Check if two arrays have the same contents (in any order)

Question

I\'m using Ruby 1.8.6 with Rails 1.2.3, and need to determine whether two arrays have the same elements, regardless of whether or not they\'re in the same order. One of the

Answer 1

combining & and size may be fast too.

require 'benchmark/ips'
require 'set'

Benchmark.ips do |x|
  x.report('sort')   { a.sort == b.sort }  
  x.report('sort!')  { a.sort! == b.sort! }  
  x.report('to_set') { a.to_set == b.to_set }  
  x.report('minus')  { ((a - b) + (b - a)).empty? }
  x.report('&.size') { a.size == b.size && (a & b).size == a.size }  
end  

Calculating -------------------------------------
                sort    896.094k (±11.4%) i/s -      4.458M in   5.056163s
               sort!      1.237M (± 4.5%) i/s -      6.261M in   5.071796s
              to_set    224.564k (± 6.3%) i/s -      1.132M in   5.064753s
               minus      2.230M (± 7.0%) i/s -     11.171M in   5.038655s
              &.size      2.829M (± 5.4%) i/s -     14.125M in   5.010414s

Answer 2

Speed comparsions

require 'benchmark/ips'
require 'set'

a = [1, 2, 3, 4, 5, 6]
b = [1, 2, 3, 4, 5, 6]

Benchmark.ips do |x|
  x.report('sort')   { a.sort == b.sort }  
  x.report('sort!')  { a.sort! == b.sort! }  
  x.report('to_set') { a.to_set == b.to_set }  
  x.report('minus')  { ((a - b) + (b - a)).empty? }  
end  

Warming up --------------------------------------
            sort    88.338k i/100ms
           sort!   118.207k i/100ms
          to_set    19.339k i/100ms
           minus    67.971k i/100ms
Calculating -------------------------------------
            sort      1.062M (± 0.9%) i/s -      5.389M in   5.075109s
           sort!      1.542M (± 1.2%) i/s -      7.802M in   5.061364s
          to_set    200.302k (± 2.1%) i/s -      1.006M in   5.022793s
           minus    783.106k (± 1.5%) i/s -      3.942M in   5.035311s

Answer 3

This doesn't require conversion to set:

a.sort == b.sort

Answer 4

One approach is to iterate over the array with no duplicates

# assume array a has no duplicates and you want to compare to b
!a.map { |n| b.include?(n) }.include?(false)

This returns an array of trues. If any false appears, then the outer include? will return true. Thus you have to invert the whole thing to determine if it's a match.

Answer 5

If you know the arrays are of equal length and neither array contains duplicates then this works too:

( array1 & array2 ) == array1

Explanation: the & operator in this case returns a copy of a1 sans any items not found in a2, which is the same as the original a1 iff both arrays have the same contents with no duplicates.

Analyis: Given that the order is unchanged, I'm guessing this is implemented as a double iteration so consistently O(n*n), notably worse for large arrays than a1.sort == a2.sort which should perform with worst-case O(n*logn).

Answer 6

If you expect [:a, :b] != [:a, :a, :b] to_set doesn't work. You can use frequency instead:

class Array
  def frequency
    p = Hash.new(0)
    each{ |v| p[v] += 1 }
    p
  end
end

[:a, :b].frequency == [:a, :a, :b].frequency #=> false
[:a, :b].frequency == [:b, :a].frequency #=> true