Ruby "count" method for hashes

Question

I have a hash in which I want to use the values as keys in a new Hash which contains a count of how many times that item appeared as a value in the original hash.

So I use:

hashA.keys.each do |i|
    puts hashA[i]
end

Example output:

And I want the new Hash to be the following:

{ 0 => 2,  1 => 4,  2 => 1 }

possible duplicate of [How to count identical string elements in a Ruby array](http://stackoverflow.com/questions/5128200/how-to-count-identical-string-elements-in-a-ruby-array). The original data structure in this question is a hash, but you're throwing away the keys, so you're effectively dealing with `hashA.values`, which is an array. — Andrew Grimm, Oct 24 '11 at 01:15
@AndrewGrimm Meh; eventually (and quickly), yes... But when searching with a "I have a map" mindset you probably won't search for stuff about arrays. — Dave Newton, Oct 24 '11 at 01:17

score 17 · Accepted Answer · answered Oct 24 '11 at 01:08

17

counts = hashA.values.inject(Hash.new(0)) do |collection, value|
  collection[value] +=1
  collection
end

answered Oct 24 '11 at 01:08

Brian Rose

1,725
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5

+1 for meaningful variable names in the block; I forget when you don't know the API, they matter. – Dave Newton Oct 24 '11 at 01:25

Dave Newton · Answer 2 · 2011-10-24T01:13:17.450

7

TL;DR: hashA.values.inject(Hash.new(0)) { |m, n| m[n] += 1; m }

> hashA = { a: 0, b: 1, c: 1, d: 2, e: 0, f: 1, g: 1 }
=> {:a=>0, :b=>1, :c=>1, :d=>2, :e=>0, :f=>1, :g=>1} 
> hashCounts = hashA.values.inject(Hash.new(0)) { |m, n| m[n] += 1; m }
=> {0=>2, 1=>4, 2=>1}

edited Oct 24 '11 at 01:13

answered Oct 24 '11 at 01:07

Dave Newton

158,873
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Ruby "count" method for hashes

2 Answers2