| Re: rebin and half pixel offset [message #39694 is a reply to message #39580] |
Mon, 07 June 2004 01:36  |
peter.julyan
Messages: 3
Registered: June 2004
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Junior Member |
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Robert Barnett <retsil@zipworld.com.au> wrote in message news:<40B68387.6020207@zipworld.com.au>...
> I wondered if anyone can verify if I understand the behaviour of rebin
> correctly. Thanks in advance for looking at this problem.
>
> I'm currently putting together a ROI drawing program which allows the
> user to draw regions on a zoomed image. Sometimes it is preferrable to
> see a bilinear interpolated image whilst at other times it is
> preferrable to see a nearest neighbour image.
> After using rebin I noticed that there was a difference between the two
> methods. Because of the way rebin works, the bilinear method offsets the
> image and hence causes my ROI's (drawn using plots) to appear offset.
>
> The offset is 0.5 pixels if you do the shifting before rebin or it may
> be zoom/2 pixels if the shifting is done after rebin
>
> I've put together a little test program to demonstrate this.
> The input array is [0,1 ... m-2,m-1]
> This array is rebined to a larger array of size (m * zoom)
> The results of using neareast neighbour and bilinear interpolation are
> printed. The difference is also printed
>
> pro testRebinOffset, m, zoom
> m = floor(m > 1.0)
> zoom = floor(zoom > 1.0)
> n = zoom * m ; The size of the output array
> input = float(indgen(m)) ; The input array
> ; use float so that rebin an do ; floating point arithmetic
>
> ; Rebin using bilinear interpolation and then apply the shift
> bi = round(shift(rebin(input,n),zoom/2))
> ; Fix up the 'wrapping' caused by the shift function
> bi[0:zoom/2] = input[0]
> print, "Bilinear Interpolation", bi
> ; Rebin using nearest neighbour method
> nn = round(rebin(input,n,/sample))
> print, "Nearest Neighbour", nn
>
> print, "Difference", nn - bi
> end
>
> ; An example usage
>
> IDL> testrebinoffset,3,4
> Bilinear Interpolation 0 0 0 0
> 1 1 1 1 2 2
> 2 2
> Nearest Neighbour 0 0 0 0
> 1
> 1 1 1 2 2 2
> 2
> Difference 0 0 0 0 0
> 0 0 0 0 0 0
> 0
>
> This test fails when the input array is anything more complicated than
> an indgen array, however, I am fairly certain that this is the best
> approximation for making coordinates in both spaces equivalent
>
> Regards, Robbie
>
> Westmead Hospital,
> Sydney
> Australia
Robert,
This seems pretty much to make sense, this is spelt out explicitly in
the manual for CONGRID where, since 5.5, there is the /CENTER option
to "shift the interpolation so that points in the input and output
arrays are assumed to lie at the midpoint of their coordinates rather
than at their lower-left corner." I came across this a while back
doing some ROI stuff and use CONGRID in this instance.
Hope this helps, Pete.
Peter Julyan, Ph.D - PET Physicist
North Western Medical Physics
Christie Hospital NHS Trust
Withington, Manchester, M20 4BX
Tel: 0161 446 3078 Fax: 0161 446 3543
e-mail: p.julyan"funny symbol"physics.org
http://www.ManPET.man.ac.uk
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