"Guruguhan N" <[EMAIL PROTECTED]> wrote in message
news:[EMAIL PROTECTED]
> Hi All,
Hello.
Please read Charles K. Clarkson's comments. They are spot on. I will not
repeat them here.
> In reply to my own posting, I have written a code like the one given
below.
> @X = (1 .. 30)
> $n_el = scalar(@X);
Scalar is unnecessary here. $n_el forces scalar context. Just use @X below
where you now have $n_el. One fewer variable is good.
:-)
> $n_row = $n_el/3; # 3 is the number of columns I want.
$n_row = @X / 3;
> for ($i=0; $i<$n_row; $i++) {
> for ( $j=$i; $j <$n_el; $j+=$n_row) {
@X
> printf ( "%4d\t", $X[$j]);
> }
> printf ( "\n");
Use print instead. You are not formatting here. print() is more efficient.
The following code works for numeric and non-numeric data. This problem
gives the appearance of being very simple. However, appearances are often
deceptive.
;-)
The following codes works:
-------BEGIN CODE-------
#!/usr/bin/perl
use warnings;
use strict;
my @data;
my @X = 1 .. 30;
my $cols = 3;
my $rows = int(@X / $cols) + (@X % $cols ? 1 : 0);
# unfortunately you have to add
# everything in @X to @data
for my $i( 0 .. $#X ) {
push @{ $data[$i % $rows] }, $X[$i];
}
foreach (@data) {
print join("\t", @$_) . "\n";
}
--------END CODE--------
Hope this helps,
ZO
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