From: Brian Candler <B.Candler@...>
Date: 2003-08-02T00:44:24+09:00
Subject: Re: A bundle of newbie queries

On Sat, Aug 02, 2003 at 12:12:39AM +0900, Gawnsoft wrote:
> On Fri, 1 Aug 2003 21:30:19 +0900, dblack@superlink.net wrote (more or
> less):
> 
> >On Fri, 1 Aug 2003, Gawnsoft wrote:
> >>...
> >>    aDictionary = Hash.new(0)
> >>
> >>    aFile.each_line { | eachLine | aDictionary[ /[0-9.]+/ ] =
> >> aDictionary[ /[0-9.]+/ ] + 1 if
> >> eachLine.include?("plastic_1.1_lite-UMLtool-fw.exe") }
> >>
> >
> >Yes, you can use a regex as a key, but in your example, you're not
> >doing anything else with it :-)  You'd get the same results with:
> >
> >   dict[/blah/] = dict[/blah/] + 1
> 
> It was in IRB, so once I had the dictionary populated, I was also then
> able to 
>    aDictionary.each_value { | entry | puts entry }
> to see how often some people had downloaded the file and the like, so
> it wasn't entrirely wasted.

It seems you are doing something strange. Consider these examples:

   h = {}
   h[/blah/] = 1
   h[/blah/] = 1
   h[/blah/] = 1
   p h             #>> {/blah/=>1, /blah/=>1, /blah/=>1}

In other words, it looks like you are populating the hash with three
different Regex objects as the key. That's not very useful.

Just to confuse things a bit:

   h = Hash.new(0)
   3.times { h[/blah/] = h[/blah/] + 1 }
   p h             #>> {/blah/=>1}

(i.e. there are only two /blah/ objects this time: one is used as the key on
the left-hand side, and one as the key on the right-hand side)

I think what you want is:

   h = Hash.new(0)
   str = "wibble 1.2.3.4 bibble"

   /([0-9.]+)/ =~ str
   h[$1] += 1  if  $1

   p h             #>> {"1.2.3.4"=>1}

You also have to beware of free-standing regular expressions, because they
have some perlish side-effects in certain circumstances (inside
conditionals, I think):

    gets            # assigns to $_ as a side-effect
    if /([0-9.]+)/  # matches against $_ by default
      h[$1] += 1
    end

This usage is uncouth and deprecated.

> >or even:
> >
> >   dict[/blah/] = dict["hello!"] + 1
> 
> Hmmm - wouldn't this result in aDictionary having an entry with a key
> of "hello!" and so through my count of IP numbers off by one?

No - unlike Perl, reading from a hash element which does not exist does not
cause that element to be created. Example:

     h = Hash.new(99)
     p h["foo"]         #>> 99      -- default value returned
     p h                #>> {}      -- but the hash was not modified

Regards,

Brian.