I have written this function (and others similar to that one) But I am not sure I am using references on their full power.

My currently concerns is if I make a huge use of memory. The subroutine recieve a reference to two files, the subroutines, like these one return a hash (except &log_time, which returns a scalar). The subroutines expect a reference, thats why I use my %current_seq = ($id_name[0] => $seqs[$j]); my %freq_seq = &hexamer_freq(\%current_seq); on the subroutine, I don't think this is a good way to do it, but I can't imagine a better way to do it.

sub comparer{
    # Compare a sequence with the reference frequency of hexamers
    # First argument the file of to analyse, second argument the file of reference

    my %score;

    #Reading arguments
    my $seq = shift;
    my $ref_seq = shift;

    # Calculating the reference log2
    my %ref_seq = &read_fasta($$ref_seq);
    my %freq_ref = &hexamer_freq(\%ref_seq);

    # Counting hexamers and frequencies for each sequence
    my %seqs = &read_fasta($seq);
    while( my ($id,  $sequen) = each %seqs){

        my @id_seq = split(/\s+/, $id);
        my @id_name = split(/\./, $id_seq[0]);
        my $max = 0;
        my $min = 999;
        for (my $i = 0; $i < 3; $i++){
            last if length $sequen <= $i;
            my $sequ = substr($sequen, $i);
            next unless (defined $sequ);
            my $rev_sequ = reverse($sequ);
            my @seqs = ($sequ, $rev_sequ);
            for (my $j = 0; $j < scalar(@seqs); $j++){
                my %current_seq = ($id_name[0] => $seqs[$j]);
                my %freq_seq = &hexamer_freq(\%current_seq);

                # Handle the sequences that are too short to contain an hexamer
                if (scalar keys %freq_seq == 0){
                    print STDERR &log_time(), "Unable to calculate the Hexamer score of $id_name[0]\n";
                # Calculate the hexamer score
                my $score = 0;
                my $n_hexamers = scalar keys %freq_seq;
                foreach my $hex (keys %freq_seq){
                    if (defined $freq_ref{$hex}){
                        $score += log2($freq_seq{$hex}/$freq_ref{$hex});
                # Store the two possible candidates of "best score"
                if ($score/$n_hexamers > $max){
                    $max = $score/$n_hexamers;
                unless ($score/$n_hexamers > $min) {
                    $min = $score/$n_hexamers
            # Store the data for each sequence
            my $key = $id_name[0] . " frame: $i";
            $key .= " FWD" if $j == 0; # The fwd + or - have the same hexamers
            $key .= " REV" if $j == 1;
            $score{$key} = [$max, $min];
    return %score;

Besides, I would like to improve the way the $min is calculated. Now the 999 is an arbitrary number, I expect it won't be reached, because the $freq_ are numbers between 0 and 1, and it is unlikely to have so big numbers (but it may happen).


1 Answer 1


I removed your comments and inserted mine.

# Documentation goes to POD.

=item comparer

Compare a sequence with the reference frequency of hexamers.

First argument the file of to analyse, second argument the file of reference


sub comparer {
    my ($seq, $ref_seq) = @_; # I like arguments being processed as the first step in the sub. No need to shift twice.

    my %score;
    my %freq_ref = hexamer_freq({ read_fasta($$ref_seq) });
    my %seqs = read_fasta($seq);

    while (my ($id,  $sequen) = each %seqs) {
        my @id_seq = split ' ', $id;
        my @id_name = split /\./, $id_seq[0];
        my ($max, $min);
        for my $i (0 .. 2) {  # No need for a C-style for.
            last if length $sequen <= $i;

            my $sequ = substr $sequen, $i;
            next unless defined $sequ;

            my $rev_sequ = reverse $sequ;
            my @seqs = ($sequ, $rev_sequ);

            for my $j (0 .. $#seqs) {  # C-style eliminated again.
                my %freq_seq = hexamer_freq({ $id_name[0] => $seqs[$j] }); # Anonymous hash.

                if (keys %freq_seq == 0) {
                    print STDERR log_time(), "Unable to calculate the Hexamer score of $id_name[0]\n";

                my $score = 0;
                my $n_hexamers = keys %freq_seq; # "scalar" not needed in scalar context.
                for my $hex (keys %freq_seq){
                    if (defined $freq_ref{$hex}){
                        $score += log2($freq_seq{$hex} / $freq_ref{$hex});

                if (! defined $max || $score / $n_hexamers > $max) {
                    $max = $score / $n_hexamers;
                if (!defined $min || $score / $n_hexamers <= $min) {
                    $min = $score / $n_hexamers

                my $key = $id_name[0] . " frame: $i";
                $key .= (' FWD', ' REV')[$j] if $j < 2;  # Poor man's "switch".
                $score{$key} = [$max, $min];
    return %score
  • There's no need to call subroutines with the & prepended.
  • Semicolons after blocks are not needed.
  • To avoid guessing the minimum, just use undef and check for that in the condition. You could also try 'INF', but it's not portable.
  • Why is the subroutine's second parameter a scalar reference? Is it a very long string?
  • \$\begingroup\$ I know I don't need & but it is easier for me to identify them (it's my second month using perl :D ) The second parameter is not a long string (~20 characters), but I thought it would be better to use reference than passing it directly. I didn't know it is C-style the way I do the loops... So in general rather than using references if I should rather use anonymous hash/arrays? Many thanks \$\endgroup\$
    – llrs
    Nov 12, 2015 at 9:21
  • 1
    \$\begingroup\$ Anonymous array is a reference to an array without a variable keeping the array. If you want to pass it to a subroutine, you must use a reference in the sub. \$\endgroup\$
    – choroba
    Nov 12, 2015 at 9:29
  • \$\begingroup\$ & for subroutine calls is a holdover from Perl 4, no longer needed in Perl 5. I also recommend dropping it, as Perl is already a punctuation-heavy language. \$\endgroup\$ Nov 12, 2015 at 9:44
  • \$\begingroup\$ @200_success Ok, I'll try to avoid it. Thanks for the edit btw. \$\endgroup\$
    – llrs
    Nov 12, 2015 at 9:55

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