Add element to an array if it's not there already

Question

I have a Ruby class

class MyClass
  attr_writer :item1, :item2
end

my_array = get_array_of_my_class() #my_array is an array of MyClass
unique_array_of_item1 = []

I want to push MyClass#item1 to unique_array_of_item1, but only if unique_array_of_item1 doesn't contain that item1 yet. There is a simple solution I know: just iterate through my_array and check if unique_array_of_item1 already contains the current item1 or not.

Is there any more efficient solution?

Vinavinaceous

Answer 1

You can use Set instead of Array.

Delineator

While it is true that the docs say that Sets are not order, they are in fact (as of Ruby 1.9) ordered. If you look at the code, the main methods you'd use to get at the order (such as Set#each and Set#to_a) delegate to @hash. And as of Ruby 1.9 Hashes are ordered. "Hashes enumerate their values in the order that the corresponding keys were inserted." ruby-doc.org/core-1.9.1/Hash.html – South

Never new there was such a thing as a set. They are awesome, thank you so much – Ruble

Answer 2

You can use | (union):

my_array | [item]

And to update my_array in place:

my_array |= [item]

Camber

or my_array |= [item] which will update my_array in place – Bighorn

What's the complexity of this? – Unlookedfor

Answer 3

You don't need to iterate through my_array by hand.

my_array.push(item1) unless my_array.include?(item1)

Edit:

As Tombart points out in his comment, using Array#include? is not very efficient. I'd say the performance impact is negligible for small Arrays, but you might want to go with Set for bigger ones.

Eighteen

you definitely don't wanna do that! array.include?(item) has complexity O(n) - so it's like iterating the whole array. have a look at this benchmark: gist.github.com/deric/4953652 – Danforth

Beautiful solution :). Readable! In my case, I cannot have a set, so I'm using this solution. – Perceval

You can also use binary search to increase the performance if the array is ordered. [1, 2, 3, 4, 5].bsearch { |e| e == 3 } – Perceval

Answer 4

You can convert item1 to array and join them:

my_array | [item1]

Utrecht

Answer 5

Important to keep in mind that the Set class and the | method (also called "Set Union") will yield an array of unique elements, which is great if you want no duplicates but which will be an unpleasant surprise if you have non-unique elements in your original array by design.

If you have at least one duplicate element in your original array that you don't want to lose, iterating through the array with an early return is worst-case O(n), which isn't too bad in the grand scheme of things.

class Array
  def add_if_unique element
    return self if include? element
    push element
  end
end

Rowlock

Answer 6

I'm not sure if it's perfect solution, but worked for me:

    host_group = Array.new if not host_group.kind_of?(Array)
    host_group.push(host)

Pteryla

This did not work for you, unless uniqueness isn't a constraint. It does not uniquely add elements to the array. >> a = [1,2,3]; => [1, 2, 3] >> a.push(3); => [1, 2, 3, 3] >> a.push(3); => [1, 2, 3, 3, 3] >> a.push(3); => [1, 2, 3, 3, 3, 3] – Esqueda