Sort Ruby Hash by order in array of keys

Question

I have a hash:

sample = { bar: 200, foo: 100, baz: 100 }

How do I sort sample using the order of keys in sort_order:

sort_order = [:foo, :bar, :baz, :qux, :quux]

Expected result:

sample #=> { foo: 100, bar: 200, baz: 100 }

All I can come up with is

new_hash = {}
sort_order.each{|k| new_hash[k] = sample[k] unless sample[k].nil? }
sample = new_hash

There's got to be a better way. Hints?

Keys without values should not be present, i.e. number of keys remain the same, which isn't the case with Sort Hash Keys based on order of same keys in array

What do you want in a better way? Shorter? Better performance? — Anand
– Anand, Commented May 24, 2016 at 6:20
sample.sort_by { |k,_| sort_order.index(k) }.to_h or sort_order.each_with_object({}) { |sym,h| h[sym] = sample[sym] if sample.key?(sym) }. — Cary Swoveland
– Cary Swoveland, Commented May 24, 2016 at 6:31
Could there be keys in sample that do not exist in sort_order? — Stefan
– Stefan, Commented May 24, 2016 at 7:02

tokland · Accepted Answer · 2016-05-24 10:29:39Z

14

A functional approach using the intersection of keys:

new_sample = (sort_order & sample.keys).map { |k| [k, sample[k]] }.to_h
#=> {:foo=>100, :bar=>200, :baz=>100}

As @Stefan noted, the abstraction Hash#slice from ActiveSupport's pretty much does the job:

require 'active_support/core_ext/hash'
new_sample = sample.slice(*sort_order)
#=> {:foo=>100, :bar=>200, :baz=>100}

edited May 24, 2016 at 10:29

answered May 24, 2016 at 8:14

tokland

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1 Comment

Stefan Over a year ago

slice checks if the keys exist (similar to Keith Bennett's answer), so you can just write sample.slice(*sort_order)

Community · Accepted Answer · 2017-05-23 11:44:06Z

3

Please, see my this answer:

sort_order = [:foo, :bar, :baz, :qux, :quux, :corge, :grault,
              :garply, :waldo, :fred, :plugh, :xyzzy, :thud]
sample = { bar: 200, foo: 100, baz: 100 }

sample.sort_by {|k, _| sort_order.index(k)}.to_h
=> {:foo=>100, :bar=>200, :baz=>100}

edited May 23, 2017 at 11:44

CommunityBot

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answered May 24, 2016 at 7:11

Ilya

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1 Comment

tokland Over a year ago

Not a problem for small inputs, but note that this is quadratic.

Keith Bennett · Accepted Answer · 2016-05-24 09:22:43Z

3

The code below does this. Note that I used has_key? because you want the output hash to contain all the keys in the input hash, even if their values are nil.

#!/usr/bin/env ruby

def sorted_hash(input_hash, key_sort_order)
  new_hash = {}
  key_sort_order.each do |key|
    if input_hash.has_key?(key)
      new_hash[key] = input_hash[key]
    end
  end
  new_hash
end

sort_order = [:foo, :bar, :baz, :qux, :quux]
sample = { bar: 200, foo: 100, baz: 100 }

puts sorted_hash(sample, sort_order)
# Outputs: {:foo=>100, :bar=>200, :baz=>100}

A simplification is to use each_with_object:

def sorted_hash_two(input_hash, key_sort_order)
  key_sort_order.each_with_object({}) do |key, result_hash|
    if input_hash.has_key?(key)
      result_hash[key] = input_hash[key]
    end
  end
end

puts sorted_hash_two(sample, sort_order)
# Outputs: {:foo=>100, :bar=>200, :baz=>100}

I like @tokland's idea of array intersection (&) better because it elmiinates the need for an if condition:

def sorted_hash_ewo_intersection(input_hash, key_sort_order)
  (key_sort_order & input_hash.keys).each_with_object({}) do |key, result_hash|
    result_hash[key] = input_hash[key]
  end
end # produces: {:foo=>100, :bar=>200, :baz=>100}

edited May 24, 2016 at 9:22

answered May 24, 2016 at 5:59

Keith Bennett

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4 Comments

absessive Over a year ago

The each_with_object looks simpler, however, I don't see how its any better than my solution.

Keith Bennett Over a year ago

As I mention in the answer, I think it's important to use has_key? rather than test for nil. As an aside, I recommend expressing this as a method; that separates the logic in question from the inputs and outputs; and your last name sample = new_hash is unnecessary.

Keith Bennett Over a year ago

I just added @tokland's approach of using array intersection (using &) but using each_with_object, and like that one the best.

Keith Bennett Over a year ago

Note that & is really set intersection; if there were duplicates then those duplicates would be stripped, e.g.: [1] & [1,1] => [1] and [1,1] & [1] => [1]. In this case, both arrays are guaranteed not to have duplicates since hash keys are always unique, but it's something to be aware of in other contexts.

Wand Maker · Accepted Answer · 2016-05-24 10:05:09Z

1

Here is one more way this can be done:

(sort_order & sample.keys).zip([nil]).to_h.merge(sample)
#=> {:foo=>100, :bar=>200, :baz=>100}

Explanation:

First we create a hash that contains only desired keys in the right order.

(sort_order & sample.keys).zip([nil]).to_h
#=> {:foo=>nil, :bar=>nil, :baz=>nil}

And then, we merge this hash with sample to get the values from sample.

edited May 24, 2016 at 10:05

answered May 24, 2016 at 9:53

Wand Maker

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1 Comment

sawa Over a year ago

Sorry, please forget about my comments that I have deleted.

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