[HELP] Finding the lowest positioned pixel in the figure.
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YJ
el 22 de Oct. de 2014
Comentada: Xiao MA
el 21 de Mayo de 2022
I have a code as follow: which is converted into binary image.
image= [pos1];
BW = im2bw(pos1,map,0.4);
imshow(pos1(:,:)),figure, imshow(BW)
image= [pos2];
BW = im2bw(pos2,map,0.4);
imshow(pos2(:,:)),figure, imshow(BW)
which will return following images
with these images, I want to find the lowest point (as circled with red) and print the location information into one matrix. (only need y axis value) = Height from the bottom of the image. (the first white pixel)
for instance if in lowest point pixel in image1 located at 123,342, and for image 2 located at 134,423
then I want to have matrix that has values of 343, and 423.
I was doing this manually for 10 images. and I gave up as there are 100s of them...
It would be really appreciate if you can help me with this.
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Image Analyst
el 22 de Oct. de 2014
Guillaume's answer is good. I just wanted to add: don't do this:
image= [pos1];
All that does it to destroy the built-in function called image(), and nothing useful whatsoever in your code. It can be eliminated without any problem. Never use "image" as the name of a variable.
By the way, another possible method to find the lowest row is
lowestRow1 = find(sum(BW1, 2)>0, 1, 'last'); % Bottom dot in image #1
lowestRow2 = find(sum(BW2, 2)>0, 1, 'last'); % Bottom dot in image #2
output = [lowestRow1, lowestRow2]; % The output array you said you wanted.
Basically you're summing the image horizontally (across columns, which gives a mean vertical profile) and finding the last (lowest) row that has something in it. However I'm not sure why you want 343 and 423 when the lowest rows are 342 and 423 - why add 1 to the first image???
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Image Analyst
el 21 de Mayo de 2022
@Xiao MA you can do this
% Find all white rows and columns in the binary image
[yRows, xColumns] = find(binaryImage);
% Find the lowest row, which will have the highest yRows value
[yRowsOfLowestPoint, indexOfLowestRow] = max(yRows);
% Get the corresponding column (x coordinate) at that location
xColumnOfLowestPoint = xColumns(indexOfLowestRow);
Xiao MA
el 21 de Mayo de 2022
@Image Analyst Thank you very much! I'm able to run it successfully and get the results I wanted. Really appreciate!!!
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