I had a 1D piddle of time values, and was trying to compute the time offset
relative to the initial values. I got some unexpected results when trying to
perform an action like '$t-=$t((0));' This is with perl 5.14.2, and the (more
or less) current PDL git, but holds all the way back to PDL-2.4.7. I run the
following script (for clarity, not in the PDL shell, but it happens there too):
**************
$ cat testme.pl
use PDL;
use PDL::NiceSlice;
print "PDL::VERSION $PDL::VERSION\n";
$t_orig = pdl(12..16);
print "orig:\t$t_orig\n\n";
for $i(0..4){
$t = $t_orig->copy;
$t-=$t(($i));
print "nslice\t$t\n";
$t = $t_orig->copy;
$t-=$t->slice("($i)");
print "slice\t$t\n";
$t = $t_orig->copy;
$t-=$t->at($i);
print "at($i)\t$t\n";
$t = $t_orig->copy;
$t-=pdl($t_orig->at($i));
print "pdl\t$t\n";
$t = $t_orig->copy;
$t = $t - $t(($i));
print "exp.nsl\t$t\n\n";
}
1;
*************
The script just defines a piddle [12 13 14 15 16] and subtracts off the 0th
element in 5 different ways, then the 1st element, etc. The first four ways
(niceslice, slice, at, and pdl) all use the -= operator, and the last one does
the explicit assignment via subtraction of a niceslice. And I get the
following output:
*************
$ perl testme.pl
PDL::VERSION 2.4.11_001
orig: [12 13 14 15 16]
nslice [0 13 14 15 16]
slice [0 13 14 15 16]
at(0) [0 1 2 3 4]
pdl [0 1 2 3 4]
exp.slc [0 1 2 3 4]
nslice [-1 0 14 15 16]
slice [-1 0 14 15 16]
at(1) [-1 0 1 2 3]
pdl [-1 0 1 2 3]
exp.slc [-1 0 1 2 3]
nslice [-2 -1 0 15 16]
slice [-2 -1 0 15 16]
at(2) [-2 -1 0 1 2]
pdl [-2 -1 0 1 2]
exp.slc [-2 -1 0 1 2]
nslice [-3 -2 -1 0 16]
slice [-3 -2 -1 0 16]
at(3) [-3 -2 -1 0 1]
pdl [-3 -2 -1 0 1]
exp.slc [-3 -2 -1 0 1]
nslice [-4 -3 -2 -1 0]
slice [-4 -3 -2 -1 0]
at(4) [-4 -3 -2 -1 0]
pdl [-4 -3 -2 -1 0]
exp.slc [-4 -3 -2 -1 0]
*************
For each group of 5 methods, I was expecting the output to be what is shown in
the 'at', 'pdl', and 'exp.slc' lines. Obviously something is not quite correct
with the 'nslice', and 'slice' lines.
So, two questions: 1) why do the nslice and slice lines act on just part of the
$t piddle, and not the whole thing? 2) why does the explicit assignment w/
nslice not produce the same output as the -= operation with slice? I thought
they were exactly equivalent, that $a-=$b was just syntactic sugar for $a =
$a-$b.
cheers,
Derek
_______________________________________________
Perldl mailing list
[email protected]
http://mailman.jach.hawaii.edu/mailman/listinfo/perldl
