Drawing a boundary box around an image?

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Warid Islam
Warid Islam 2022 年 9 月 29 日
コメント済み: Warid Islam 2022 年 9 月 30 日
I want to draw a boundary box around an image. I tried the code below, but it is throwing me an error. Any suggestions would be appreciated.
m=imread('bb.png')
props = regionprops(m, 'BoundingBox');
bbx = rectangle('Position',props.BoundingBox,'EdgeColor','b','LineWidth',3);
I am getting the following error message.
Error using rectangle
Input arguments must be parameter-value pairs. Each parameter name must be followed by a corresponding value.
  1 件のコメント
KSSV
KSSV 2022 年 9 月 29 日
props is a structire....you cannot use that like you have used. What exactly you want to do?

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採用された回答

Walter Roberson
Walter Roberson 2022 年 9 月 29 日
My goal is to create a bounding box around the non-zero pixel value region of my image.
filename = 'https://www.mathworks.com/matlabcentral/answers/uploaded_files/1140570/box.png';
I = imread(filename);
BW = imbinarize(I);
hmask = any(BW,1);
vmask = any(BW,2);
firstrow = find(hmask, 1); lastrow = find(hmask, 1, 'last');
firstcol = find(vmask, 1); lastcol = find(vmask, 1, 'last');
imshow(I);
rectangle('Position', [firstrow, firstcol, lastrow-firstrow, lastcol-firstcol], 'EdgeColor', 'r');
What happened? Well... it turns out that your image has a white border around it. White borders are non-zero so the entire image needs to be enclosed in the border according to your stated requirements.
BW = any(I < 240, 3) & any(I > 10,3);
hmask = any(BW,1);
vmask = any(BW,2);
firstrow = find(hmask, 1); lastrow = find(hmask, 1, 'last');
firstcol = find(vmask, 1); lastcol = find(vmask, 1, 'last');
imshow(I);
rectangle('Position', [firstrow, firstcol, lastrow-firstrow, lastcol-firstcol], 'EdgeColor', 'r');
  1 件のコメント
Warid Islam
Warid Islam 2022 年 9 月 30 日
Thank you very much.

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その他の回答 (2 件)

KSSV
KSSV 2022 年 9 月 29 日
I = imread('https://in.mathworks.com/matlabcentral/answers/uploaded_files/1139790/bb.png');
props = regionprops(rgb2gray(I), 'BoundingBox');
imshow(I)
hold on
for i = 1:length(props)
rectangle('Position',props(i).BoundingBox,'EdgeColor','b','LineWidth',3);
end
  4 件のコメント
Warid Islam
Warid Islam 2022 年 9 月 29 日
Hi @KSSV,
Thank you for your response. I don't get any error message. However, after using imcrop, the size of my image reduces which may not be the ideal case. I am using images of the same size which I would multiply later. If the above image size decreases, then the later operations would give me an error message. My goal is to create a bounding box around the non-zero pixel value region of my image.I would perform some algorithm later on which I would like to restrict within the bounding box. I was wondering if it is possible to create a bounding box without reducing the size of the image.
Warid Islam
Warid Islam 2022 年 9 月 29 日
Hi @KSSV,
I tried to implement your code above.
I = imread('https://in.mathworks.com/matlabcentral/answers/uploaded_files/1139790/bb.png');
props = regionprops(rgb2gray(I), 'BoundingBox');
imshow(I)
hold on
for i = 1:length(props)
rectangle('Position',props(i).BoundingBox,'EdgeColor','b','LineWidth',3);
end
The region inside the box is filled with Blue color. I just want the boundary to be Blue and the inside region remain visible to me. I would implement some algorithms later which would be limited inside the box.

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Image Analyst
Image Analyst 2022 年 9 月 29 日
You can use imclearborder to get rid of the white frame, then use regionprops on the blue channel.
rgbImage = imread('box.png');
subplot(2, 1, 1);
imshow(rgbImage);
impixelinfo;
title('Original Image')
% Get blue box. Use imclear border to get rid of the white outside
% padding/frame.
mask = imclearborder(rgbImage(:, :, 3) > 0);
mask = bwareafilt(mask, 1);
subplot(2, 1, 2);
imshow(mask);
title('Mask Image')
% Get the bounding box amd dos[;au ot pver the original image in red.
props = regionprops(mask, 'BoundingBox');
subplot(2, 1, 1);
hold on;
rectangle('Position', props.BoundingBox, 'EdgeColor', 'r', 'LineWidth', 2)
  1 件のコメント
Warid Islam
Warid Islam 2022 年 9 月 30 日
Thank you very much. The bounding box works.

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