About "projective" command

조회 수: 5 (최근 30일)
Alex Molina
Alex Molina 2012년 5월 7일
I have a problem, i have an imagen and then i applied "imtransform" like this tutorial ( http://www.mathworks.com/help/toolbox/images/ref/imtransform.html). I used tform = maketform('projective'...). So the question: Is there anyway to known where goes any pixel and known what is it his transformation in the new coordinates?. I hope you can help me
Thanks.
Alex

답변 (1개)

Alex Taylor
Alex Taylor 2012년 5월 7일
Yes,
The functions tformfwd and tforminv are what you are looking for.
help tformfwd
help tforminv
You can calculate where the forward transformation T maps in the input point (1,2) using the following example:
tform = maketform('projective',[2 1 0; 2 1 2; 0.1 0 2]);
[X,Y] = tformfwd(tform,1,2)
Hope that helps,
Alex.
  댓글 수: 2
Alex Molina
Alex Molina 2012년 5월 10일
Thanks a lot Alex, it´s works perfectly
Yanliang
Yanliang 2013년 3월 27일
Hey Alex,
I am doing imtransform, yet it scale the output image automatically,
And I can not find the destination of a specific feature point.
Could you help me out?
Related codes for checking:
[ imgDouble{1}, ybound, xbound ] = imtransform( imgDouble{1}, tform );
Ap = tformfwd( tform, inlier_array );
Ap = mean(Ap);
ApCord = tformfwd( tform, [ [1, 1]; [height, 1]; [1, width]; [height, width] ] );
ApCord = [ min( ApCord ); max( ApCord ) ];
AppH = xbound(2) - xbound(1) + 1;
ApCordH = ApCord( 2, 1 ) - ApCord( 1, 1 ) + 1;
App(1) = ( Ap(1) - ApCord( 1, 1 ) + 1 ) * AppH / ApCordH + xbound(1) - 1;
AppW = ybound(2) - ybound(1) + 1;
ApCordW = ApCord( 2, 2 ) - ApCord( 1, 2 ) + 1;
App(2) = ( Ap(2) - ApCord( 1, 2 ) + 1 ) * AppW / ApCordW + ybound(1) - 1;
B_Location = mean( dest_array );
A_Location = App - [ xbound(1), ybound(1) ] + [1, 1];
offset_AB = A_Location - B_Location + [1, 1];
putative_off_B = offset_AB ;

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