| Re: Creating a new image from an image input in IDL [message #70798 is a reply to message #70796] |
Wed, 05 May 2010 06:37   |
bcubeb3
Messages: 3
Registered: May 2010
|
Junior Member |
|
|
On May 5, 8:01 am, Jeremy Bailin <astroco...@gmail.com> wrote:
> On May 5, 2:45 am, bcubeb3 <barry.brian.barr...@gmail.com> wrote:
>
>
>
>> After typing this line of code:
>> IMAGE=READ_TIFF(FILEPATH('/bin/butterfly.tiff'))
>> help, IMAGE
>
>> I get the output
>> IMAGE BYTE = Array[3, 4800, 6000]
>
>> Now I want to write a computer program to systematically loop through
>> each of the n x n pixels of the image and to use a coordinate system
>> in pixel units to compute new coordinates based on the formula theta_s
>> = theta - (size parameter of your choosing in units of
>> pixels)*theta_hat.
>
>> The vector theta_s tells me where to look in the original image to
>> extract intensity information which will then store in my image array.
>> I will use a bilinear scheme when assigning new intensity values that
>> will be stored for my newly created image array theta. Now I have no
>> idea how to even begin. I was looking for stuff online and I was
>> looking at help manuals but all efforts proved futile. Let me know of
>> your suggestions and I greatly appreciate your help on this.
>
>> -Barry
>
> So I think what you're saying (please correct me if I've
> misunderstood!) is that you have a simple transformation between the
> pixel coordinates of the new image and the pixel coordinates of the
> old image. So there's two steps here:
>
> 1) Calculate the transformation for each of the pixel locations of the
> new image
> 2) Copy the image values over
>
> For 1) I would do something like this:
>
> ; size of image:
> imagesize = size(image,/dimen)
> nchan=imagesize[0]
> nx=imagesize[1]
> ny=imagesize[2]
> npix = nx*ny
>
> ; get a 2D list of all pixel coordinates. This is going to take a LOT
> of memory
> ; for a 4800x6000 image!
> newcoords_2d = array_indices([nx,ny], lindgen(npix), /dimen)
>
> ; apply the transformation. I don't really understand your formula,
> ; but hopefully this example will help you.
> ; reform is needed to flatten it instead of being 1xNPIX
> ; X coordinate in old image, as a linear combination of X and Y
> ; coordinates of new image (oldX = A*newX + B*newY)
> oldcoordsX = reform( A * newcoords[0,*] + B * newcoords[1,*], npix)
> ; Y coordinates in new image, as a linear combination of X and Y
> ; coordinates of new image (oldY = C*newX + D*newY)
> oldcoordsY = reform( C * newcoords[0,*] + D * newcoords[1,*], npix)
>
> Once you've done that, copying it over is easy (though again, very
> memory-intensive):
>
> newimage = bytarr(imagesize)
> for chan=0,2 do newimage[chan,*,*] = oldimage[replicate(chan,npix),
> oldcoordsX, oldcoordsY]
>
> (NOTE: COMPLETELY UNTESTED)
>
> -Jeremy.
THIS IS RIGHT you understood my problem. I ran your code and it
works.
The only question I have is how would you use a bilinear interpolation
then. It is important that I take that into account for the after
picture after the transformation. I am reading in wikipedia what it
is. I am assuming bilinear interpolation is needed since the
transformation move pixels to fractional coordinates ??? But in any
case how would that be done. I can tweak the transformation, but I
think you gave me a great start. I am new so it took a while to
understand what your code did. I am still not clear what this line
does:
newcoords = array_indices([nx,ny], lindgen(npix), /dimen)
It initializes an array to lindgen(npix) of the same size of the array
of original pic???
Also for the transformation,
i.e.,
oldcoordsX = reform( A * newcoords[0,*] + B * newcoords[1,*], npix)
is oldcoordsX a known or is newcoords a known. From this fragment of
code, it leads me to believe that newcoords is given and oldcoordsX is
being solved by the transformation which to me means that oldcoords is
really the new coordinates after the transformation. Not entirely sure
if I am understanding your code,but it ran and I saw a before and
after picture.
best,
Barry
|
|
|
|
comp.lang.idl-pvwave archive
Members
Search
Help
Login
Home
