From: Serabe <serabe@...>
Date: 2008-07-15T05:24:52+09:00
Subject: Re: Factorials

2008/7/14 Victor Reyes <victor.reyes@gmail.com>:

> The "Rubish" version of your factorial solutions appear to be much faster!
> Although I find it more confusing than the "traditional" recursive solution.
> I guess my problem is understanding the role of the *inject* method.
> I use 100 instead of 5 to make it run a bit longer.

Copied from ri documentation for Enumerable#inject:

------------------------------------------------------ Enumerable#inject
     enum.inject(initial) {| memo, obj | block }  => obj
     enum.inject          {| memo, obj | block }  => obj
------------------------------------------------------------------------
     Combines the elements of _enum_ by applying the block to an
     accumulator value (_memo_) and each element in turn. At each step,
     _memo_ is set to the value returned by the block. The first form
     lets you supply an initial value for _memo_. The second form uses
     the first element of the collection as a the initial value (and
     skips that element while iterating).

        # Sum some numbers
        (5..10).inject {|sum, n| sum + n }              #=> 45
        # Multiply some numbers
        (5..10).inject(1) {|product, n| product * n }   #=> 151200

        # find the longest word
        longest = %w{ cat sheep bear }.inject do |memo,word|
           memo.length > word.length ? memo : word
        end
        longest                                         #=> "sheep"

        # find the length of the longest word
        longest = %w{ cat sheep bear }.inject(0) do |memo,word|
           memo >= word.length ? memo : word.length
        end
        longest                                         #=> 5

Regards,

Serabe

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