Problem
Several months ago, I asked Program to count vowels, which was asking for a code review of a vowel counting function in implemented in Python.
Since I’ve been trying to learn Ruby, I decided to reimplement the entire program in it. I’m mainly looking for critiques on:
- Naming conventions
- Best practices for string iteration
do/end
vs{}
- Coding standards and best practices
Of course there are some slight changes in how I implemented this in ruby vs Python. The biggest being simply using a Hash
since I couldn’t find a suitable replacement for Python’s tuples. On a unrelated side note, is there a replacement for Python’s tuples in ruby?
def count_vowels(string)
# Create a function that will receive a string. The
# function will count each vowel in the string and
# return a count of all the vowels, whether found or not,
# each being a `key, value` in a Hash.
vowels = {"a" => 0, "e" => 0, "i" => 0, "o" => 0, "u" => 0}
string.each_char do |chr|
if vowels.has_key?(chr)
vowels[chr] += 1
end
end
return vowels
end
Solution
You could use Enumerable#count
:
string.chars.count { |char| vowels.has_key?(char) }
You very rarely need to declare a variable (i
in this case), and then modify it form inside a block; there’s almost always a better way in Ruby.
So while calling count
and passing a block is the most direct, you could also have done something like:
string.chars.select { |char| vowels.has_key?(char) }.count
Same idea, just written out in separate filter/count steps.
You can also skip the return
since that’s implied at the end of a method.
I’d also use a plain array for the vowels. Looking up hash keys is faster, sure, but does it really matter that much? Nah. “Premature optimization is the root of all evil” and all that.
vowels = %w(a e i o u) # same as ["a", "e", "i", ...]
and then the block becomes:
vowels.include?(char)
If you keep the vowels array sorted, you can use bsearch
if you want to speed it up a little. Or you can construct the hash in code, instead of typing all that => 0
:
list = %w(a e i o u)
vowels = Hash[list.zip(list)] # => { "a" => "a", "e" => "e", ... }
Of course you could also do something funky like this: Remove all the vowels and see how much shorter the string is afterward:
def count_vowels(string)
string.length - string.gsub(/[aeiou]/, '').length
end
This is also easily made case-insensitive:
string.length - string.gsub(/[aeiou]/i, '').length
Or instead of gsub
you can use tr
:
string.length - string.tr('aeiou', '').length
(or 'aeiouAEIOU'
for case-insensitivity.)
Alternatively, you could extend/monkey-patch the String
class a little. I wouldn’t recommend it for production code, though. This is just illustrate some Ruby features:
class String
def vowel?
self =~ /A[aeiou]z/ ? true : false
end
end
def count_vowels(string)
string.chars.count(&:vowel?)
end
The &x
syntax is a shorthand meaning “invoke x
on each element in the collection”, so it’s equivalent to { |e| e.x }
.
Again, this is mostly for fun. I just thought it made the count_vowels
method nice and short. And of course, count_vowels
could itself be monkey-patched onto String
, so you could just call some_string.count_vowels
. But again, monkey-patching – while fun – shouldn’t be the first thing you reach for. Just illustrating the principle.
Try this
def count_vowels(string)
vowels = { "a" => 0, "e" => 0, "i" => 0, "o" => 0, "u" => 0 }
string.each_char { |char| vowels[char] += 1 if vowels.include?(char) }
vowels
end
What did I change?
- Use two space indentation
- Don’t abbr var nms, spell out
char
- Use trailing
if
statement - Use one-line loop with
{}
given the loop body is a single line now - Omit
return
, the last statement is always returned
And if you want a oneliner try
vowels = string.scan(/[aeiou]/).group_by(&:itself).transform_values(&:count)