in reply to PDL: efficient selfreferencing math?
I haven't used PDL at all, but maybe you have enough dimensions in your piddle to make the following approach efficient:
 Create a matrix to calculate the shifted, negative piddle (can be cached)
 Multiply the source piddle by the matrix to create the shifted negative source piddle
 Add the source piddle and the new piddle
The matrix would be (or, would be)
( 0 0 0 ...)
( 1 0 0 ...)
( 0 1 0 ...)
( 0 0 1 ...)
...
Reading from the PDL documentation, basically the following code should work (while PDL is installing):
# 4 dimensions
my $matrix = pdl([[0,0,0,0],[1,0,0,0],[0,1,0,0],[0,0,1,0]]);
my $pdl = pdl(...);
my $diff = $pdl x $matrix; # not sure about whether to use left or ri
+ghtmultiplication here
my $res = $pdl + $diff;
Re^2: PDL: efficient selfreferencing math? by BrowserUk (Pope) on Sep 04, 2011 at 08:30 UTC 
That's an interesting approach.
My piddle is just a 1D vector so maybe if I create a second 1D vector that is a copy of the first minus the first element, I can just subtract one from the other and produce the result I need?
Hm. Need to play.
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