How to count identical string elements in a Ruby array

Ruby Problem Overview

I have the following Array = ["Jason", "Jason", "Teresa", "Judah", "Michelle", "Judah", "Judah", "Allison"]

How do I produce a count for each identical element?

Where:
"Jason" = 2, "Judah" = 3, "Allison" = 1, "Teresa" = 1, "Michelle" = 1?

or produce a hash Where:

Where: hash = { "Jason" => 2, "Judah" => 3, "Allison" => 1, "Teresa" => 1, "Michelle" => 1 }

Ruby Solutions

Solution 1 - Ruby

Ruby v2.7+ (latest)

As of ruby v2.7.0 (released December 2019), the core language now includes Enumerable#tally - a new method, designed specifically for this problem:

names = ["Jason", "Jason", "Teresa", "Judah", "Michelle", "Judah", "Judah", "Allison"]

names.tally
#=> {"Jason"=>2, "Teresa"=>1, "Judah"=>3, "Michelle"=>1, "Allison"=>1}

Ruby v2.4+ (currently supported, but older)

The following code was not possible in standard ruby when this question was first asked (February 2011), as it uses:

Object#itself, which was added to Ruby v2.2.0 (released December 2014).
Hash#transform_values, which was added to Ruby v2.4.0 (released December 2016).

These modern additions to Ruby enable the following implementation:

names = ["Jason", "Jason", "Teresa", "Judah", "Michelle", "Judah", "Judah", "Allison"]

names.group_by(&:itself).transform_values(&:count)
#=> {"Jason"=>2, "Teresa"=>1, "Judah"=>3, "Michelle"=>1, "Allison"=>1}

Ruby v2.2+ (deprecated)

If using an older ruby version, without access to the above mentioned Hash#transform_values method, you could instead use Array#to_h, which was added to Ruby v2.1.0 (released December 2013):

names.group_by(&:itself).map { |k,v| [k, v.length] }.to_h
#=> {"Jason"=>2, "Teresa"=>1, "Judah"=>3, "Michelle"=>1, "Allison"=>1}

For even older ruby versions (<= 2.1), there are several ways to solve this, but (in my opinion) there is no clear-cut "best" way. See the other answers to this post.

Solution 2 - Ruby

names.inject(Hash.new(0)) { |total, e| total[e] += 1 ;total}

gives you

{"Jason"=>2, "Teresa"=>1, "Judah"=>3, "Michelle"=>1, "Allison"=>1}

Solution 3 - Ruby

names = ["Jason", "Jason", "Teresa", "Judah", "Michelle", "Judah", "Judah", "Allison"]
counts = Hash.new(0)
names.each { |name| counts[name] += 1 }
# => {"Jason" => 2, "Teresa" => 1, ....

Solution 4 - Ruby

Now using Ruby 2.2.0 you can leverage the itself method.

names = ["Jason", "Jason", "Teresa", "Judah", "Michelle", "Judah", "Judah", "Allison"]
counts = {}
names.group_by(&:itself).each { |k,v| counts[k] = v.length }
# counts > {"Jason"=>2, "Teresa"=>1, "Judah"=>3, "Michelle"=>1, "Allison"=>1}

Solution 5 - Ruby

Ruby 2.7+

Ruby 2.7 is introducing Enumerable#tally for this exact purpose. There's a good summary here.

In this use case:

array.tally
# => { "Jason" => 2, "Judah" => 3, "Allison" => 1, "Teresa" => 1, "Michelle" => 1 }

Docs on the features being released are here.

Solution 6 - Ruby

There's actually a data structure which does this: MultiSet.

Unfortunately, there is no MultiSet implementation in the Ruby core library or standard library, but there are a couple of implementations floating around the web.

