| Re: Cleaver 2d reverse indicies? [message #63292] |
Tue, 04 November 2008 07:48 |
Jeremy Bailin
Messages: 618
Registered: April 2008
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Senior Member |
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On Nov 4, 4:56 am, Chris <beaum...@ifa.hawaii.edu> wrote:
>> In fact, come to think of it, you don't even need to do the
>> transformation. IDL just converts array[x,y,z] back into array[i]
>> anyways - this could save some extra time.
>
> IN FACT, maybe we don't need loops at all...
>
> t0 = systime(/seconds)
> mean3 = fltarr(size(nd,/dim))
>
> ind = ulindgen(nx * ny - 1) + 1 ;- don't include 0- do it manually
> bad = where(ri[ind] eq ri[ind+1], ct) + 1 ;-add 1 because ind starts
> at 1
> newRI = ri[0:nx * ny] - ri[0]
>
> runningSum = total(pxxm[ri[ri[0]:ri[nx * ny] - 1]], /cumulative)
>
> mean3[ind] = (runningSum[newRi[ind+1] - 1] - runningSum[newRi[ind] -
> 1]) / (newRi[ind+1] - newRi[ind])
> if ct ne 0 then mean3[bad] = 0 ;- fix empty bins
>
> ;-manually fill in first element
> if newRI[1] ne 0 then $
> mean3[0] = runningSum[newRI[1] - 1] / newRI[1]
>
> print, 'time: ',systime(/seconds) - t0
>
> All of the adding and subtracting of 1s is super ugly, but it runs
> about 30x faster for me. Also, the /CUMULATIVE keyword for total seems
> to be unstable - the errors between this method and the earlier method
> grow with the index number. That seems bizarre, but the errors were
> minor (.01%) for the input I used.
>
> chris
Try using /INTEGER with it.
-Jeremy.
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| Re: Cleaver 2d reverse indicies? [message #63297 is a reply to message #63292] |
Tue, 04 November 2008 01:56  |
Chris[6]
Messages: 84
Registered: July 2008
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Member |
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>
> In fact, come to think of it, you don't even need to do the
> transformation. IDL just converts array[x,y,z] back into array[i]
> anyways - this could save some extra time.
IN FACT, maybe we don't need loops at all...
t0 = systime(/seconds)
mean3 = fltarr(size(nd,/dim))
ind = ulindgen(nx * ny - 1) + 1 ;- don't include 0- do it manually
bad = where(ri[ind] eq ri[ind+1], ct) + 1 ;-add 1 because ind starts
at 1
newRI = ri[0:nx * ny] - ri[0]
runningSum = total(pxxm[ri[ri[0]:ri[nx * ny] - 1]], /cumulative)
mean3[ind] = (runningSum[newRi[ind+1] - 1] - runningSum[newRi[ind] -
1]) / (newRi[ind+1] - newRi[ind])
if ct ne 0 then mean3[bad] = 0 ;- fix empty bins
;-manually fill in first element
if newRI[1] ne 0 then $
mean3[0] = runningSum[newRI[1] - 1] / newRI[1]
print, 'time: ',systime(/seconds) - t0
All of the adding and subtracting of 1s is super ugly, but it runs
about 30x faster for me. Also, the /CUMULATIVE keyword for total seems
to be unstable - the errors between this method and the earlier method
grow with the index number. That seems bizarre, but the errors were
minor (.01%) for the input I used.
chris
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| Re: Cleaver 2d reverse indicies? [message #63300 is a reply to message #63297] |
Mon, 03 November 2008 16:40 |
Chris[6]
Messages: 84
Registered: July 2008
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Member |
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On Nov 3, 11:26 am, Brian Larsen <balar...@gmail.com> wrote:
> Chris,
>
> This works like a charm and is noticeably faster (~1.5 times). Gotta
> love this usenet group.
>
> Thanks much.
>
> Say you were to move to 3d (or 4d) does this extend in the same manner
> with just one total loop or can you just kill one? Brain is a bit
> slow on this today (it is Monday after all).
>
> Brian
>
> ------------------------------------------------------------ --------------
> Brian Larsen
> Boston University
> Center for Space Physicshttp://people.bu.edu/balarsen/Home/IDL
Yes, I think one loop through all of the indices is always possible.
In fact, come to think of it, you don't even need to do the
transformation. IDL just converts array[x,y,z] back into array[i]
anyways - this could save some extra time.
chris
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| Re: Cleaver 2d reverse indicies? [message #63308 is a reply to message #63300] |
Mon, 03 November 2008 13:26 |
Brian Larsen
Messages: 270
Registered: June 2006
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Senior Member |
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Chris,
This works like a charm and is noticeably faster (~1.5 times). Gotta
love this usenet group.
Thanks much.
Say you were to move to 3d (or 4d) does this extend in the same manner
with just one total loop or can you just kill one? Brain is a bit
slow on this today (it is Monday after all).
Brian
------------------------------------------------------------ --------------
Brian Larsen
Boston University
Center for Space Physics
http://people.bu.edu/balarsen/Home/IDL
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| Re: Cleaver 2d reverse indicies? [message #63310 is a reply to message #63308] |
Mon, 03 November 2008 13:15 |
Chris[6]
Messages: 84
Registered: July 2008
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Member |
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> bins_mean = fltarr(size(nd, /dim))
> nx = (size(nd, /dim))[0]
> ny = (size(nd, /dim))[1]
> FOR i = 0UL, nx-1 DO BEGIN
> FOR j = 0UL, ny-1 DO BEGIN
> ind_ri = [i+nx*j]
> IF ri[ind_ri] EQ ri[ind_ri+1] THEN CONTINUE ; nothing to do in
> this iteration
> ri_sel = ri[ri[ind_ri]:ri[ind_ri+1]-1]
> bins_mean[i, j] = mean(pxxm(ri_sel), /nan)
> ENDFOR
> ENDFOR
Well, it seems like you can definitely eliminated the nested loop:
FOR i = 0UL, nx * ny - 1, DO BEGIN
if ri[i] eq ri[i+1] then continue ; nothing to do here
ri_sel = ri[ri[i] : ri[i+1] - 1]
bins_mean[( i mod nx), (i / nx)] = mean(pxxm(ri_sel),/nan)
endfor
I've always wondered if there is a way to eliminated the first loop
when using reverse indices, but haven't thought of / seen a way to
yet.
chris
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