Can pdl efficiently(*) perform the equivalent of the following operation on a piddle? 

$pdl[ $_ -1 ] -= $pdl[ $_ ] for reverse 1 .. $#pdl;

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(*)Ie. Without converting every element to a perl var, perform the math and then convert the result back, and assign it back.

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In reply to PDL: efficient self-referencing math? by BrowserUk

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