in reply to Grouping one piddle based on ranges of another

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]

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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]

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Replies are listed 'Best First'.
Re^2: Grouping one piddle based on ranges of another
by astroman (Novice) on Aug 27, 2015 at 01:23 UTC

    My apologies; I had modified the appearance to try to make it more clear what I was attempting. I've fixed it now and the code runs as written.

    Yes, you are correct about the intermediate piddles (the ones of which I am finding the median) being different lengths. I was thinking perhaps there's some way to expand into extra dimensions, where each dimension would contain the values in a given $b range, and then take medover the entire array to condense it into the form of $d. Is this possible?

    Like this, only without having to manually type all of the $mask(..) .= 1 lines:

    
pdl> $e = $a(,,*3)->copy

pdl> p $e

[
 [
  [ 0  6 18  7 19  3 10  2 12  4  8  9  1 15 11 11 19 17  0  9]
 ]
 [
  [ 0  6 18  7 19  3 10  2 12  4  8  9  1 15 11 11 19 17  0  9]
 ]
 [
  [ 0  6 18  7 19  3 10  2 12  4  8  9  1 15 11 11 19 17  0  9]
 ]
]

pdl> $mask = $e->zeroes

pdl> $mask(,,0)->where(($b>0)*($b<=5)) .= 1

pdl> $mask(,,1)->where(($b>5)*($b<=10)) .= 1

pdl> $mask(,,2)->where(($b>10)*($b<=15)) .= 1

pdl> p $mask

[
 [
  [0 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
 ]
 [
  [0 0 0 0 0 0 1 1 1 1 1 0 0 0 0 0 0 0 0 0]
 ]
 [
  [0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1]
 ]
]

pdl> $mask = $mask->setbadif($mask==0)

pdl> p $mask

[
 [
  [BAD   1   1   1   1   1 BAD BAD BAD BAD BAD BAD BAD BAD BAD BAD BAD
+ BAD BAD BAD]
 ]
 [
  [BAD BAD BAD BAD BAD BAD   1   1   1   1   1 BAD BAD BAD BAD BAD BAD
+ BAD BAD BAD]
 ]
 [
  [BAD BAD BAD BAD BAD BAD BAD BAD BAD BAD BAD   1   1   1   1   1   1
+   1   1   1]
 ]
]


pdl> $f = $e*$mask

pdl> p $f

[
 [
  [BAD   6  18   7  19   3 BAD BAD BAD BAD BAD BAD BAD BAD BAD BAD BAD
+ BAD BAD BAD]
 ]
 [
  [BAD BAD BAD BAD BAD BAD  10   2  12   4   8 BAD BAD BAD BAD BAD BAD
+ BAD BAD BAD]
 ]
 [
  [BAD BAD BAD BAD BAD BAD BAD BAD BAD BAD BAD   9   1  15  11  11  19
+  17   0   9]
 ]
]

pdl> $g = $f->medover

pdl> p $g

[
 [ 7]
 [ 8]
 [11]
]

    [download]

    Thank you very much for your help!

[reply]
[d/l]

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