Comparing two binary images: if a pixel is 1 in both images, write a 1 into a new image, otherwise, write a 0.

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Julian
Julian 2012-10-22
评论： Walter Roberson 2022-2-18
Hello everyone,
I have a somewhat noisy image and I would like to reduce or clear the noise.
To do so, I performed and opening. Since I don't want to lose any information of the objects' surface, I think it is easiest, to "compare" the two pictures, the original and the opened one. Meaning, when I have a 1 in the original image and a 1 in the opened image, I want to write a 1 into a third, new image. When I have a 1 in the original picture and a 0 in the opened picture I want to write a 0 in the new, third image. And when there are two 0s, I of couse want a 0 in the new image.
If you click on the link, you see the original image on the left and the opened image in the middle. It should look like a mixture of the first to images with all the surface information...
I tried to multipy the images (result=ImgOpen*ImgOrg), which in my opinion should lead to the desired result (1*1=1; 1*0=0), but the result looks like the image on the right...?!
http://s14.directupload.net/file/d/3051/29ym8bbc_jpg.htm
Do you understand what I mean? It's a little hard to describe that in another language. Sorry!
Thank you! Thymes

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采纳的回答

Walter Roberson
Walter Roberson 2012-10-22
result = ImgOpen .* ImgOrg;
The * operator is mathematical matrix multiplication, not element-by-element multiplication.
Note: the answer given for your other question is still correct,
result = ImgOpen & ImgOrg;
  2 个评论
Walter Roberson
Walter Roberson 2022-2-18
@Falak Memon
You should open a new question about that; it is not related to the current topic.
Note: "compare" of two images is not well-defined unless you define it precisely. "FInd 12 mistakes in the second image" is a quite different technique than "Decide which kind of brain tumour this patient has" or "Figure out where in the world this picture was taken".

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更多回答（2 个）

Julian
Julian 2012-10-23
Hello and thank you!
I tried it with the element-by-element multiplication with .* and the result is again on the right.
http://s7.directupload.net/file/d/3052/y94k2vch_jpg.htm
Do I think wrong, or shouldn't the surface of the main structure be much more like the one of the original picture on the right? I thought that this might be due to the size and shape of the sructuring element, but when I increased it's size it looked even worse...
Any ideas?
Julian
Julian 2012-10-23
Ok, I found the mistake in my thoughts. imopen performs dilation and erosion with the same structuring element. I chose the wrong size of it, deleting to many informations in the first place. I added a second structuring element with a bigger size and perform the dilation with it.
That works :)

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