This is a great example of how the choice of a data structure can simplify an algorithm. In fact, in this particular example, the algorithm even completely goes away. It's literally just:

Multiset.new(*names)

And that's it. Example, using https://GitHub.Com/Josh/Multimap/:

require 'multiset'

names = %w[Jason Jason Teresa Judah Michelle Judah Judah Allison]

histogram = Multiset.new(*names)
# => #<Multiset: {"Jason", "Jason", "Teresa", "Judah", "Judah", "Judah", "Michelle", "Allison"}>

histogram.multiplicity('Judah')
# => 3

Example, using http://maraigue.hhiro.net/multiset/index-en.php:

require 'multiset'

names = %w[Jason Jason Teresa Judah Michelle Judah Judah Allison]

histogram = Multiset[*names]
# => #<Multiset:#2 'Jason', #1 'Teresa', #3 'Judah', #1 'Michelle', #1 'Allison'>

Solution 7 - Ruby

Enumberable#each_with_object saves you from returning the final hash.

names.each_with_object(Hash.new(0)) { |name, hash| hash[name] += 1 }

Returns:

=> {"Jason"=>2, "Teresa"=>1, "Judah"=>3, "Michelle"=>1, "Allison"=>1}

Solution 8 - Ruby

This works.

arr = ["Jason", "Jason", "Teresa", "Judah", "Michelle", "Judah", "Judah", "Allison"]
result = {}
arr.uniq.each{|element| result[element] = arr.count(element)}

Solution 9 - Ruby

The following is a slightly more functional programming style:

array_with_lower_case_a = ["Jason", "Jason", "Teresa", "Judah", "Michelle", "Judah", "Judah", "Allison"]
hash_grouped_by_name = array_with_lower_case_a.group_by {|name| name}
hash_grouped_by_name.map{|name, names| [name, names.length]}
=> [["Jason", 2], ["Teresa", 1], ["Judah", 3], ["Michelle", 1], ["Allison", 1]]

One advantage of group_by is that you can use it to group equivalent but not exactly identical items:

another_array_with_lower_case_a = ["Jason", "jason", "Teresa", "Judah", "Michelle", "Judah Ben-Hur", "JUDAH", "Allison"]
hash_grouped_by_first_name = another_array_with_lower_case_a.group_by {|name| name.split(" ").first.capitalize}
hash_grouped_by_first_name.map{|first_name, names| [first_name, names.length]}
=> [["Jason", 2], ["Teresa", 1], ["Judah", 3], ["Michelle", 1], ["Allison", 1]]

Solution 10 - Ruby

names = ["Jason", "Jason", "Teresa", "Judah", "Michelle", "Judah", "Judah", "Allison"]
Hash[names.group_by{|i| i }.map{|k,v| [k,v.size]}]
# => {"Jason"=>2, "Teresa"=>1, "Judah"=>3, "Michelle"=>1, "Allison"=>1}

Solution 11 - Ruby

a = [1, 2, 3, 2, 5, 6, 7, 5, 5]
a.each_with_object(Hash.new(0)) { |o, h| h[o] += 1 }

# => {1=>1, 2=>2, 3=>1, 5=>3, 6=>1, 7=>1}

Credit Frank Wambutt

Solution 12 - Ruby

Lots of great implementations here.

But as a beginner I would consider this the easiest to read and implement

names = ["Jason", "Jason", "Teresa", "Judah", "Michelle", "Judah", "Judah", "Allison"]

name_frequency_hash = {}

names.each do |name|
  count = names.count(name)
  name_frequency_hash[name] = count  
end
#=> {"Jason"=>2, "Teresa"=>1, "Judah"=>3, "Michelle"=>1, "Allison"=>1}

The steps we took:

we created the hash
we looped over the names array
we counted how many times each name appeared in the names array
we created a key using the name and a value using the count

It may be slightly more verbose (and performance wise you will be doing some unnecessary work with overriding keys), but in my opinion easier to read and understand for what you want to achieve

Solution 13 - Ruby

This is more a comment than an answer, but a comment wouldn't do it justice. If you do Array = foo, you crash at least one implementation of IRB:

