Update 1:

just noted you want a threaded way of doing this; this isn't it

Update 2:

There's a fundamental problem with threading this. If your $b piddle is typical, your data subsets won't have the same number of matching elements in them, which makes it impossible to thread over. 

# calculate the number of elements in each chunk
pdl> $idx = ($b/5)->floor
pdl> $idx -= ($b - $idx * 5) == 0

pdl> p [$qidx->hist(-1,4,1)]->[1]
[1 5 5 9 0]

[download]

Original Response:

Hi,

Note:Your updated code doesn't compile and when fixed doesn't give the stated results, e.g.: 

pdl> p pdl($a(($b>0)*($b<=5);?)->medover, $a(($b>5)*($b<=10);?)->medov
+er, $a(($b>10)*($b<=15);?)->medover)
[7 8 11]

[download]
How about this: 
pdl> $start = 0;
pdl> $step = 5;
pdl> $d = pdl(  map { $a(($b > $start + $step *$_ )*( $b <= $start + $
+step*($_+1) );?)->medover } 0..3 )
pdl> p $d
[7 8 11 13]

[download]

In reply to Re: Grouping one piddle based on ranges of another by djerius
in thread Grouping one piddle based on ranges of another by astroman

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