frequency of a letter in a string - ruby

Question

When trying to find the frequency of letters in 'fantastic' I am having trouble understanding the given solution:
def letter_count(str)
counts = {}
str.each_char do |char|
next if char == " "
counts[char] = 0 unless counts.include?(char)
counts[char] += 1
end
counts
end
I tried deconstructing it and when I created the following piece of code I expected it to do the exact same thing. However it gives me a different result.
blah = {}
x = 'fantastic'
x.each_char do |char|
next if char == " "
blah[char] = 0
unless
blah.include?(char)
blah[char] += 1
end
blah
end
The first piece of code gives me the following
puts letter_count('fantastic')
>
{"f"=>1, "a"=>2, "n"=>1, "t"=>2, "s"=>1, "i"=>1, "c"=>1}
Why does the second piece of code give me
puts blah
>
{"f"=>0, "a"=>0, "n"=>0, "t"=>0, "s"=>0, "i"=>0, "c"=>0}
Can someone break down the pieces of code and tell me what the underlying difference is. I think once I understand this I'll be able to really understand the first piece of code. Additionally if you want to explain a bit about the first piece of code to help me out that'd be great as well.

Answer 1

You can't split this line...
counts[char] = 0 unless counts.include?(char)
... over multiple line the way you did it. The trailing conditional only works on a single line.
If you wanted to split it over multiple lines you would have to convert to traditional if / end (in this case unless / end) format.
unless counts.include?(char)
counts[char] = 0
end
Here's the explanation of the code...
# we define a method letter_count that accepts one argument str
def letter_count(str)
# we create an empty hash
counts = {}
# we loop through all the characters in the string... we will refer to each character as char
str.each_char do |char|
# we skip blank characters (we go and process the next character)
next if char == " "
# if there is no hash entry for the current character we initialis the
# count for that character to zero
counts[char] = 0 unless counts.include?(char)
# we increase the count for the current character by 1
counts[char] += 1
# we end the each_char loop
end
# we make sure the hash of counts is returned at the end of this method
counts
# end of the method
end

Answer 2

Now that #Steve has answered your question and you have accepted his answer, perhaps I can suggest another way to count the letters. This is just one of many approaches that could be taken.
Code
def letter_count(str)
str.downcase.each_char.with_object({}) { |c,h|
(h[c] = h.fetch(c,0) + 1) if c =~ /[a-z]/ }
end
Example
letter_count('Fantastic')
#=> {"f"=>1, "a"=>2, "n"=>1, "t"=>2, "s"=>1, "i"=>1, "c"=>1}
Explanation
Here is what's happening.
str = 'Fantastic'
We use String#downcase so that, for example, 'f' and 'F' are treated as the same character for purposes of counting. (If you don't want that, simply remove .downcase.) Let
s = str.downcase #=> "fantastic"
In
s.each_char.with_object({}) { |c,h| (h[c] = h.fetch(c,0) + 1) c =~ /[a-z]/ }
the enumerator String#each_char is chained to Enumerator#with_index. This creates a compound enumerator:
enum = s.each_char.with_object({})
#=> #<Enumerator: #<Enumerator: "fantastic":each_char>:with_object({})>
We can view what the enumerator will pass to the block by converting it to an array:
enum.to_a
#=> [["f", {}], ["a", {}], ["n", {}], ["t", {}], ["a", {}],
# ["s", {}], ["t", {}], ["i", {}], ["c", {}]]
(Actually, it only passes an empty hash with 'f'; thereafter it passes the updated value of the hash.) The enumerator with_object creates an empty hash denoted by the block variable h.
The first element enum passes to the block is the string 'f'. The block variable c is assigned that value, so the expression in the block:
(h[c] = h.fetch(c,0) + 1) if c =~ /[a-z]/
evaluates to:
(h['f'] = h.fetch('f',0) + 1) if 'f' =~ /[a-z]/
Now
c =~ /[a-z]/
is true if and only if c is a lowercase letter. Here
'f' =~ /[a-z]/ #=> true
so we evaluate the expression
h[c] = h.fetch(c,0) + 1
h.fetch(c,0) returns h[c] if h has a key c; else it returns the value of Hash#fetch's second parameter, which here is zero. (fetch can also take a block.)
Since h is now empty, it becomes
h['f'] = 0 + 1 #=> 1
The enumerator each_char then passes 'a', 'n' and 't' to the block, resulting in the hash becoming
h = {'f'=>1, 'a'=>1, 'n'=>1, 't'=>1 }
The next character passed in is a second 'a'. As h already has a key 'a',
h[c] = h.fetch(c,0) + 1
evaluates to
h['a'] = h['a'] + 1 #=> 1 + 1 => 2
The remainder of the string is processed the same way.