C:\Documents and Settings\a.grimm>irb
irb(main):001:0> Array = nil
(irb):1: warning: already initialized constant Array
=> nil
C:/Ruby19/lib/ruby/site_ruby/1.9.1/rbreadline.rb:3177:in `rl_redisplay': undefined method `new' for nil:NilClass (NoMethodError)
        from C:/Ruby19/lib/ruby/site_ruby/1.9.1/rbreadline.rb:3873:in `readline_internal_setup'
        from C:/Ruby19/lib/ruby/site_ruby/1.9.1/rbreadline.rb:4704:in `readline_internal'
        from C:/Ruby19/lib/ruby/site_ruby/1.9.1/rbreadline.rb:4727:in `readline'
        from C:/Ruby19/lib/ruby/site_ruby/1.9.1/readline.rb:40:in `readline'
        from C:/Ruby19/lib/ruby/1.9.1/irb/input-method.rb:115:in `gets'
        from C:/Ruby19/lib/ruby/1.9.1/irb.rb:139:in `block (2 levels) in eval_input'
        from C:/Ruby19/lib/ruby/1.9.1/irb.rb:271:in `signal_status'
        from C:/Ruby19/lib/ruby/1.9.1/irb.rb:138:in `block in eval_input'
        from C:/Ruby19/lib/ruby/1.9.1/irb/ruby-lex.rb:189:in `call'
        from C:/Ruby19/lib/ruby/1.9.1/irb/ruby-lex.rb:189:in `buf_input'
        from C:/Ruby19/lib/ruby/1.9.1/irb/ruby-lex.rb:103:in `getc'
        from C:/Ruby19/lib/ruby/1.9.1/irb/slex.rb:205:in `match_io'
        from C:/Ruby19/lib/ruby/1.9.1/irb/slex.rb:75:in `match'
        from C:/Ruby19/lib/ruby/1.9.1/irb/ruby-lex.rb:287:in `token'
        from C:/Ruby19/lib/ruby/1.9.1/irb/ruby-lex.rb:263:in `lex'
        from C:/Ruby19/lib/ruby/1.9.1/irb/ruby-lex.rb:234:in `block (2 levels) in each_top_level_statement'
        from C:/Ruby19/lib/ruby/1.9.1/irb/ruby-lex.rb:230:in `loop'
        from C:/Ruby19/lib/ruby/1.9.1/irb/ruby-lex.rb:230:in `block in each_top_level_statement'
        from C:/Ruby19/lib/ruby/1.9.1/irb/ruby-lex.rb:229:in `catch'
        from C:/Ruby19/lib/ruby/1.9.1/irb/ruby-lex.rb:229:in `each_top_level_statement'
        from C:/Ruby19/lib/ruby/1.9.1/irb.rb:153:in `eval_input'
        from C:/Ruby19/lib/ruby/1.9.1/irb.rb:70:in `block in start'
        from C:/Ruby19/lib/ruby/1.9.1/irb.rb:69:in `catch'
        from C:/Ruby19/lib/ruby/1.9.1/irb.rb:69:in `start'
        from C:/Ruby19/bin/irb:12:in `<main>'

C:\Documents and Settings\a.grimm>

That's because Array is a class.

Solution 14 - Ruby

arr = ["Jason", "Jason", "Teresa", "Judah", "Michelle", "Judah", "Judah", "Allison"]

arr.uniq.inject({}) {|a, e| a.merge({e => arr.count(e)})}

Time elapsed 0.028 milliseconds

interestingly, stupidgeek's implementation benchmarked:

Time elapsed 0.041 milliseconds

and the winning answer:

Time elapsed 0.011 milliseconds

Solution 15 - Ruby

With ruby 2.6 you can do:

names.to_h{ |name| [name, names.count(name)] }

gives you:

{"Jason"=>2, "Teresa"=>1, "Judah"=>3, "Michelle"=>1, "Allison"=>1}

Content Type	Original Author	Original Content on Stackoverflow
Question	user398520	View Question on Stackoverflow
Solution 1 - Ruby	Tom Lord	View Answer on Stackoverflow
Solution 2 - Ruby	Mauricio	View Answer on Stackoverflow
Solution 3 - Ruby	Dylan Markow	View Answer on Stackoverflow
Solution 4 - Ruby	Ahmed Fahmy	View Answer on Stackoverflow
Solution 5 - Ruby	SRack	View Answer on Stackoverflow
Solution 6 - Ruby	Jörg W Mittag	View Answer on Stackoverflow
Solution 7 - Ruby	Anconia	View Answer on Stackoverflow
Solution 8 - Ruby	Shreyas	View Answer on Stackoverflow
Solution 9 - Ruby	Andrew Grimm	View Answer on Stackoverflow
Solution 10 - Ruby	Arup Rakshit	View Answer on Stackoverflow
Solution 11 - Ruby	narzero	View Answer on Stackoverflow
Solution 12 - Ruby	Sami Birnbaum	View Answer on Stackoverflow
Solution 13 - Ruby	Andrew Grimm	View Answer on Stackoverflow
Solution 14 - Ruby	Alex Moore-Niemi	View Answer on Stackoverflow
Solution 15 - Ruby	Akash Kinwad	View Answer on Stackoverflow