PERLLOL(1)       Perl Programmers Reference Guide      PERLLOL(1)

       perlLoL - Manipulating Lists of Lists in Perl

DDeeccllaarraattiioonn aanndd AAcccceessss ooff LLiissttss ooff LLiissttss
       The simplest thing to build is a list of lists (sometimes
       called an array of arrays).  It's reasonably easy to
       understand, and almost everything that applies here will
       also be applicable later on with the fancier data

       A list of lists, or an array of an array if you would, is
       just a regular old array @LoL that you can get at with two
       subscripts, like $LoL[3][2].  Here's a declaration of the

           # assign to our array a list of list references
           @LoL = (
                  [ "fred", "barney" ],
                  [ "george", "jane", "elroy" ],
                  [ "homer", "marge", "bart" ],

           print $LoL[2][2];

       Now you should be very careful that the outer bracket type
       is a round one, that is, a parenthesis.  That's because
       you're assigning to an @list, so you need parentheses.  If
       you wanted there not to be an @LoL, but rather just a
       reference to it, you could do something more like this:

           # assign a reference to list of list references
           $ref_to_LoL = [
               [ "fred", "barney", "pebbles", "bambam", "dino", ],
               [ "homer", "bart", "marge", "maggie", ],
               [ "george", "jane", "elroy", "judy", ],

           print $ref_to_LoL->[2][2];

       Notice that the outer bracket type has changed, and so our
       access syntax has also changed.  That's because unlike C,
       in perl you can't freely interchange arrays and references
       thereto.  $ref_to_LoL is a reference to an array, whereas
       @LoL is an array proper.  Likewise, $LoL[2] is not an
       array, but an array ref.  So how come you can write these:


       instead of having to write these:


       Well, that's because the rule is that on adjacent brackets
       only (whether square or curly), you are free to omit the
       pointer dereferencing arrow.  But you cannot do so for the
       very first one if it's a scalar containing a reference,
       which means that $ref_to_LoL always needs it.

GGrroowwiinngg YYoouurr OOwwnn
       That's all well and good for declaration of a fixed data
       structure, but what if you wanted to add new elements on
       the fly, or build it up entirely from scratch?

       First, let's look at reading it in from a file.  This is
       something like adding a row at a time.  We'll assume that
       there's a flat file in which each line is a row and each
       word an element.  If you're trying to develop an @LoL list
       containing all these, here's the right way to do that:

           while (<>) {
               @tmp = split;
               push @LoL, [ @tmp ];

       You might also have loaded that from a function:

           for $i ( 1 .. 10 ) {
               $LoL[$i] = [ somefunc($i) ];

       Or you might have had a temporary variable sitting around
       with the list in it.

           for $i ( 1 .. 10 ) {
               @tmp = somefunc($i);
               $LoL[$i] = [ @tmp ];

       It's very important that you make sure to use the [] list
       reference constructor.  That's because this will be very

           $LoL[$i] = @tmp;

       You see, assigning a named list like that to a scalar just
       counts the number of elements in @tmp, which probably
       isn't what you want.

       If you are running under use strict, you'll have to add
       some declarations to make it happy:

           use strict;
           my(@LoL, @tmp);
           while (<>) {
               @tmp = split;
               push @LoL, [ @tmp ];

       Of course, you don't need the temporary array to have a
       name at all:

           while (<>) {
               push @LoL, [ split ];

       You also don't have to use push().  You could just make a
       direct assignment if you knew where you wanted to put it:

           my (@LoL, $i, $line);
           for $i ( 0 .. 10 ) {
               $line = <>;
               $LoL[$i] = [ split ' ', $line ];

       or even just

           my (@LoL, $i);
           for $i ( 0 .. 10 ) {
               $LoL[$i] = [ split ' ', <> ];

       You should in general be leery of using potential list
       functions in a scalar context without explicitly stating
       such.  This would be clearer to the casual reader:

           my (@LoL, $i);
           for $i ( 0 .. 10 ) {
               $LoL[$i] = [ split ' ', scalar(<>) ];

       If you wanted to have a $ref_to_LoL variable as a
       reference to an array, you'd have to do something like

           while (<>) {
               push @$ref_to_LoL, [ split ];

       Now you can add new rows.  What about adding new columns?
       If you're dealing with just matrices, it's often easiest
       to use simple assignment:

           for $x (1 .. 10) {
               for $y (1 .. 10) {
                   $LoL[$x][$y] = func($x, $y);

           for $x ( 3, 7, 9 ) {
               $LoL[$x][20] += func2($x);

       It doesn't matter whether those elements are already there
       or not: it'll gladly create them for you, setting
       intervening elements to undef as need be.

