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Re: Adding x,y events to a 2d array (quickly) [message #86433 is a reply to message #86432] Thu, 07 November 2013 14:20 Go to previous messageGo to previous message
Dick Jackson is currently offline  Dick Jackson
Messages: 347
Registered: August 1998
Senior Member
Phillip Bitzer wrote, On 2013-11-07, 12:16pm:
> On Thursday, November 7, 2013 1:27:00 PM UTC-6, Dick Jackson wrote:
>
>> I seem to recall someone explaining this behaviour before, and thanks to
>>
>> Russell, I realize one good way of getting *part* of what you (reasonably!) want
>>
>> to do. If all of your 'e' values were equal, then you can find how many counts
>>
>> of each (x,y) pair exist by using Hist_ND:
>>
>> (http://tir.astro.utoledo.edu/idl/hist_nd.pro)
>>
>> IDL> Print, Hist_ND(Transpose([[1,1,2],[1,1,2]]), 1, Min=0)
>>
>> But, in general, to add a varying set of 'e' values to those (x,y) locations...
>>
>> I have to think a bit...
>>
>
> I've got you covered....
>
> Oliver, reverse indices are your friend here, as Russell alluded to. Get the two-dimensional histogram, slightly modified from Dick's version:
>
> h = HIST_ND( [ TRANSPOSE(x), TRANSPOSE(y) ], 1, MIN=0, REVERSE_INDICES=ri )
>
> Since you said you have large arrays, I transpose each individually, and then concatenate.
>
> Now, go through the reverse indices:
>
> totalE = FLTARR(SIZE(h, /DIM))
> FOR i=0, N_ELEMENTS(h)-1 do if h[i] GT 0 THEN totalE[i]= TOTAL( e[ri[ri[i]:ri[i+1]-1]])
>
> print, totalE
> 0.00000 0.00000 0.00000
> 0.00000 20.0000 0.00000
> 0.00000 0.00000 10.0000
>
> This is the basic idea. It can be sped up by only looping over the elements of h with non-zero counts (as opposed to "skipping" them as I did here).
>
> Here's some highly recommended reading on histograms: http://www.idlcoyote.com/tips/histogram_tutorial.html

Histograms and reverse-indices are amazingly powerful and the right way to go in
many tough problems, but I think Oliver is looking for a solution avoiding loops
(I am too!). If a loop solution were OK, the last block here would be more
direct, with no need for histograms:

x=[1,1,2]
y=[1,1,2]
e=[10,11,12]

counts=fltarr(3,3)
counts(x,y)++
Print, 'counts:'
Print, counts ; Shows that three increments by 1 were done

totalenergy=fltarr(3,3)
totalenergy(x,y)+=e
Print, 'totalenergy:'
Print, totalenergy ; It appears that only two increments by 10 were done

totalenergy2=fltarr(3,3)
FOR i=0, N_Elements(x)-1 DO totalenergy2(x[i],y[i])+=e[i]
Print, 'totalenergy2:'
Print, totalenergy2 ; All three increments were done

... which gives us:

counts:
0.000000 0.000000 0.000000
0.000000 2.00000 0.000000
0.000000 0.000000 1.00000
totalenergy:
0.000000 0.000000 0.000000
0.000000 11.0000 0.000000
0.000000 0.000000 12.0000
totalenergy2:
0.000000 0.000000 0.000000
0.000000 21.0000 0.000000
0.000000 0.000000 12.0000

Still looking for the "IDL way" (read: "ideal way") to do this...

--

Cheers,
-Dick

Dick Jackson Software Consulting
Victoria, BC, Canada
www.d-jackson.com

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