From: Robert Klemme <shortcutter@...>
Date: 2008-12-19T17:15:22+09:00
Subject: Re: Show newest file in directory?

2008/12/19 Mike Cargal <mike@cargal.net>:
> Just a pet peeve.... but... Sorts are expensive and should be avoided when
> possible.  Finding min and max values are prime examples.
>
> So...
>
> newest = Dir.entries(dir).max {|a,b| (File.mtime(File.join(dir,a)) <=>
> File.mtime(File.join(dir,b)))}

While we're nitpicking... :-)  In this case File.mtime is most likely
the most expensive operation as it does IO (or at least has to travel
into system call land). #max will invoke File.mtime multiple times for
the same file.

So, efficiency wise an explicit one pass solution is probably best:

max = Time.at 0
file = nil

Dir.open(dir) do |d|
  d.each do |f|
    mt = File.mtime(File.join(dir, f))
    if mt > max
      max = mt
      file = f
    end
  end
end

puts file

Btw, there is another nice short solution, which unfortunately also
suffers the multiple #mtime per File issue (you can easily see this by
placing "p f;" before the "File.mtime" in the block):

puts Dir.open(dir) {|d| d.max_by {|f| File.mtime(File.join(dir, f))}}

Kind regards

robert

-- 
remember.guy do |as, often| as.you_can - without end