       If you wanted just to append to a row, you'd have to do
       something a bit funnier looking:

           # add new columns to an existing row
           push @{ $LoL[0] }, "wilma", "betty";

       Notice that I couldn't say just:

           push $LoL[0], "wilma", "betty";  # WRONG!

       In fact, that wouldn't even compile.  How come?  Because
       the argument to push() must be a real array, not just a
       reference to such.

AAcccceessss aanndd PPrriinnttiinngg
       Now it's time to print your data structure out.  How are
       you going to do that?  Well, if you want only one of the
       elements, it's trivial:

           print $LoL[0][0];

       If you want to print the whole thing, though, you can't

           print @LoL;         # WRONG

       because you'll get just references listed, and perl will
       never automatically dereference things for you.  Instead,
       you have to roll yourself a loop or two.  This prints the
       whole structure, using the shell-style for() construct to
       loop across the outer set of subscripts.

           for $aref ( @LoL ) {
               print "\t [ @$aref ],\n";

       If you wanted to keep track of subscripts, you might do

           for $i ( 0 .. $#LoL ) {
               print "\t elt $i is [ @{$LoL[$i]} ],\n";

       or maybe even this.  Notice the inner loop.

           for $i ( 0 .. $#LoL ) {
               for $j ( 0 .. $#{$LoL[$i]} ) {
                   print "elt $i $j is $LoL[$i][$j]\n";

       As you can see, it's getting a bit complicated.  That's
       why sometimes is easier to take a temporary on your way

           for $i ( 0 .. $#LoL ) {
               $aref = $LoL[$i];
               for $j ( 0 .. $#{$aref} ) {
                   print "elt $i $j is $LoL[$i][$j]\n";

       Hmm... that's still a bit ugly.  How about this:

           for $i ( 0 .. $#LoL ) {
               $aref = $LoL[$i];
               $n = @$aref - 1;
               for $j ( 0 .. $n ) {
                   print "elt $i $j is $LoL[$i][$j]\n";

       If you want to get at a slice (part of a row) in a
       multidimensional array, you're going to have to do some
       fancy subscripting.  That's because while we have a nice
       synonym for single elements via the pointer arrow for
       dereferencing, no such convenience exists for slices.
       (Remember, of course, that you can always write a loop to
       do a slice operation.)

       Here's how to do one operation using a loop.  We'll assume
       an @LoL variable as before.

           @part = ();
           $x = 4;
           for ($y = 7; $y < 13; $y++) {
               push @part, $LoL[$x][$y];

       That same loop could be replaced with a slice operation:

           @part = @{ $LoL[4] } [ 7..12 ];

       but as you might well imagine, this is pretty rough on the

       Ah, but what if you wanted a two-dimensional slice, such
       as having $x run from 4..8 and $y run from 7 to 12?
       Hmm... here's the simple way:

           @newLoL = ();
           for ($startx = $x = 4; $x <= 8; $x++) {
               for ($starty = $y = 7; $y <= 12; $y++) {
                   $newLoL[$x - $startx][$y - $starty] = $LoL[$x][$y];

       We can reduce some of the looping through slices

           for ($x = 4; $x <= 8; $x++) {
               push @newLoL, [ @{ $LoL[$x] } [ 7..12 ] ];

       If you were into Schwartzian Transforms, you would
       probably have selected map for that

           @newLoL = map { [ @{ $LoL[$_] } [ 7..12 ] ] } 4 .. 8;

       Although if your manager accused of seeking job security
       (or rapid insecurity) through inscrutable code, it would
       be hard to argue. :-) If I were you, I'd put that in a

           @newLoL = splice_2D( \@LoL, 4 => 8, 7 => 12 );
           sub splice_2D {
               my $lrr = shift;        # ref to list of list refs!
               my ($x_lo, $x_hi,
                   $y_lo, $y_hi) = @_;

               return map {
                   [ @{ $lrr->[$_] } [ $y_lo .. $y_hi ] ]
               } $x_lo .. $x_hi;

       perldata(1), perlref(1), perldsc(1)

       Tom Christiansen <>

       Last update: Thu Jun  4 16:16:23 MDT 1998

27/Mar/1999            perl 5.005, patch 03                     1