Ruby : How to sort an array of hash in a given order of a particular key

Question

I have an array of hashes, id being one of the keys in the hashes. I want to sort the array elements according to a given order of ID values.

Suppose my array(size=5) is:

[{"id"=>1. ...}, {"id"=>4. ...}, {"id"=>9. ...}, {"id"=>2. ...}, {"id"=>7. ...}]

I want to sort the array elements such that their id s are in the following order:

[1,3,5,7,9,2,4,6,8,10]

So the expected result is:

[{'id' => 1},{'id' => 7},{'id' => 9},{'id' => 2},{'id' => 4}]

Answer 1

Here is a solution for any custom index:

def my_index x 
  # Custom code can be added here to handle items not in the index.
  # Currently an error will be raised if item is not part of the index.
  [1,3,5,7,9,2,4,6,8,10].index(x) 
end

my_collection = [{"id"=>1}, {"id"=>4}, {"id"=>9}, {"id"=>2}, {"id"=>7}]
p my_collection.sort_by{|x| my_index x['id'] } #=> [{"id"=>1}, {"id"=>7}, {"id"=>9}, {"id"=>2}, {"id"=>4}]

Then you can format it in any way you want, maybe this is prettier:

my_index = [1,3,5,7,9,2,4,6,8,10]
my_collection.sort_by{|x| my_index.index x['id'] }

Answer 2

General note on sorting. Use #sort_by method of the ruby 's array class:

[{'id' => 1},{'id'=>3},{'id'=>2}].sort_by {|x|x['id'] }
# => [{"id"=>1}, {"id"=>2}, {"id"=>3}]

Or with usage #values method as a callback:

[{'id' => 1},{'id'=>3},{'id'=>2}].sort_by(&:values)
# => [{"id"=>1}, {"id"=>2}, {"id"=>3}]

or you can use more obvious version with #sort method:

[{'id' => 1},{'id'=>3},{'id'=>2}].sort {|x,y| x['id'] <=> y['id'] }
# => [{"id"=>1}, {"id"=>2}, {"id"=>3}]

For your case, to sort with extended condition use #% to split even and odd indexes:

[{'id' => 1},{'id'=>4},{'id'=>9},{'id'=>2},{'id'=>7}].sort do |x,y|
    u = y['id'] % 2 <=> x['id'] % 2
    u == 0 && y['id'] <=> x['id'] || u 
end
# => [{"id"=>1}, {"id"=>7}, {"id"=>9}, {"id"=>2}, {"id"=>4}]

For your case, to sort with extended condition use #% to split according the index, even id value is absent in the index array:

index = [1,3,5,7,4,2,6,8,10] # swapped 2 and 4, 9 is absent

[{'id' => 1},{'id'=>4},{'id'=>9},{'id'=>2},{'id'=>7}].sort do |x,y|
   !index.rindex( x[ 'id' ] ) && 1 || index.rindex( x[ 'id' ] ) <=> index.rindex( y[ 'id' ] ) || -1 
end
# => [{"id"=>1}, {"id"=>7}, {"id"=>4}, {"id"=>2}, {"id"=>9}]

Answer 3

I would map the hash based on the values like so:

a = [{"id"=>1}, {"id"=>4}, {"id"=>9}, {"id"=>2}, {"id"=>7}]

[1,3,5,7,9,2,4,6,8,10].map{|x| a[a.index({"id" => x})] }.compact

#=> [{"id"=>1}, {"id"=>7}, {"id"=>9}, {"id"=>2}, {"id"=>4}]

Answer 4

Why not just sort ?

def doit(arr, order)
  arr.sort { |h1,h2| order.index(h1['id']) <=> order.index(h2['id']) }
end

order = [1,3,5,7,9,2,4,6,8,10]
arr = [{'id' => 1}, {'id' => 4}, {'id' => 9}, {'id' => 2}, {'id' => 7}]     

doit(arr, order)
  #=> [{'id' => 1}, {'id' => 7}, {'id' => 9}, {'id' => 2}, {'id' => 4}]     

Answer 5

a= [{"id"=>1}, {"id"=>4}, {"id"=>9}, {"id"=>2}, {"id"=>7}]
b=[1,3,5,7,9,2,4,6,8,10]
a.sort_by{|x| b.index (x['id'])}