Crop an image using coordinates
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I am getting a grid for an image using vertical projection.
I want to crop the image part, within the 2 of the points as one cropping image.
Can someone advice me on this?
Thanks!
4 Comments
Accepted Answer
Image Analyst
on 2 Nov 2019
I would use regionprops() to find the start and stop of each non-zero segment, then crop. Something like (untested)
props = regionprops(verticalProfile, 'PixelIdxList');
for k = 1 : length(props)
row1 = props(k).PixelIdxList(1, 1)
row2 = props(k).PixelIdxList(1, end)
croppedImage = grayImage(row1:row2, :)
imshow(croppedImage);
drawnow;
end
11 Comments
Image Analyst
on 4 Nov 2019
It should only get to that line if it's an RGB image. What does this say?
ndims(grayImage)
whos grayImage
[rows, columns, numberOfColorChannels] = size(grayImage)
Don't use semicolons and look to see what it reports to the command window.
More Answers (1)
darova
on 2 Nov 2019
Here is an attempt
clc,clear
I = imread('Capture.jpeg');
I1 = im2bw(I); % convert to binary
I2 = bwareaopen(~I1,100); % remove numbers (small areas)
[row,col] = find(I2);
I3 = I2*0;
h = 1; % crop thickness
ii = row(1)+h:row(end)-h;
jj = col(1)+h:col(end)-h;
I3(ii,jj) = I2(ii,jj); % crop
I4 = bwareaopen(I3,200); % remove small areas
[L,n] = bwlabel(I4); % label image
for i = 1:4
subplot(2,2,i)
I5 = L == i; % find part of a curve
I6 = imdilate(I5,ones(5)); % highlight the part
II = cat(3,~I6,I1,I1)*255;
imshow(II) % show original image and part of a curve
end
im assuming the original image has no those circles
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