Cylinder unwrapping with imtransform
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Elisa
am 18 Sep. 2014
Beantwortet: Elisa
am 19 Sep. 2014
Hello everybody, I am trying to make the unwrapping of a photo of a cylindrical object (only one side), I applied this code I wrote:
if true
% code
I = imread('2258.jpg');
axis on
box on
imshow(I)
ndims_in = 2;
ndims_out = 2;
forward_mapping = []
f = @(x, unused) arcsin(1/x);
inverse_mapping = f;
tdata = [];
tform = maketform('custom', ndims_in, ndims_out, ...
forward_mapping, inverse_mapping, tdata);
udata = [-1 1];
vdata = [-1 1];
xdata = [-1.57 1.57];
ydata = [-1 1];
I2 = imtransform(I, tform, 'UData', udata, 'VData', vdata, ...
'XData', xdata, 'YData', ydata);
subplot(1,2,1)
imshow(I)
subplot(1,2,2)
imshow(I2)
end
but I received lots of errors: ??? Error using ==> mldivide Matrix dimensions must agree.
Error in ==> @(x,unused)arcsin(1/x)
Error in ==> maketform>inv_composite at 592 U = feval(t.tdata(i).inverse_fcn, U, t.tdata(i));
Error in ==> tform at 56 X = feval( t.(f.fwd_fcn), U, t );
Error in ==> tforminv at 68 varargout = tform('inv', nargout, varargin{:});
Error in ==> tformarray at 241 M = tforminv(G,T);
Error in ==> imtransform at 275 B = tformarray(args.A, args.tform, args.resampler, tdims_a, tdims_b, ...
Where is the error for you? How could I solve it? Thank you a lot, Phalaen
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Akzeptierte Antwort
David Young
am 18 Sep. 2014
Bearbeitet: David Young
am 18 Sep. 2014
I haven't tested this, but here's a quick thought: try replacing
f = @(x, unused) arcsin(1/x);
with
f = @(x, unused) asin(1./x);
as you probably want the element-by-element division rather than the matrix inverse here.
[Edit: arcsin replaced with asin to take note of Youssef Khmou's helpful comment]
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Weitere Antworten (2)
Elisa
am 19 Sep. 2014
1 Kommentar
David Young
am 19 Sep. 2014
Hard to know - maybe if you attach the image as suggested by Image Analyst it might be possible to figure it out.
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