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image registration

Alexia werk
Posted 10 years ago

can someone has idea how to start with two totally different image for registration. Both image has few points which I marked. Then I want to make registration.

coordinate-img1 = {{125, 205}, {266, 190}, {102, 49}, {191, 32}, {258, 38}};
coordinate-img2 = {{280, 250}, {103, 251}, {91, 56}, {200, 44}, {284, 57}};

are position of the markers in both images landmark based registration or pointwise registration ? No idea how to strat with.
POSTED BY: Alexia werk
9 Replies
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Hi Alexia,

I think your problem can nicely be solved: The function FindGeometricTransform is made for this kind of task. The crucial thing is to re-oder one of the sets of points, so that there is a correspondence between the points of both sets - see attached notebook. Hope that helps!

Regards -- Henrik

Attachments:
POSTED BY: Henrik Schachner
Alexia werk
Alexia werk
Posted 10 years ago

Thank you so much.
POSTED BY: Alexia werk
Alexia werk
Alexia werk
Posted 10 years ago

With highlight function can we also make the crossmarker solid. The marker does not looks so solid. I tried but it seems its not avilable with this.
POSTED BY: Alexia werk

Method->"CrossMarkers" in HighlightImage are solid (or, at least, supposed to be). If you see this differently, please send us an example.
POSTED BY: Shadi Ashnai
Alexia werk
Alexia werk
Posted 10 years ago

see the attachment.

HighlightImage[img1, coordimg1, "HighlightColor" -> Green, Method -> {"CrossMarkers", 8}];

Attachments:
POSTED BY: Alexia werk

Here the markers are solid. Try this:

In[82]:= Union[
 PixelValue[image, 
  PixelValuePositions[ColorDistance[image, Green], Black, .1]]]

Out[82]= {{0., 0.976471, 0., 1.}}
POSTED BY: Shadi Ashnai
Alexia werk
Alexia werk
Posted 10 years ago

Thanks all. Not working for these points

coordimg1 = {{309., 174.}, {252., 98.}, {113., 116.}, {181., 161.}}

coordimg2 = {{314., 145.}, {267., 120.}, {159., 129.}, {215., 161.}}
POSTED BY: Alexia werk

Hi Alexia,

Not working for these points ...

This is not a very specific description of you problem. It depends entirely on what you want: A glance over the documentation on FindGeometricTransform reveals that one has different choices of transformation, using the option TransformationClass (see in "Details and Options"). Obviously only with the setting "Perspective" (adopted here by the algorithm automatically) it is possible to meet the conditions for all your four points . But that leads apparently to a unwanted or unexpected result. The transformation as such is working, see my attached notebook.

Regards -- Henrik

Attachments:
POSTED BY: Henrik Schachner
Alexia werk
Alexia werk
Posted 10 years ago

I have seen the attached file, my output is almost same. I am just thinking is such results can be true. Because they are looking very strange. This was one reason so I still think its not working.

Attachments:
POSTED BY: Alexia werk
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