On 8/5/05 1:55 PM, in article
1123264545.837438.93790@g47g2000cwa.googlegroups.com, "rechoncho@yahoo.com"
<rechoncho@yahoo.com> wrote:
> I'm trying to figure out an IDL-efficient way to resample a series of
> images. I know how to do this is a ruinously laborious fashion using
> loops but I know there's any easier way.
>
> Consider the following 4x4 array:
>
> x = [$
> [0,3,4,5],$
> [1,2,7,0],$
> [3,2,9,0],$
> [7,0,5,6]]
>
> I want to resample this to y, a 2x2 array. Each element would contain
> the maximum value of the corresponding 4 pixels. y would then look like
>
> [$
> [3,7],$
> [7,9]]
>
> So element [0,0] in y is max(x[0:1,0:1]) etc. As I understand it,
> rebin/congrid won't do this. Each image is about 5000x2000 and there
> are several hundred to process.
>
> Thanks!
>
Try this (I don't know how it scales with image size, but give it a try!)
The basic idea is to make four rebinned copies of the image with the upper
left, upper right, lower left, and lower right pixels in each 2x2 pair and
to replace the array elements by the largest value as you compare each of
the rebinned images in turn. Probably you can cut down on memory use by
using TEMPORARY() and there may be other increases in efficiency as well.
Good luck!
Dick French
x=[$
[0,3,4,5,3,7],$
[1,2,7,0,2,9],$
[3,2,9,0,4,2],$
[7,0,5,6,1,5]]
Print,x
; get size
size=size(x,/DIM)
nx=size[0]
ny=size[1]
nx2=nx/2
ny2=ny/2
; get indices of upper left element of each 2x2 cell
l=rebin(nx*2#lindgen(ny2),nx2,ny2)+rebin(lindgen(nx2)#2,nx2, ny2)
; get indices of ul,ur, ll, lr of each 2x2 cell
offsets=[0,1,nx,nx+1]
for n=0,3 do begin
xtemp=rebin(x[l+offsets[n]],nx,ny,/sample)
x=x+((xtemp-x)>0) ; replace x by current largest value
endfor
xfinal=rebin(x,nx2,ny2)
print
print,xfinal
End
IDL> .run trythis
0 3 4 5 3 7
1 2 7 0 2 9
3 2 9 0 4 2
7 0 5 6 1 5
3 7 9
7 9 5
|
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