Well, the docs
http://pdl.perl.org/PDLdocs/Core.html#sclr
state "The sclr method is useful to turn a piddle into a normal Perl
scalar."
If at works and sclr doesn't than maybe something is wrong with the sclr code?
The docs say they should have the same output with the convenience that sclr
doesn't require you to specify a position.
So, lets see what happens when I use at...
Ok, I now see this in the profiler output
@s_t_stats=sort {$b <=> $a} @t_stats;
# spent 3.92ms making 4 calls to CORE:sort, avg 980µs/call
So, ok, at wins!
Thanks for the advice!
Finally though, does this indicate a problem with sclr?
Date: Thu, 2 Feb 2012 10:31:53 -0600
Subject: Re: [Perldl] how to sort a piddle ???
From: [email protected]
To: [email protected]
CC: [email protected]
Hmm, looks to me like "sclr" may not be doing what you mean. I'm not sure,
though, because I've never noticed sclr before. At any rate, it looks like the
sort is calling the spaceship operator on piddles, but it should be using the
built-in spaceship operator because it should be sorting scalars.
What happens to your sort if you use "at" instead of "sclr"?
David
On Thu, Feb 2, 2012 at 10:01 AM, Adam Russell <[email protected]> wrote:
This thread reminded me of a pdl sort issue I was having.
I hope I not too off topic...
So, I am using pdl from within a larger body of Perl code.
I am using PDL::Stats to perform a t_test on a bunch of data I store in pdls.
Once I get done with this I sort the t-statistics and then throw away the pdls.
I noticed my code was running somewhat slower than I would have thought it
shoud.
So, I ran the code under the NYT profiler. My code was spending the majority of
its time (~80% of total execution time!)
on the last line below(yeah, I uncreatively named the pdls "pdls"):
foreach my $dim_n (0..$self->{dimension}-1){
my ($t, $df) = t_test($pdls[$dim_n][$cat_n][0],
$pdls[$dim_n][$cat_n][1]);
$t_stats[$dim_n]=$t->abs->sclr;
}
@s_t_stats=sort {$b <=> $a} @t_stats;
Here is what profiler output for that line looks like:
@s_t_stats=sort {$b <=> $a} @t_stats;
# spent 5.18s making 4 calls to CORE:sort, avg 1.29s/call
# spent 3.43s making 91451 calls to PDL::string, avg 38µs/call
# spent 939ms making 91451 calls to PDL::spaceship, avg 10µs/call
# spent 44µs making 15 calls to PDL::DESTROY, avg 3µs/call
1.29 seconds for each call to sort is very long time! From the calls made when
that line is executed it seems that for some reason
it is doing some sort of string conversion? But why? Surely $t->abs->sclr is
returning a numeric, right?
The code currently takes about 10 seconds to run. If I take care of this sort
problem I could probably get runs in
under 3 seconds.
Any advice on why this sort is so slow?
Best Regards,
Adam
> Date: Thu, 2 Feb 2012 07:07:24 -0600
> From: [email protected]
> To: [email protected]
> CC: [email protected]
> Subject: Re: [Perldl] how to sort a piddle ???
>
> I agree with Matt that you are probably looking for `qsort`.
>
> As to what
>
> @e = pdl(3,2,6,4,8,6);
> @r = sort{$a <=> $b} @e;
>
> is doing, its working perfectly; its just not doing what you mean. @e
> is a one element Perl-level array, its one element is a PDL object.
> Any sort on a one element array will return the same order, what else
> could it do.
>
> You have to remember that a PDL object is just another scalar in
> Perl's eyes, as are all objects.
>
> Here is another example
>
> @e = (pdl(3,2,6,4,8,6), pdl(5,6,2,1));
> @r = sort{$a <=> $b} @e;
>
> Here @e has two PDL object. When you sort objects numerically ( using
> <=> ), what you will actually sort on is not their contents, but their
> address in memory.
>
> The take-away message is this: PDL overloads many of the Perl
> operators, and it can feel like PDL and Perl are fully integrated, but
> in truth a PDL object is still an object, that is a scalar reference
> with methods and overloads. PDL tries to Do What You Mean when it can,
> this is not one of those times.
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