How to check if a value exists in an array in Ruby

There is an in? method in ActiveSupport (part of Rails) since v3.1, as pointed out by @campaterson. So within Rails, or if you require 'active_support', you can write:

'Unicorn'.in?(['Cat', 'Dog', 'Bird']) # => false

OTOH, there is no in operator or #in? method in Ruby itself, even though it has been proposed before, in particular by Yusuke Endoh a top notch member of ruby-core.

As pointed out by others, the reverse method include? exists, for all Enumerables including Array, Hash, Set, Range:

['Cat', 'Dog', 'Bird'].include?('Unicorn') # => false

Note that if you have many values in your array, they will all be checked one after the other (i.e. O(n)), while that lookup for a hash will be constant time (i.e O(1)). So if you array is constant, for example, it is a good idea to use a Set instead. E.g:

require 'set'
ALLOWED_METHODS = Set[:to_s, :to_i, :upcase, :downcase
                       # etc
                     ]

def foo(what)
  raise "Not allowed" unless ALLOWED_METHODS.include?(what.to_sym)
  bar.send(what)
end

A quick test reveals that calling include? on a 10 element Set is about 3.5x faster than calling it on the equivalent Array (if the element is not found).

A final closing note: be wary when using include? on a Range, there are subtleties, so refer to the doc and compare with cover?...

Try

['Cat', 'Dog', 'Bird'].include?('Dog')

Use Enumerable#include:

a = %w/Cat Dog Bird/

a.include? 'Dog'

Or, if a number of tests are done,¹ you can get rid of the loop (that even include? has) and go from O(n) to O(1) with:

h = Hash[[a, a].transpose]
h['Dog']

If you want to check by a block, you could try any? or all?.

%w{ant bear cat}.any? {|word| word.length >= 3}   #=> true  
%w{ant bear cat}.any? {|word| word.length >= 4}   #=> true  
[ nil, true, 99 ].any?                            #=> true  

See Enumerable for more information.

My inspiration came from "evaluate if array has any items in ruby"

Ruby has eleven methods to find elements in an array.

The preferred one is include? or, for repeated access, creat a Set and then call include? or member?.

Here are all of them:

array.include?(element) # preferred method
array.member?(element)
array.to_set.include?(element)
array.to_set.member?(element)
array.index(element) > 0
array.find_index(element) > 0
array.index { |each| each == element } > 0
array.find_index { |each| each == element } > 0
array.any? { |each| each == element }
array.find { |each| each == element } != nil
array.detect { |each| each == element } != nil

They all return a trueish value if the element is present.

include? is the preferred method. It uses a C-language for loop internally that breaks when an element matches the internal rb_equal_opt/rb_equal functions. It cannot get much more efficient unless you create a Set for repeated membership checks.

VALUE
rb_ary_includes(VALUE ary, VALUE item)
{
  long i;
  VALUE e;

  for (i=0; i<RARRAY_LEN(ary); i++) {
    e = RARRAY_AREF(ary, i);
    switch (rb_equal_opt(e, item)) {
      case Qundef:
        if (rb_equal(e, item)) return Qtrue;
        break;
      case Qtrue:
        return Qtrue;
    }
  }
  return Qfalse;
}

member? is not redefined in the Array class and uses an unoptimized implementation from the Enumerable module that literally enumerates through all elements:

static VALUE
member_i(RB_BLOCK_CALL_FUNC_ARGLIST(iter, args))
{
  struct MEMO *memo = MEMO_CAST(args);

  if (rb_equal(rb_enum_values_pack(argc, argv), memo->v1)) {
    MEMO_V2_SET(memo, Qtrue);
    rb_iter_break();
  }
  return Qnil;
}

static VALUE
enum_member(VALUE obj, VALUE val)
{
  struct MEMO *memo = MEMO_NEW(val, Qfalse, 0);

  rb_block_call(obj, id_each, 0, 0, member_i, (VALUE)memo);
  return memo->v2;
}

Translated to Ruby code this does about the following:

def member?(value)
  memo = [value, false, 0]
  each_with_object(memo) do |each, memo|
    if each == memo[0]
      memo[1] = true 
      break
    end
  memo[1]
end

Both include? and member? have O(n) time complexity since the both search the array for the first occurrence of the expected value.

We can use a Set to get O(1) access time at the cost of having to create a Hash representation of the array first. If you repeatedly check membership on the same array this initial investment can pay off quickly. Set is not implemented in C but as plain Ruby class, still the O(1) access time of the underlying @hash makes this worthwhile.

Here is the implementation of the Set class:

module Enumerable
  def to_set(klass = Set, *args, &block)
    klass.new(self, *args, &block)
  end
end

class Set
  def initialize(enum = nil, &block) # :yields: o
    @hash ||= Hash.new
    enum.nil? and return
    if block
      do_with_enum(enum) { |o| add(block[o]) }
    else
      merge(enum)
    end
  end

  def merge(enum)
    if enum.instance_of?(self.class)
      @hash.update(enum.instance_variable_get(:@hash))
    else
      do_with_enum(enum) { |o| add(o) }
    end
    self
  end

  def add(o)
    @hash[o] = true
    self
  end

  def include?(o)
    @hash.include?(o)
  end
  alias member? include?

  ...
end

As you can see the Set class just creates an internal @hash instance, maps all objects to true and then checks membership using Hash#include? which is implemented with O(1) access time in the Hash class.

I won't discuss the other seven methods as they are all less efficient.

There are actually even more methods with O(n) complexity beyond the 11 listed above, but I decided to not list them since they scan the entire array rather than breaking at the first match.

Don't use these:

# bad examples
array.grep(element).any? 
array.select { |each| each == element }.size > 0
...