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| Re: Anyway to avoid this last for loop [message #64579 is a reply to message #64572] |
Thu, 08 January 2009 09:31   |
hldevil
Messages: 6
Registered: September 2008
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Junior Member |
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On 8 Jan., 15:28, Tom McGlynn <t...@milkyway.gsfc.nasa.gov> wrote:
> On Jan 8, 4:59 am, hlde...@gmx.de wrote:
>
>> ;AClength is how many frames are thrown out after one AC Hit ->
>> multiple entries can fall into one bin
>
>> detHist=histogram(data,REVERSE_INDICES=ri, /L64, binSize=AClength)
>> acHist=histogram(ACdeletes, /L64, binSize=AClength)
>
>> ;renormalize to one
>> dI=where(detHist GT 1, cntD)
>> aI=where(acHist GT 1, cntA)
>> IF cntD NE 0 THEN detHist[dI]=1
>> IF cntA NE 0 THEN acHist[aI]=1
>
> Can't these last four lines be replaced by
> detHist = detHist<1
> acHist = acHist<1
>
> The < and > operators are useful though not necessarily intuitive.
>
>
>
>
>
>> ;subtract the two histograms. All detector frames which have
>> corresponding AC frame should now have value 0, if AC frame exists but
>> not detector frame then value is -1. If only detector frame exists
>> (the ones we want) value stays 1!!!
>> diffHist=detHist-acHist
>
>> ;keep the ones with one
>> keep=where(diffHist EQ 1, cnt)
>
>> ;And now my problem. I need the indices of the keep-frames but there
>> can be more than one in a bin so I'm stuck with this loop:
>
>> FOR k=0L, cnt-1 DO BEGIN
>> keepIndices=[keepIndices,ri[ri[keep[k]]:ri[keep[k]+1]-1]]
>> ENDFOR
>
> To do it all at once, I think you can just go back to the data array
> as below:
>
> keepHist = where (diffHist EQ 1, cnt) ; From the original
> if (cnt gt 0) then begin
> keepers = intarr(n_elements(detHist))
> keepers[keepHist] = 1;
> keepIndices = where(keepers[data/ACLENGTH] eq 1)
> endif else begin
> keepIndices = [-1]
> endelse
>
> Haven't tested it, but it seems like it or something like it should
> work reasonably efficiently.
>
> Good luck,
> Tom McGlynn
Seems reasonable. I'll try it and see if I get the same results as
with the for loop.
Thanks,
Steffen
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| Re: Anyway to avoid this last for loop [message #64594 is a reply to message #64589] |
Thu, 08 January 2009 06:28 |
Tom McGlynn
Messages: 13
Registered: January 2008
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Junior Member |
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On Jan 8, 4:59 am, hlde...@gmx.de wrote:
> ;AClength is how many frames are thrown out after one AC Hit ->
> multiple entries can fall into one bin
>
> detHist=histogram(data,REVERSE_INDICES=ri, /L64, binSize=AClength)
> acHist=histogram(ACdeletes, /L64, binSize=AClength)
>
> ;renormalize to one
> dI=where(detHist GT 1, cntD)
> aI=where(acHist GT 1, cntA)
> IF cntD NE 0 THEN detHist[dI]=1
> IF cntA NE 0 THEN acHist[aI]=1
Can't these last four lines be replaced by
detHist = detHist<1
acHist = acHist<1
The < and > operators are useful though not necessarily intuitive.
>
> ;subtract the two histograms. All detector frames which have
> corresponding AC frame should now have value 0, if AC frame exists but
> not detector frame then value is -1. If only detector frame exists
> (the ones we want) value stays 1!!!
> diffHist=detHist-acHist
>
> ;keep the ones with one
> keep=where(diffHist EQ 1, cnt)
>
> ;And now my problem. I need the indices of the keep-frames but there
> can be more than one in a bin so I'm stuck with this loop:
>
> FOR k=0L, cnt-1 DO BEGIN
> keepIndices=[keepIndices,ri[ri[keep[k]]:ri[keep[k]+1]-1]]
> ENDFOR
>
To do it all at once, I think you can just go back to the data array
as below:
keepHist = where (diffHist EQ 1, cnt) ; From the original
if (cnt gt 0) then begin
keepers = intarr(n_elements(detHist))
keepers[keepHist] = 1;
keepIndices = where(keepers[data/ACLENGTH] eq 1)
endif else begin
keepIndices = [-1]
endelse
Haven't tested it, but it seems like it or something like it should
work reasonably efficiently.
Good luck,
Tom McGlynn
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| Re: Anyway to avoid this last for loop [message #64598 is a reply to message #64594] |
Thu, 08 January 2009 05:54 |
R.Bauer
Messages: 1424
Registered: November 1998
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Senior Member |
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There is no PEP8 for idl somewhere defined, or isn't it. And if so I
probably have to clean also my code
The example is difficult to read.
Reimar
hldevil@gmx.de schrieb:
> Hi Everybody,
>
> after having worked my way through dfannings and jdsmiths histogram
> tips I still haven't found a way to avoid one last (and time
> consuming) FOR-loop. Maybe someone has an idea.
>
> I have to lists (Millions of entries): one containing detector hit-
> times, one containing anticoincidence hit-times. Only those detector
> hits which do not have a AC event at the same time are supposed to be
> kept.
>
> I used a histogram approach:
>
> ;AClength is how many frames are thrown out after one AC Hit ->
> multiple entries can fall into one bin
>
> detHist=histogram(data,REVERSE_INDICES=ri, /L64, binSize=AClength)
> acHist=histogram(ACdeletes, /L64, binSize=AClength)
>
>
> ;renormalize to one
> dI=where(detHist GT 1, cntD)
> aI=where(acHist GT 1, cntA)
> IF cntD NE 0 THEN detHist[dI]=1
> IF cntA NE 0 THEN acHist[aI]=1
>
> ;subtract the two histograms. All detector frames which have
> corresponding AC frame should now have value 0, if AC frame exists but
> not detector frame then value is -1. If only detector frame exists
> (the ones we want) value stays 1!!!
> diffHist=detHist-acHist
>
> ;keep the ones with one
> keep=where(diffHist EQ 1, cnt)
>
> ;And now my problem. I need the indices of the keep-frames but there
> can be more than one in a bin so I'm stuck with this loop:
>
> FOR k=0L, cnt-1 DO BEGIN
> keepIndices=[keepIndices,ri[ri[keep[k]]:ri[keep[k]+1]-1]]
> ENDFOR
>
> Is there any way I can avoid this, especially because it scales with
> the number of hits which are kept.
>
> Best Regards,
>
> Steffen
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| Re: Anyway to avoid this last for loop [message #64668 is a reply to message #64594] |
Fri, 09 January 2009 10:28 |
Jeremy Bailin
Messages: 618
Registered: April 2008
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Senior Member |
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> Can't these last four lines be replaced by
> detHist = detHist<1
> acHist = acHist<1
>
> The < and > operators are useful though not necessarily intuitive.
For that matter, those operators are valid for the op= construction,
so you can write the very efficient (in both performance and typing
time):
detHist <= 1
but which will confuse the hell out of people who normally think in
most other languages where <= is a comparison. ;-)
-Jeremy.
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