根据数组中相同键的顺序对哈希键进行排序

Question

I need to sort an Hash according to order of keys present in other array:

hash = { a: 23, b: 12 }
array = [:b, :a]
required_hash #=> { b: 12, a: 23 } 

Is there any way to do it in single line?

Answer 1

hash = { a: 23, b: 12 }
array = [:b, :a]

array.zip(hash.values_at(*array)).to_h
  #=> {:b=>12, :a=>23}

The steps:

v = hash.values_at(*array)
  #=> [12, 23]
a = array.zip(v) 
  #=> [[:b, 12], [[:a, 23]]
a.to_h
  #=> {:b=>12, :a=>23}

Answer 2

You can try something like this:

array = [:b, :a] 
{ a: 23, b: 12 }.sort_by { |k, _| array.index(k) }.to_h
#=> {:b=>12, :a=>23}

Answer 3

If

hash = { a:23, b:12 }
array = [:b, :a]

As a one liner:

Hash[array.map{|e| [e, hash[e]]}]

As for ordering, it was some time ago, when ruby changed the implementation where the hash keys are ordered according to insert order. (Before that there was no guarantee that keys would be sorted one way or another).

Answer 4

I do not see any sorting, I think essentially you are doing this:

hash = { a:23, b:12 }
array = [:b, :a]
res = {}
array.each { |key| res[key] = hash[key] } && res # `&& res` is just syntax sugar to return result in single line
#=> {:b=>12, :a=>23}