| Re: Simple question in IDL, looking for solution, thank you [message #81758] |
Mon, 22 October 2012 14:37  |
Danxia
Messages: 10
Registered: September 2012
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Junior Member |
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Hey Jeremy,
I haven't read this before, but this is deadly helpful. Thanks a lot.
Danxia
On Monday, October 22, 2012 2:10:16 PM UTC-4, Jeremy Bailin wrote:
> On 10/22/12 7:55 AM, Heinz Stege wrote:
>
>> Hi Danxia,
>
>>
>
>> you didn't ask for a solution without a loop. So here is my simple
>
>> answer:
>
>>
>
>> arr=[5,2,3,1,8,3,1,2,3]
>
>> bcg=[1,2,3,2,1,4,2,3,5]
>
>> sum=intarr(max(arr)+1)
>
>> for i=0,n_elements(bcg)-1 do sum[arr[i]]+=bcg[i]
>
>> print,sum[1:*]
>
>>
>
>> Cheers, Heinz
>
>>
>
>
>
> And of course, if you need a very efficient implementation of this (i.e.
>
> if your arrays have millions of elements), then read the "chunk
>
> indexing" section of JD's HISTOGRAM tutorial
>
> http://www.idlcoyote.com/tips/histogram_tutorial.html (you HAVE read
>
> JD's HISTOGRAM tutorial, right???)
>
>
>
> -Jeremy.
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| Re: Simple question in IDL, looking for solution, thank you [message #81759 is a reply to message #81758] |
Mon, 22 October 2012 14:36  |
Danxia
Messages: 10
Registered: September 2012
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Junior Member |
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Hey Heinz,
Thanks for the reply.
Danxia
On Monday, October 22, 2012 8:01:19 AM UTC-4, Heinz Stege wrote:
> Hi Danxia,
>
>
>
> you didn't ask for a solution without a loop. So here is my simple
>
> answer:
>
>
>
> arr=[5,2,3,1,8,3,1,2,3]
>
> bcg=[1,2,3,2,1,4,2,3,5]
>
> sum=intarr(max(arr)+1)
>
> for i=0,n_elements(bcg)-1 do sum[arr[i]]+=bcg[i]
>
> print,sum[1:*]
>
>
>
> Cheers, Heinz
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| Re: Simple question in IDL, looking for solution, thank you [message #81762 is a reply to message #81759] |
Mon, 22 October 2012 11:10 |
Jeremy Bailin
Messages: 618
Registered: April 2008
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Senior Member |
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On 10/22/12 7:55 AM, Heinz Stege wrote:
> Hi Danxia,
>
> you didn't ask for a solution without a loop. So here is my simple
> answer:
>
> arr=[5,2,3,1,8,3,1,2,3]
> bcg=[1,2,3,2,1,4,2,3,5]
> sum=intarr(max(arr)+1)
> for i=0,n_elements(bcg)-1 do sum[arr[i]]+=bcg[i]
> print,sum[1:*]
>
> Cheers, Heinz
>
And of course, if you need a very efficient implementation of this (i.e.
if your arrays have millions of elements), then read the "chunk
indexing" section of JD's HISTOGRAM tutorial
http://www.idlcoyote.com/tips/histogram_tutorial.html (you HAVE read
JD's HISTOGRAM tutorial, right???)
-Jeremy.
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| Re: Simple question in IDL, looking for solution, thank you [message #81766 is a reply to message #81762] |
Mon, 22 October 2012 04:55 |
Heinz Stege
Messages: 189
Registered: January 2003
|
Senior Member |
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Hi Danxia,
you didn't ask for a solution without a loop. So here is my simple
answer:
arr=[5,2,3,1,8,3,1,2,3]
bcg=[1,2,3,2,1,4,2,3,5]
sum=intarr(max(arr)+1)
for i=0,n_elements(bcg)-1 do sum[arr[i]]+=bcg[i]
print,sum[1:*]
Cheers, Heinz
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| Re: Simple question in IDL, looking for solution, thank you [message #81849 is a reply to message #81762] |
Tue, 23 October 2012 11:07 |
Heinz Stege
Messages: 189
Registered: January 2003
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Senior Member |
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On Mon, 22 Oct 2012 14:10:18 -0400, Jeremy Bailin wrote:
> On 10/22/12 7:55 AM, Heinz Stege wrote:
>> Hi Danxia,
>>
>> you didn't ask for a solution without a loop. So here is my simple
>> answer:
>>
>> arr=[5,2,3,1,8,3,1,2,3]
>> bcg=[1,2,3,2,1,4,2,3,5]
>> sum=intarr(max(arr)+1)
>> for i=0,n_elements(bcg)-1 do sum[arr[i]]+=bcg[i]
>> print,sum[1:*]
>>
>> Cheers, Heinz
>>
>
> And of course, if you need a very efficient implementation of this (i.e.
> if your arrays have millions of elements), then read the "chunk
> indexing" section of JD's HISTOGRAM tutorial
> http://www.idlcoyote.com/tips/histogram_tutorial.html (you HAVE read
> JD's HISTOGRAM tutorial, right???)
>
> -Jeremy.
Hi Jeremy,
I suppose you mean something like the following:
h=histogram(total(bcg,/cumulative,/integer)-1,/binsize,min=0 ,reverse_indices=ri)
i=ri[0:n_elements(h)-1]-ri[0]
print,histogram(arr[i],min=1)
The histogram methods in general are very smart. The above code is
significantly faster than my, which contains the loop. However, from
my point of view, this is not a good solution.
In case of very many elements within arr (and bcg) and/or big numbers
within bcg the reverse indices array ri gets very large. The size of
ri is always greater than total(bcg). IDL may run out of memory.
So I would say, the loop may compete with the reverse indices.
When I wrote "simple answer", I had in mind that there must be another
solution. One without a loop. It is more the "IDL-style". But it is a
little bit more complex:
ii=sort(arr)
sarr=arr[ii]
tot=total(bcg[ii],/cumulative,/integer)
;
ii=where(sarr ne shift(sarr,-1),count)
if count eq 0 then ii=[n_elements(sarr)-1]
tot=tot[ii]
if count ge 2 then tot[1:*]-=tot
;
sum=lonarr(sarr[n_elements(sarr)-1]+1)
sum[sarr[ii]]=tot
;
print,sum[1:*]
This code has a moderate memory consumption and seems to be a true
alternative to both, the loop-method and the reverse-indices-method.
A word to the developers of IDL: What about a WEIGHT keyword in the
histogram function?
print,histogram(arr,weight=bcg,/integer,min=1)
This would be nice. By the way, when I type the line above, IDL
(Version 8.0.1) says:
% Keyword INTEGER not allowed in call to: HISTOGRAM
% Error occurred at: $MAIN$
% Execution halted at: $MAIN$
No integer keyword allowed in the histogram function? Strange! ;-)
Cheers, Heinz
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