I can initialise a large 2D piddle using zeroes and then convert that to a byte piddle: 

perldl> $a = zeroes 1000, 1000
perldl> $b = byte $a
perldl> help vars
PDL variables in package main::

Name         Type   Dimension       Flow  State          Mem
----------------------------------------------------------------
$a           Double D [1000,1000]          P            7.63Mb
$b             Byte D [1000,1000]          P            0.95Mb

[download]

I can then undef $a to discard the default Double piddle and recover space 

perldl> undef $a
perldl> help vars
PDL variables in package main::

Name         Type   Dimension       Flow  State          Mem
----------------------------------------------------------------
$b             Byte D [1000,1000]          P            0.95Mb

[download]

And I can do that all in one line: 

perldl> $c = byte zeroes 1000,1000
perldl> help vars
PDL variables in package main::

Name         Type   Dimension       Flow  State          Mem
----------------------------------------------------------------
$c             Byte D [1000,1000]          P            0.95Mb
$b             Byte D [1000,1000]          P            0.95Mb

[download]

but it appears as if the latter still creates the 8-times bigger intermediary before automatically discarding it.

Is there a way to create a large 2D byte piddle directly?

I'm just starting to find my way around the PDL docs and I seem to be missing this example.

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Lingua non convalesco, consenesco et abolesco. -- Rule 1 has a caveat! -- Who broke the cabal?
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In reply to PDL How to create/initialise a large byte piddle directly. by BrowserUk

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