Comparing two arrays ignoring element order in Ruby

Question

I need to check whether two arrays contain the same data in any order. Using the imaginary compare method, I would like to do:

arr1 = [1,2,3,5,4         


        

Answer 1

You can actually implement this #compare method by monkey patching the Array class like this:

class Array
  def compare(other)
    sort == other.sort
  end
end

Keep in mind that monkey patching is rarely considered a good practice and you should be cautious when using it.

There's probably is a better way to do this, but that's what came to mind. Hope it helps!

Answer 2

You can open array class and define a method like this.

class Array
  def compare(comparate)
    to_set == comparate.to_set
  end
end

arr1.compare(arr2)
irb => true

OR use simply

arr1.to_set == arr2.to_set
irb => true

Answer 3

Here is a version that will work on unsortable arrays

class Array
  def unordered_hash
    unless @_compare_o && @_compare_o == hash
      p = Hash.new(0)
      each{ |v| p[v] += 1 }
      @_compare_p = p.hash
      @_compare_o = hash
    end
    @_compare_p
  end
  def compare(b)
    unordered_hash == b.unordered_hash
  end
end

a = [ 1, 2, 3, 2, nil ]
b = [ nil, 2, 1, 3, 2 ]
puts a.compare(b)

Answer 4

The most elegant way I have found:

arr1 = [1,2,3,5,4]
arr2 = [3,4,2,1,5]
arr3 = [3,4,2,1,5,5]


(arr1 - arr2).empty? 
=> true

(arr3 - arr2).empty?
=> false

Answer 5

Sorting the arrays prior to comparing them is O(n log n). Moreover, as Victor points out, you'll run into trouble if the array contains non-sortable objects. It's faster to compare histograms, O(n).

You'll find Enumerable#frequency in Facets, but implement it yourself, which is pretty straightforward, if you prefer to avoid adding more dependencies:

require 'facets'
[1, 2, 1].frequency == [2, 1, 1].frequency 
#=> true

Answer 6

If you know that there are no repetitions in any of the arrays (i.e., all the elements are unique or you don't care), using sets is straight forward and readable:

Set.new(array1) == Set.new(array2)