3 perlLoL - Manipulating Lists of Lists in Perl
7 =head1 Declaration and Access of Lists of Lists
9 The simplest thing to build is a list of lists (sometimes called an array
10 of arrays). It's reasonably easy to understand, and almost everything
11 that applies here will also be applicable later on with the fancier data
14 A list of lists, or an array of an array if you would, is just a regular
15 old array @LoL that you can get at with two subscripts, like C<$LoL[3][2]>. Here's
16 a declaration of the array:
18 # assign to our array a list of list references
21 [ "george", "jane", "elroy" ],
22 [ "homer", "marge", "bart" ],
28 Now you should be very careful that the outer bracket type
29 is a round one, that is, parentheses. That's because you're assigning to
30 an @list, so you need parentheses. If you wanted there I<not> to be an @LoL,
31 but rather just a reference to it, you could do something more like this:
33 # assign a reference to list of list references
35 [ "fred", "barney", "pebbles", "bambam", "dino", ],
36 [ "homer", "bart", "marge", "maggie", ],
37 [ "george", "jane", "alroy", "judy", ],
40 print $ref_to_LoL->[2][2];
42 Notice that the outer bracket type has changed, and so our access syntax
43 has also changed. That's because unlike C, in perl you can't freely
44 interchange arrays and references thereto. $ref_to_LoL is a reference to an
45 array, whereas @LoL is an array proper. Likewise, C<$LoL[2]> is not an
46 array, but an array ref. So how come you can write these:
51 instead of having to write these:
56 Well, that's because the rule is that on adjacent brackets only (whether
57 square or curly), you are free to omit the pointer dereferencing arrow.
58 But you cannot do so for the very first one if it's a scalar containing
59 a reference, which means that $ref_to_LoL always needs it.
61 =head1 Growing Your Own
63 That's all well and good for declaration of a fixed data structure,
64 but what if you wanted to add new elements on the fly, or build
65 it up entirely from scratch?
67 First, let's look at reading it in from a file. This is something like
68 adding a row at a time. We'll assume that there's a flat file in which
69 each line is a row and each word an element. If you're trying to develop an
70 @LoL list containing all these, here's the right way to do that:
77 You might also have loaded that from a function:
80 $LoL[$i] = [ somefunc($i) ];
83 Or you might have had a temporary variable sitting around with the
91 It's very important that you make sure to use the C<[]> list reference
92 constructor. That's because this will be very wrong:
96 You see, assigning a named list like that to a scalar just counts the
97 number of elements in @tmp, which probably isn't what you want.
99 If you are running under C<use strict>, you'll have to add some
100 declarations to make it happy:
109 Of course, you don't need the temporary array to have a name at all:
112 push @LoL, [ split ];
115 You also don't have to use push(). You could just make a direct assignment
116 if you knew where you wanted to put it:
118 my (@LoL, $i, $line);
121 $LoL[$i] = [ split ' ', $line ];
128 $LoL[$i] = [ split ' ', <> ];
131 You should in general be leery of using potential list functions
132 in a scalar context without explicitly stating such.
133 This would be clearer to the casual reader:
137 $LoL[$i] = [ split ' ', scalar(<>) ];
140 If you wanted to have a $ref_to_LoL variable as a reference to an array,
141 you'd have to do something like this:
144 push @$ref_to_LoL, [ split ];
147 Actually, if you were using strict, you'd have to declare not only
148 $ref_to_LoL as you had to declare @LoL, but you'd I<also> having to
149 initialize it to a reference to an empty list. (This was a bug in
150 perl version 5.001m that's been fixed for the 5.002 release.)
154 push @$ref_to_LoL, [ split ];
157 Ok, now you can add new rows. What about adding new columns? If you're
158 dealing with just matrices, it's often easiest to use simple assignment:
162 $LoL[$x][$y] = func($x, $y);
167 $LoL[$x][20] += func2($x);
170 It doesn't matter whether those elements are already
171 there or not: it'll gladly create them for you, setting
172 intervening elements to C<undef> as need be.
