One immediate problem is that you are only reading 1 file

#your code
#my $infile = $ARGV[0];  
#my $infile2 = $ARGV[0];  <--- $infile == $infile2
#corrected code
my $infile = $ARGV[0];
my $infile2 = $ARGV[1];

[download]

Update: found another issue

In the following code block you never initialize the value to 0 if the key does not exist.

#yourcode
while (<INFILE>){
    chomp;
    my @line = split ('\t', $_);
    my $key1 = 'A' . $line[0] . '.' . 'A' . $line[1]; ##first key
    my $key2 = 'A' . $line[1] . '.' . 'A' . $line[0]; ##key the other 
+way
    ##check to see if the key exists in the hash
     ##if it doesn't there is data in your infile, not in you names ar
+ray
    if (exists $diplotypes{$key1} && $line[0] <= $line[1]) {
        $diplotypes{$key1} += $line[2];
        }
    elsif (exists $diplotypes{$key2} && $line[0] >= $line[1]) {
        $diplotypes{$key2} += $line[2];
        }
    else{##world is out to get you
        print STDERR "No key for $key1 or $key2\n";
        next;
    }
}

[download]

Something along the lines of the following might help your issue.

while (<INFILE>){
    chomp;
    my @line = split ('\t', $_);
    my $key1 = 'A' . $line[0] . '.' . 'A' . $line[1]; ##first key
    my $key2 = 'A' . $line[1] . '.' . 'A' . $line[0]; ##key the other 
+way
    ##check to see if the key exists in the hash
    ##if it doesn't there is data in your infile, not in you names arr
+ay
    ##new logic
    if($line[0] <= $line[1]) {
        if(exists $diplotypes{$key1}) {
            $diplotypes{$key1} += $line[2];
        }
        else { 
            #key doesnt exist so add it
            $diplotypes{$key1} = $line[2];
        }
    }
    else {
        if(exists $diplotypes{$key2}) {
            $diplotypes{$key2} += $line[2];
        }
        else { 
            #key doesnt exist so add it
            $diplotypes{$key2} = $line[2];
        }
    }
}}

[download]

2nd Update: I did not notice that you had an initHashes function. That makes my 1st update unnecessary however what I posted is a more compact way of acheiving the same result. Sorry for the confusion.

Hope this helps.

In reply to Re: How to divide the value of a hash key by the value of another hash key (when the keys are equivalent)? by zek152
in thread How to divide the value of a hash key by the value of another hash key (when the keys are equivalent)? by aquinom

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