PERLLOL(1)
NNAAMMEE
perlLoL - Manipulating Lists of Lists in Perl
DDEESSCCRRIIPPTTIIOONN
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
structures.
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
array:
# assign to our array a list of list references
@LoL = (
[ "fred", "barney" ],
[ "george", "jane", "elroy" ],
[ "homer", "marge", "bart" ],
);
print $LoL[2][2];
bart
Now you should be very careful that the outer bracket type
is a round one, that is, parentheses. 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", "alroy", "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:
$LoL[2][2]
$ref_to_LoL->[2][2]
instead of having to write these:
$LoL[2]->[2]
$ref_to_LoL->[2]->[2]
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
wrong:
$LoL[$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 $ref_to_LoL variable as a
reference to an array, you'd have to do something like
this:
while (<>) {
push @$ref_to_LoL, [ split ];
}
Actually, if you were using strict, you'd have to declare
not only $ref_to_LoL as you had to declare @LoL, but you'd
also having to initialize it to a reference to an empty
list. (This was a bug in perl version 5.001m that's been
fixed for the 5.002 release.)
my $ref_to_LoL = [];
while (<>) {
push @$ref_to_LoL, [ split ];
}
Ok, 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
say
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
this:
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
through:
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";
}
}
SSlliicceess
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
reader.
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
function:
@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;
}
SSEEEE AALLSSOO
perldata(1), perlref(1), perldsc(1)
AAUUTTHHOORR
Tom Christiansen lt;tchrist@perl.com
Last udpate: Sat Oct 7 19:35:26 MDT 1995