How to sort an array of hashes into hashes with multiple values for a key?

Question

I am working on Ruby on Rails project using rails4.

Scenario:

I have an array of hashes. An array contains hashes where keys are the same.

 a = [{132=>[1000.0]}, {132=>[4000.0]}, {175=>[1000.0]}, {175=>[1000.0]}, {133=>[1200.0]}]
 h = a.each {|key,value| key.each {|k| value}}
#{132=>[1000.0]}
#{132=>[4000.0]}
#{175=>[1000.0]}
#{175=>[1000.0]}
#{133=>[1200.0]}

Problem:

How to get rid off duplicate keys but with values added to unique keys like this:

 {132=>[1000,4000]}
 {175=>[1000,1000]}
 {133=>[1200]}

Thank you.

BroiSatse · Accepted Answer · 2014-08-29 16:42:05Z

7

That would do it:

a.inject({}) {|sum, hash| sum.merge(hash) {|_, old, new| old + new }}

answered Aug 29, 2014 at 16:42

BroiSatse

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Comments

Alexander Kireyev · Accepted Answer · 2014-08-30 08:06:03Z

6

This works for me:

p a.each_with_object(Hash.new([])) { |e, h| e.each { |k, v| h[k] += v } }

# => {132=>[1000.0, 4000.0], 175=>[1000.0, 1000.0], 133=>[1200.0]}

edited Aug 30, 2014 at 8:06

answered Aug 29, 2014 at 16:40

Alexander Kireyev

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3 Comments

MBJH Over a year ago

Thank you, I will think it through.

Alexander Kireyev Over a year ago

Do you mean that you like the possibility to avoid a condition by set a default hash value?=). Or you are hinting that it could be done another way?

Cary Swoveland Over a year ago

I didn't express myself well, but what I meant is that I liked Hash.new([]) and h[k] += v better than what I thought to be the main alternative: Hash.new {|h,k| h[k]=[]} and h[k] << v.first. On reflection, however, the former may be less efficient, as a new array is created when each element is added. Hmmm. Would you mind doing an edit, so I can roll back my +1?. Actually, (h[k] ||= []) <<... probably would be quite a bit faster yet. (Nix the edit request. :-) ).

Cary Swoveland · Accepted Answer · 2014-08-29 19:25:13Z

2

Another way:

a.each_with_object({}) do |g,h|
  k, v = g.to_a.flatten
  (h[k] ||= []) << v
end
  #=> {132=>[1000.0, 4000.0], 175=>[1000.0, 1000.0], 133=>[1200.0]}

or

a.each_with_object(Hash.new { |h,k| h[k]=[] }) do |g,h|
  k, v = g.to_a.flatten
  h[k] << v
end
  #=> {132=>[1000.0, 4000.0], 175=>[1000.0, 1000.0], 133=>[1200.0]}

edited Aug 29, 2014 at 19:25

answered Aug 29, 2014 at 19:08

Cary Swoveland

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1 Comment

Alexander Kireyev Over a year ago

Nice approach as usual.

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