From: Ben Ben <argh@...>
Date: 2009-12-26T05:58:37+09:00
Subject: Re: Making a counter for each word's occurrences in a string

Ben Ben wrote:
> By looking at this method below, I couldn't understand a few things.
> Let me list the method first!
> 
> 1) def count_frequency(word_list)
> 2)   counts = Hash.new(0)
> 3)   for word in word_list
> 4)     counts[word] += 1
> 5)   end
> 6)   counts
> 7) end
> 
> p count_frequency(["sparky", "the", "cat", "sat", "on", "the", "mat"])
> 
> ----
> that code above produces {"sparky" => 1, "the" => 2, "cat"=>1, "on"=>1,
> "mat"=>1}
> 
> What I don't understand is that why line 6 is there.  OK, like 4, it
> seems counts[word] combines to pin point which word and add 1 to the
> counter of that word, right?  Because I'm confuse how ruby works as it's
> so compact and nice, and under the hood I have no idea.  So line 2 is to
> create an empty hash with 0 item and deposits as counts (object)?  Using
> for loop to go through each word in word_list, adding 1 to counts[word],
> but I thought counts[word] produces a position of a word and not a
> counter.  OK, you can say I'm completely confused about the whole code
> above.
> 
> Help?
> 
> Thanks in advance.

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.
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