On 8/24/07, Kirk Wythers <[EMAIL PROTECTED]> wrote:
>
> On Aug 23, 2007, at 11:17 PM, Gunnar Hjalmarsson wrote:
>
> > Kirk Wythers wrote:
> >> I don't see how $totals{$year}{$month}{count} ++; is holding the
> >> count.
> >
> > Read about the auto-increment operator in "perldoc perlop".
>
> OK. I'll try and be more clear to the degree of my ignorance. First,
> I do not understand the use of $totals in both the sum of the scalar
> tmax, and the incremented count.
snip
> I think what I am confused by is the relationship between the scalar
> $tmax and the (whatever you call it) $totals{$year}{$month}{tmax}
snip
The $tmax variable is the current value the fourth field for each line you read.
The name of the totals variable is %totals. It is a hash. A hash is
like an array, but instead of using numbers as offsets it uses strings
as keys (it is also unordered). Elements in the hash are accessed by
giving the hash a key like this:
my %hash = (
foo => 10;
bar => 20;
);
my $scalar = $hash{foo}; #scalar is now 10
Hashes can only contain scalar values, but a reference to a hash (or
an array) is a scalar, so Perl mimics multi-dimensional data
structures using references. You can create hash references in many
ways. One common way is to use an anonymous hash reference:
my $hashref = {
foo => 10,
bar => 20
};
You can get at the values stored in a hash ref with the arrow operator:
my $scalar = $hashref->{bar}; #$scalar is now 20
So if I want a multi-dimensional hash I can declare it like this:
my %hash = (
foo => $hashref
bar => {
foo => 30,
bar => 40
}
);
I could get the first level like this
my $ref = $hash{foo}; #$ref is now a reference to $hashref
and the second like this
my $scalar = $ref->{bar}; #$scalar is now 20
You could also do it more directly like this
my $scalar $hash{foo}->{bar}; #$scalar is now 20
Now, this happens often enough and is fairly unambiguous, so Perl has
some syntactic sugar that lets drop the arrow operator between {} and
[] operators. So you could write the last example more clearly as
my $scalar $hash{foo}{bar}; #$scalar is now 20
In addition, Perl also has something called auto-vivification that
causes hash references to auto-magically come into existence when
needed, so I can say
my %hash;
$hash{foo}{bar} = 10;
and the hash reference needed to store the second level will
auto-magically come into existence. This is how
$total{$year}{$month}{tmax} += $tmax;
works. It is creating a three level deep hash and storing the totals
in the last level.
snip
> Therefore, if I want the sum of tmax, I'm not sure what is holding
> the total of tmax. Is it $tmax, or $totals{$year}{$month}{tmax}?
> The same goes for count. What is the holding the count?
snip
After the loop finishes the sum of all of the tmax fields for a given
year and month will be in
$total{$year}{$month}{tmax}
where $year is the year you want and $month is the month you want.
> I guess I'm just used to the very simple $ as defining the scalar.
$ does is the sigil associated with scalar values, which is why, in
Perl 5, you also see them when referring single element (which is a
scalar) in an array or hash (you see @ when referring to multiple
values, as my solution did frequently).
snip
> SImilarly, if I want the average of $tmax, then I would want to
> divide $totals{$year}{$month}{tmax} by $totals{$year}{$month}{count}
snip
Yes.
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