From: Phillip Gawlowski <pg@...>
Date: 2009-12-26T06:39:17+09:00
Subject: Re: Making a counter for each word's occurrences in a string

On 25.12.2009 21:58, Ben Ben wrote:
> Also I forgot to add that how on earth ruby in the end knows to produce
> sparky =>1, the =>2, cat =>1, etc... even though the code above doesn't
> seem to be relate the counter with the word anywhere, and yet using p
> command it knows to relate each word with its occurrence's counts.

I modified your script a little:

def count_frequency(word_list)
	counts = Hash.new(0)
	for word in word_list
	counts[word] += 1
	end
	puts "counts' class: #{counts.class}"
	puts "inspect counts: #{counts.inspect}"
	puts "counts' Hash keys: #{counts.keys.join("; ")}"
	counts
end

p count_frequency(["sparky", "the", "cat", "sat", "on", "the", "mat"])

Output:
c:\Scripts>ruby word_freq.rb
counts' class: Hash
inspect counts: {"mat"=>1, "cat"=>1, "sat"=>1, "the"=>2, "on"=>1, 
"sparky"=>1}
counts' Hash keys: mat; cat; sat; the; on; sparky
{"mat"=>1, "cat"=>1, "sat"=>1, "the"=>2, "on"=>1, "sparky"=>1}


The mystery is solved in line 4:
	counts[word] += 1
which tells Ruby to use "word" as the name for the key. If the key 
doesn't exist, it is created, with the count of "1". Further ocurrences 
increment the count (obviously enough).

--
Phillip Gawlowski
Wishing everyone a merry Christmas, and happy holidays!