Enumerating Enumerable: Enumerable#count

by Joey deVilla on July 2, 2008

Welcome to the fourth installment of Enumerating Enumerable, a series of articles in which I challenge myself to do a better job of documenting Ruby’s Enumerable module than RubyDoc.org does. In this article, I’ll cover Enumerable#count, one of the new methods added to Enumerable in Ruby 1.9.

In case you missed the earlier installments, they’re listed (and linked) below:

  1. all?
  2. any?
  3. collect / map

Enumerable#count Quick Summary

Graphic representation of the Enumberable#count method in Ruby

In the simplest possible terms How many items in the collection meet the given criteria?
Ruby version 1.9 only
Expects Either:

  • An argument to be matched against the items in the collection
  • A block containing an expression to test the items in the collection
Returns The number of items in the collection that meet the given criteria.
RubyDoc.org’s entry Enumerable#count

Enumerable#count and Arrays

When used on an array and an argument is provided, count returns the number of times the value of the argument appears in the array:

# How many instances of "zoom" are there in the array?
["zoom", "schwartz", "profigliano", "zoom"].count("zoom")
=> 2

# Prior to Ruby 1.9, you'd have to use this equivalent code:
["zoom", "schwartz", "profigliano", "zoom"].select {|word| word == "zoom"}.size
=> 2

When used on an array and a block is provided, count returns the number of items in the array for which the block returns true:

# How many members of "The Mosquitoes" (a Beatles-esque band that appeared on
# "Gilligan's Island") have names following the "B*ngo" format?
["Bingo", "Bango", "Bongo", "Irving"].count {|bandmate| bandmate =~ /B[a-z]ngo/}
=> 3

# Prior to Ruby 1.9, you'd have to use this equivalent code:
["Bingo", "Bango", "Bongo", "Irving"].select {|bandmate| bandmate =~ /B[a-z]ngo/}.size

RubyDoc.org says that when count is used on an array without an argument or a block, it simply returns the number of items in the array (which is what the length/size methods do). However, when I’ve tried it in irb and ruby, I got results like this:

[1, 2, 3, 4].count
=> #<Enumerable::Enumerator:0x189d784>

Enumerable#count and Hashes

As with arrays, when used on a hash and an argument is provided, count returns the number of times the value of the argument appears in the hash. The difference is that for the comparison, each key/value pair is treated as a two-element array, with the key being element 0 and the value being element 1.

# Here's a hash where the names of recent movies are keys
# and their metacritic.com ratings are the corresponding values.
movie_ratings = {"Get Smart" => 53, "Kung Fu Panda" => 88, "The Love Guru" => 15,
"Sex and the City" => 51, "Iron Man" => 93}
=> {"Get Smart"=>53, "Kung Fu Panda"=>88, "The Love Guru"=>15, "Sex and the City"=>51, "Iron Man"=>93}

# This statement will return a count of 0, since there is no item in movie_ratings
# that's just plain "Iron Man".
movie_ratings.count("Iron Man")
=> 0

# This statement will return a count of 1, since there is an item in movie_ratings
# with the key "Iron Man" and the corresponding value 93.
movie_ratings.count(["Iron Man", 93])
=> 1

# This statement will return a count of 0. There's an item in movie_ratings
# with the key "Iron Man", but its corresponding value is NOT 92.
movie_ratings.count(["Iron Man", 92])
=> 0

count is not useful when used with a hash and an argument. It will only ever return two values:

  • 1 if the argument is a two-element array and there is an item in the hash whose key matches element [0] of the array and whose value matches element [1] of the array.
  • 0 for all other cases.

When used with a hash and a block, count is more useful. count passes each key/value pair in the hash to the block, which you can “catch” as either:

  1. A two-element array, with the key as element 0 and its corresponding value as element 1, or
  2. Two separate items, with the key as the first item and its corresponding value as the second item.

Each key/value pair is passed to the block and count returns the number of items in the hash for which the block returns true.

# Once again, a hash where the names of recent movies are keys
# and their metacritic.com ratings are the corresponding values.
movie_ratings = {"Get Smart" => 53, "Kung Fu Panda" => 88, "The Love Guru" => 15,
 "Sex and the City" => 51, "Iron Man" => 93}
=> {"Get Smart"=>53, "Kung Fu Panda"=>88, "The Love Guru"=>15, "Sex and the City"=>51, "Iron Man"=>93}

# How many movie titles in the collection start
# in the first half of the alphabet?
# (Using a one-parameter block)
movie_ratings.count {|movie| movie[0] <= "M"}
=> 3

# How many movie titles in the collection start
# in the first half of the alphabet?
# (This time using a two-parameter block)
movie_ratings.count {|title, rating| title <= "M"}
=> 3

# Here's how you'd do it in pre-1.9 Ruby:
movie_ratings.select {|movie| movie[0] <= "M"}.size
=> 3
# or...
movie_ratings.select {|title, rating| title <= "M"}.size
=> 3

# How many movies in the collection had a rating
# higher than 80?
# (Using a one-parameter block)
movie_ratings.count {|movie| movie[1] > 80}
=> 2

# How many movies in the collection had a rating
# higher than 80?
# (This time using a two-parameter block)
movie_ratings.count {|title, rating| rating > 80}
=> 2

# Here's how you'd do it in pre-1.9 Ruby:
movie_ratings.select {|title, rating| rating > 80}.size
=> 2


# How many movies in the collection have both:
# - A title starting in the second half of the alphabet?
# - A rating less than 50?
# (Using a one-parameter block)
movie_ratings.count {|movie| movie[0] >= "M" && movie[1] < 50}
=> 1

# How many movies in the collection have both:
# - A title starting in the second half of the alphabet?
# - A rating less than 50?
# (This time using a two-parameter block)
movie_ratings.count {|title, rating| title >= "M" && rating < 50}
=> 1

# Here's how you'd do it in pre-1.9 Ruby:
movie_ratings.select {|title, rating| title >= "M" && rating < 50}.size
=> 1

(You should probably skip The Love Guru completely, or at least until it gets aired on TV for free.)

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