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Re: How to represent the spatial distribution of a parameter [message #64247 is a reply to message #64172] Thu, 11 December 2008 07:36 Go to previous messageGo to previous message
Jeremy Bailin is currently offline  Jeremy Bailin
Messages: 618
Registered: April 2008
Senior Member
On Dec 11, 10:34 am, Jeremy Bailin <astroco...@gmail.com> wrote:
> On Dec 10, 8:05 am, "Jean H." <jghas...@DELTHIS.ucalgary.ANDTHIS.ca>
> wrote:
>
>
>
>> Hi,
>
>> I did something similar a while ago... here is part of it:
>
>> newX = Xdata / CellSizeX    ;Agregate the data
>> newY = Ydata / CellSizeY
>
>> nbCol = ceil((maxX+1.0) / cellSizeX)
>> nbRow = ceil((maxY+1.0) / cellSizeY)
>> nPoint = n_elements(newX)
>
>> image = lonarr(nbCol, nbRow)
>> nbPointsXY = lonarr(nbCol, nbRow)
>
>> for i = 0L, nPoint-1 do begin
>>      image[newX[i], newY[i]] += v[i]
>>      nbPointsXY[newX[i], newY[i]] += 1
>> endfor
>
>> image /= nbPointsXY   ;do the average
>
>> tvscl, image
>
>> Jean
>
> If there are on average lots of points per grid cell, it would be more
> efficient to use the reverse indices in David's HIST_ND:
>
> histimage = hist_nd(transpose([[X],[Y]], [binX,binY], $
>  min=[minX,minY], max=[maxX,maxY], reverse_indices=hiri)
> meanimage = fltarr(size(histimage,/dimen))
> for i=0l,n_elements(histimage)-1 do if histimage[i] gt 0 then $
>   meanimage[i] = mean(V[hiri[hiri[i]:hiri[i+1]-1]])
>
> If performance is still a serious issue, you should go readhttp://www.dfanning.com/code_tips/drizzling.html(it's effectively the
> same thing - you just need to divide by histimage at the end). In
> fact, go read it anyway - it's very illuminating. :-)=
>
> -Jeremy.

...although, to answer the original poster's question, I'd agree with
Giorgio - your easiest option is probably to use GRIDDATA.

-Jeremy.
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