174 If you wanted just to append to a row, you'd have
175 to do something a bit funnier looking:
177 # add new columns to an existing row
178 push @{ $LoL[0] }, "wilma", "betty";
180 Notice that I I<couldn't> say just:
182 push $LoL[0], "wilma", "betty"; # WRONG!
184 In fact, that wouldn't even compile. How come? Because the argument
185 to push() must be a real array, not just a reference to such.
187 =head1 Access and Printing
189 Now it's time to print your data structure out. How
190 are you going to do that? Well, if you want only one
191 of the elements, it's trivial:
195 If you want to print the whole thing, though, you can't
200 because you'll get just references listed, and perl will never
201 automatically dereference things for you. Instead, you have to
202 roll yourself a loop or two. This prints the whole structure,
203 using the shell-style for() construct to loop across the outer
207 print "\t [ @$aref ],\n";
210 If you wanted to keep track of subscripts, you might do this:
212 for $i ( 0 .. $#LoL ) {
213 print "\t elt $i is [ @{$LoL[$i]} ],\n";
216 or maybe even this. Notice the inner loop.
218 for $i ( 0 .. $#LoL ) {
219 for $j ( 0 .. $#{$LoL[$i]} ) {
220 print "elt $i $j is $LoL[$i][$j]\n";
224 As you can see, it's getting a bit complicated. That's why
225 sometimes is easier to take a temporary on your way through:
227 for $i ( 0 .. $#LoL ) {
229 for $j ( 0 .. $#{$aref} ) {
230 print "elt $i $j is $LoL[$i][$j]\n";
234 Hmm... that's still a bit ugly. How about this:
236 for $i ( 0 .. $#LoL ) {
240 print "elt $i $j is $LoL[$i][$j]\n";
246 If you want to get at a slice (part of a row) in a multidimensional
247 array, you're going to have to do some fancy subscripting. That's
248 because while we have a nice synonym for single elements via the
249 pointer arrow for dereferencing, no such convenience exists for slices.
250 (Remember, of course, that you can always write a loop to do a slice
253 Here's how to do one operation using a loop. We'll assume an @LoL
258 for ($y = 7; $y < 13; $y++) {
259 push @part, $LoL[$x][$y];
262 That same loop could be replaced with a slice operation:
264 @part = @{ $LoL[4] } [ 7..12 ];
266 but as you might well imagine, this is pretty rough on the reader.
268 Ah, but what if you wanted a I<two-dimensional slice>, such as having
269 $x run from 4..8 and $y run from 7 to 12? Hmm... here's the simple way:
272 for ($startx = $x = 4; $x <= 8; $x++) {
273 for ($starty = $y = 7; $y <= 12; $y++) {
274 $newLoL[$x - $startx][$y - $starty] = $LoL[$x][$y];
278 We can reduce some of the looping through slices
280 for ($x = 4; $x <= 8; $x++) {
281 push @newLoL, [ @{ $LoL[$x] } [ 7..12 ] ];
284 If you were into Schwartzian Transforms, you would probably
285 have selected map for that
287 @newLoL = map { [ @{ $LoL[$_] } [ 7..12 ] ] } 4 .. 8;
289 Although if your manager accused of seeking job security (or rapid
290 insecurity) through inscrutable code, it would be hard to argue. :-)
291 If I were you, I'd put that in a function:
293 @newLoL = splice_2D( \@LoL, 4 => 8, 7 => 12 );
295 my $lrr = shift; # ref to list of list refs!
300 [ @{ $lrr->[$_] } [ $y_lo .. $y_hi ] ]
307 perldata(1), perlref(1), perldsc(1)
311 Tom Christiansen <F<tchrist@perl.com>>
313 Last udpate: Sat Oct 7 19:35:26 MDT 1995