating"=>3, "id"=>19985}]

See the third paragraph of the doc Array#<=> for an explanation of how Ruby orders arrays. When Ruby orders results[0] and results[2] she compares [-6, 0] with [-6, 2]. Because the first elements of these arrays are equal (-6), Ruby compares 0 with 2 to break the tie, so results[0] is ordered before results[2].

Note that when the method Enumerable#sort_by has no block, it returns an enumerator. Here that enumerator is chained to Enumerator#with_index to produce another enumerator.

Ruby - sorting hashes into descending order by value

You are kinda mixing the sort and filtering out phases. My solution

1. Filter results with value >= 60
2. Sort for values (descending, -v)

3. Extract the first element for every array (the language name)

   def my_languages(results)
     results.select { |k, v| v >= 60 }.sort_by { |(k,v)| -v }.map(&:first)
   end
   

Sorting Hash by value

Enumerable#sort_by never promised to sort the enumerable inplace. It returns the result:

puts mapa.sort_by {|a,b| -b}
#⇒ sorted

or, in more idiomatic manner:

mapa.sort_by(&:last).reverse.each do |k, v|
  puts ...
end

Please note, that when examining the output in irb, the order might not correspond to the real order of elements in a hash, since irb (with print formatters installed) has it’s own ideas on how to output hashes.

descending sorting on hash values

it seems sorted alphabetically. convert to numeric first.

h.sort_by {|_key, value| -value.to_f}

the minus sign could avoid the using of reverse

How to sort a Ruby Hash by number value?

No idea how you got your results, since it would not sort by string value... You should reverse a1 and a2 in your example

Best way in any case (as per Mladen) is:

metrics = {"sitea.com" => 745, "siteb.com" => 9, "sitec.com" => 10 }
metrics.sort_by {|_key, value| value}
  # ==> [["siteb.com", 9], ["sitec.com", 10], ["sitea.com", 745]]

If you need a hash as a result, you can use to_h (in Ruby 2.0+)

metrics.sort_by {|_key, value| value}.to_h
  # ==> {"siteb.com" => 9, "sitec.com" => 10, "sitea.com", 745}

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