Also, if you already have a working code, please post it so we can test our solutions against it or propose enhancements.

in the following arrays qw/name pos loc age/ and qw/car dog age/ the "age" field is present in the first array already, so I don't want to add it again, I want to add the in the "age" (=last) position in the second array to the already existing "age" position in the final combined arref..

I can post the code I built up which doesn't look nice, though it does the job.. I'd prefer to have something nicer if possible

my $arr = [[qw/name pos loc age/],
       [qw/ike boss 12 44/],
       [qw/mat slave 22 21/],
       [qw/jill sec 15 32/],];

my $add = [[qw/car dog age/],
       [qw/a1 grr 3/],
       [qw/s2 miew 7/],];

my ($arrfld,$addfld);
for ($i=0;$i<@{$arr->[0]};$i++){ $arrfld->{ $arr->[0]->[$i] } = $i }
for ($i=0;$i<@{$add->[0]};$i++){
  unless ( $arrfld->{ $add->[0]->[$i] } ){ 
    $arrfld->{ $add->[0]->[$i] } = scalar(keys %$arrfld);
  }
  $addfld->{ $arrfld->{ $add->[0]->[$i] } } = $i;
}
foreach my $row (@$arr){
  for($x=scalar(@$row);$x<scalar(keys %$arrfld);$x++){push @$row,""}
}
shift @$add;
foreach my $row (@$add){
  my $nr;
  for ($x=0;$x<scalar(keys %$arrfld);$x++){
    $nr->[$x] = defined($addfld->{$x})?$row->[$addfld->{$x}]:""; 
  }
  push @$arr,$nr;
}
[download]

Note that you replied to yourself instead of replying to me. Therefore, I did not notice your reply immediately.

Your code does not do what you claim. No arrays get merged when I run it. Please try to give more details: should the arrays be merged only if the last member is the same, or if any member is the same (i.e., should the script merge qw/a b key c/ with qw/x key y z/)? What would you do if more than one array from the second list could be merged into an array from the first one (i.e. both qw/car dog age/ and qw/train cat age/ in the second list)?

لսႽ† ᥲᥒ⚪⟊Ⴙᘓᖇ Ꮅᘓᖇ⎱ Ⴙᥲ𝇋ƙᘓᖇ

I've been staring at the screen for too long..

It's funny you say it doesn't merge, I've just copied the code from the post and used Data::Dumper to dump $arr at the end and I can see it merged as it should be..

No, the initial position of the member doesn't matter. If we look at an arbitrary first row of two arrefs:

$a->[0] = [qw/f0 f1 f2 f3/];
$b->[0] = [qw/e0 f2 e1 f0 e2];
[download]

If the value of the field in $b->[0] matches one already present in $a->[0] then all values in the matching field in the $b->1..$ arrays should be set in the already matched field. So the final array would be like:

$m = [
[qw/f0 f1 f2 f3 e0 e1 e2],
[qw/a0 a1 a2 a3 "" "" ""/],# ..and so on all of $a
[qw/b3 "" b1 "" b0 b2 b4/],# ..and so on for $b->[1..$]
[download]

a0,1,2.. being the first,second,third element of a $a->[x] array and likewise for the b0,1,2 of the $b->[x] array.

Does this make it any clearer?

You could use nested slices of position informations of the column-heads

  DB<158> @head1= @{$arr1->[0]}
 => ("name", "pos", "loc", "age")

  DB<159> @head2= @{$arr2->[0]}
 => ("car", "dog", "age")

  DB<161> %pos=();$p=0

  DB<162> for ( @head1 , @head2 )  {
           $pos{$_}=$p++ unless exists $pos{$_}
          }

  DB<163> %pos
 => ("dog", 5, "loc", 2, "car", 4, "name", 0, "pos", 1, "age", 3)
[download]

  DB<168> @n= ("") x $p    #default
 => ("", "", "", "", "", "")

  DB<169> @n[ @pos{@head1} ] = @{ $arr1->[1] }
 => ("ike", "boss", 12, 44)

  DB<170> @n
 => ("ike", "boss", 12, 44, "", "")

  DB<171> @n= ("") x $p    #default
 => ("", "", "", "", "", "")

  DB<172> @n[ @pos{@head2} ] = @{ $arr2->[1] }
 => ("a1", "grr", 3)

  DB<173> @n
 => ("", "", "", 3, "a1", "grr")
[download]