Determining whether one array contains the contents of another array in ruby

Question

In ruby, how do I test that one array not only has the elements of another array, but contain them in that particular order?

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


        

Answer 1

I think it can be done simply.

class Array
  def contain? other; (self & other) == other end
end

correct_combination = [1, 2, 3, 4, 5]
[1, 5, 8, 2, 3, 4, 5].contain?(correct_combination) # => false
[8, 10, 1, 2, 3, 4, 5, 9].contain?(correct_combination) # => true

Answer 2

A really quick way to do this is to simply subtract one array from the other and test for an empty array.

correct_combination = [1, 2, 3, 4, 5]
yep = [8, 10, 1, 2, 3, 4, 5, 9]
nope = [1, 8, 2, 3, 4]
if correct_combination - yep == []
  puts "yep has all the values"
end
if correct_combination - nope == []
  puts "nope has all the values"
end

This approach does not care about position so delete away!

Sorry... I missed the point to the question as well. Didn't realize you were looking for order of precedence. I came across this when looking for a solution to evaluating if one large array contained all the entries of another large array. The .all?/include? approach takes a really long time to complete. Good luck!

Answer 3

You could simply compare the arrays as strings:

correct_combination = [1, 2, 3, 4, 5]
yep = [8, 10, 1, 2, 3, 4, 5, 9]
nope = [1, 5, 8, 2, 3, 4, 5]
if yep.to_s.include?(correct_combination.to_s)
  puts "yep"
end
if nope.to_s.include?(correct_combination.to_s)
  puts "nope"
end

Answer 4

If you want to ignore the order, (as I did when I came across this post), you could use Array.sort and <=> http://ruby-doc.org/core-1.8.7/classes/Array.html#M000316

a = [1, 2, 3, 4, 5]
b = [2, 1, 5, 4, 3]
a.sort <=> b.sort

You then need to check the output value equals 0.

Answer 5

Maybe <=> is what you are looking for.

Comparison—Returns an integer (-1, 0, or +1) if this array is less than, equal to, or greater than other_array

a = [1, 2, 3, 4, 5]
b = [1, 5, 8, 2, 3, 4, 5]
c = [8, 10, 1, 2, 3, 4, 5, 9]

puts a <=> b # => -1
puts a <=> c # => -1
puts a <=> a # => 0

Update: nevermind, just noted it doesn't care about position.

puts a <=> a.reverse # => -1

Answer 6

Not exactly the best solution possible, but at least it's brief

(',' + [1, 5, 8, 2, 3, 4, 5].join(',') + ',').include?(',' + correct_combination.join(',') + ',')

The best solution possible would be to employ one of string searching algorithms on array but you would have to code it yourself, I don't think there's standard solution.