2
$\begingroup$

I remember that there was some thread of this sort, but I could not find it, so I hope it's ok to open a new thread.

I want to segment pictures of the brain, i.e. I only want the area within the skull without the bones (all the grey matter/pixels. White parts are bones of the skull).

I have the following two pictures. In the first picture I also want all the grey parts, but you see that there are some white parts intermingled between. So basically the algorithm should segment four different images. In the second pictures it seems to be easier.

Can you guys give me a hint to a function or some examples or threads?

enter image description here

enter image description here

$\endgroup$
2
  • 1
    $\begingroup$ The second one is super easy. You could do it with just Binarize or your could do something like DeleteSmallComponents@MorphologicalBinarize[img, {0.8, 0.9}]. The first one is another beast because those four segments aren't properly enclosed. $\endgroup$
    – C. E.
    Commented Jun 12, 2014 at 9:55
  • 1
    $\begingroup$ Try DeleteBorderComponents@ ImageSubtract[i1, Closing[MorphologicalBinarize[i1, {0.9, .99}], 5]] on the first... where i1 is the image. $\endgroup$
    – ciao
    Commented Jun 12, 2014 at 10:00

1 Answer 1

4
$\begingroup$

Copying the first image into symbol i1:

DeleteBorderComponents@ImageSubtract[i1, Closing[MorphologicalBinarize[i1, {0.9, .99}], 5]]

enter image description here

Second is pretty trivial, see Pickett's comment...

$\endgroup$
2
  • $\begingroup$ ahh MorphologicalBinarize was what I was looking for. Thanks! $\endgroup$
    – holistic
    Commented Jun 12, 2014 at 10:07
  • 2
    $\begingroup$ @holistic: There's probably a dozen ways to do this, when you have the time, explore the various image-related functions, and play with them - that's how to get a good feel for them... $\endgroup$
    – ciao
    Commented Jun 12, 2014 at 10:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.