What is the best way to determine the highest peak of each plant in a cup?

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B.ray 2023-3-16

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评论： Image Analyst 2023-3-17

Hi,

Am trying to detect the highest point of each plant in the cup, I have around 30 small plants (see attached Image)

Can anyone suggest, how to do so. what algorithms to be used to solve it anf if there is something already builtin in MatLab.

Thanks

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Answer 1

Image Analyst 2023-3-16

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编辑：Image Analyst 2023-3-16

It would be best if you could plant them in a line and view them from the side with a white uniform background. Then it would be very straightforward.

If you insist on taking a top-down photo then you need to get a 3-D camera or profilometer that can give you a matrix/image where the pixel values are the heights.

https://www.3dsourced.com/rankings/best-3d-camera/

See my Image Segmentation Tutorial in my File Exchange:

Image Analyst's File Exchange

It's a generic, general purpose demo of how to threshold an image to find blobs, and then measure things about the blobs, and extract certain blobs based on their areas or diameters.

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B.ray 2023-3-17

Actually that's a good idea to threshold the depth first, that will make less overlapping between the plants. Though, there are different cups with different plants hights but I can set a minimum depth.

you mean this function right!!

stats = regionprops("table",bw,"Centroid", ...

"MajorAxisLength","MinorAxisLength");

I don't have a depth map at hand now. But it is changing from cup to cup and I will be using multiple cups.

Image Analyst 2023-3-17

在 MATLAB Online 中打开

No, I meant MaxIntensity

props = regionprops('table', mask, heightImage, 'MaxIntensity', 'Centroid');

This is because with a depth image, the pixel value represents either the height above the floor (use MaxIntensity) or the distance to the camera (use MinIntensity).

An alternative is to use imfill on the gray scale image to make a local background, then subtract that from the original, and then threshold. I have no idea if that will be any good or not but it might be worth a try.

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