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The code you provided does not compile due to a few syntax errors:

• Missing semicolon after the 1st my statement.
• Do not use the % sigil to assign values to a hash; use the $ sigil. For example: $high_temp_days{$person} =

Is it possible for you to show a little of the structure and contents of BodyTemp? You could do this with:

use Data::Dumper;
print Dumper(\%BodyTemp);
[download]

You may also find HASHES OF HASHES helpful in figuring out how to access this type of data structure.

Also, it is a good practice to use the strictures:

use warnings;
use strict;
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Toolic, I had already read the link you provided (and I am using #strict in my code), but didn't find what I needed. Let me try to ask again, using the examples provided in the link (with a few values changes). Please understand also that I've consulted about six Perl books (3 Llama ones) in search of the answer, so I've spent a fair bit of time on this.) I want to count the number of members of each family named jane. I tried the following code:
#to make this easier to read, HoH is defined below.
my ($group);
for $group ("simpsons","jetsons" ) {
my $janemembers=grep {$_ eq "jane"} values $HoH{$group};
}
The error given is that "Type of arg1 to values must be hash (not hash element). What's going wrong??
%HoH = (
flintstones => {
lead => "fred",
pal => "barney",
},
jetsons => {
lead => "george",
wife => "jane",
kid => "jane",
},
simpsons => {
lead => "homer",
wife => "marge",
kid => "bart",
},
);

The error message "Type of arg1 to values must be hash (not hash element)" is giving you precisely the fix you need.

$HoH{$group} is a hash element. The keys function you have in front of it wants a hash, not a hash element. Since that hash element is a reference to the actual hash, you must dereference the reference to get to the hash. In this case, wrap %{ ... } around it, as in %{$HoH{$group}}, and put keys in front of that.

I'm surprised you checked "3 llamas". The llama doesn't talk about references (deliberately, because there's only so much we can cover in a week). You want The Alpaca instead, which covers this topic (and more!).

-- Randal L. Schwartz, Perl hacker

When posting code, please use "code" tags, not "br" tags, as described in Writeup Formatting Tips.

Updated: $HoH{$group} is a reference to a hash. I refactored your code as follows:

#!/usr/bin/env perl

use warnings;
use strict;

my %HoH = (
    flintstones => {
                        lead => "fred",
                        pal => "barney",
    },
    jetsons => {
                        lead => "george",
                        wife => "jane",
                        kid => "jane",
    },
    simpsons => {
                        lead => "homer",
                        wife => "marge",
                        kid => "bart",
    },
);

my $total_jane_count = 0;
for my $group ("simpsons" ,"jetsons" ) {    
    my %family_hash = %{ $HoH{$group} };
    for my $name (values %family_hash) {
        if ($name eq 'jane') { $total_jane_count++ }
    }
} 
print "total_jane_count = $total_jane_count\n";
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This prints:

total_jane_count = 2
[download]

Don't try to “bum code” too much: there is no return-on-investment in that. The basic structure of this routine is going to be to loop through each person-ID, then inner-loop through each person's values, and within that nested loop test the value and use the key to populate a separate hash. It's going to be a dozen source-lines at most. So, just find a correct way to do it and don't worry about it being an elegant way to do it. “Correct” is always “elegant.”

Thank you for the quick response. I know that I can use a foreach loop, and that's how I've been doing it so far. But, for future reference, if I want to refer to a row of a hash in grep, how would I do it? Or alternatively, what would be the syntax for assigning the row of the two dimensional hash to a new variable (itself a one dimensional hash).