In my quest to learn PDL, I am trying out simple problems. Recently
someone asked on Perlmonks the following question -- given a 3 x 100K
array of arrays, how to 'push' a fourth value into each array
calculated from the 1st and second elements. That is, given
[
[x1, y1, z1],
[x2, y2, z2]
..
[x100K, y100K, z100K]
]
get
[
[x1, y1, z1, n1],
[x2, y2, z2, n2]
..
[x100K, y100K, z100K, n100K]
]
where nn = (xn ** 2 + yn ** 2) ** 0.5
So, I piped in and suggested PDL. Given $a, a 3 x 100K piddle
# add a fourth column to hold the calculated values
$b = $a->append( zeros(1, 100000) );
# calc the value
$b->slice('3,:') .= (($b->slice('0,:') ** 2) + ($b->slice('1,:') ** 2)) ** 0.5;
Then I was curious if PDL brought any speed advantage to this
calculation. So, I wrote a little script
for my $c (10000, 100000, 1000000, 2000000, 3000000) {
my @dat = ();
for (1 .. $c) {
push @dat, [int(rand(10)), int(rand(10)), int(rand(10))];
}
by_pdl(\...@dat, $c);
#by_arr(\...@dat, $c);
}
sub by_arr {
my $t0 = time();
my ($dat, $count) = @_;
for (@$dat) {
push @$_, (($_->[0] ** 2) + ($_->[1] ** 2)) ** 0.5;
}
my $t1 = time();
print "arr $count: " . ($t1 - $t0) . " secs\n";
}
sub by_pdl {
my $t0 = time();
my ($dat, $count) = @_;
my $a = pdl @$dat;
my $b = $a->append( zeros(1, $count) );
$b->slice('3,:') .= (($b->slice('0,:') ** 2) + ($b->slice('1,:')
** 2)) ** 0.5;
my $t1 = time();
print "pdl $count: " . ($t1 - $t0) . " secs\n";
}
Here is what I got
punk...@lucknow ~$perl arr_v_pdl.pl
arr 10000: 0.00679993629455566 secs
arr 100000: 0.0664570331573486 secs
arr 1000000: 0.673020124435425 secs
arr 2000000: 1.33305907249451 secs
arr 3000000: 2.01763200759888 secs
punk...@lucknow ~$perl arr_v_pdl.pl
pdl 10000: 0.00865507125854492 secs
pdl 100000: 0.102356910705566 secs
pdl 1000000: 1.04948401451111 secs
pdl 2000000: 2.03625702857971 secs
pdl 3000000: 13.5644547939301 secs
As potentially embarrassing as it may be for me, I ask you -- am I
doing something very wrong and silly? Otherwise please explain why PDL
is so much slower than Perl.
--
Puneet Kishor
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