How to scan an image to find out the row or column that will be the first background pixel?

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Zara Khan 2018-2-19

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编辑： Zara Khan 2018-2-21

P1_G3_5.png

I have a binary image. I have to scan the image from centroid to its right to find out row consisting of first background pixels ,then centroid to left side and from centroid to top. in each case I have to find out the distance between two columns and rows and find out the max distance.

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Guillaume 2018-2-19

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编辑：Guillaume 2018-2-19

在 MATLAB Online 中打开

As pointed out by Walter, background is normally black, so you need to invert the image. Finding out the centroid is trivial with regionprops and from there it's also easy to find the boundary pixels.

bwimg = ~yourimage;  %invert image so that background is black
props = regionpros(bwimg, 'Centroid');
assert(numel(props) == 1, 'more than one object found in image');
centroid = round(props.Centroid);
left = find(bwimg(centroid(1), :), 1)
right = find(bwimg(centroid(1), :), 1, 'last');
top = find(bwimg(:, centroid(2)), 1);
bottom = find(bwimg(:, centroid(2)), 1, 'last');

edit: capitalisation of Centroid and rounding of value

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Zara Khan 2018-2-21

编辑：Zara Khan 2018-2-21

Yes that I have done. Now in your second code you have helped me in finding out the upper left square box. Similar box I want for upper right image.imgside = max(props.Centroid(1) - left, props.Centroid(2) - top ); upperleftimage = bwimage(props.Centroid(2) - imgside : props.Centroid(2), props.centroid(1) - imgside : props.centroid(1).That was your codes. Now I want same like for upper right part. Plz help me in doing this. I already have attached a pic that how the output will look like. Thanks in advance

Guillaume 2018-2-21

在 MATLAB Online 中打开

This is really basic matrix indexing so I don't understand why you can't work it out on your own

subimage = image(toprow:bottomrow, leftrow:rightrow);

replace toprow, bottomrow, leftrow and rightrow by whichever coordinate you're interested in. In this latter case,

subimage = image(top:centroid(2), centroid(1):right);

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Walter Roberson 2018-2-19

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The distance in each case is 0. The centroid is inside the background area, so the distance to background is zero.

A note in this regard: in binary images, 0 is background and 1 is foreground.

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Zara Khan 2018-2-19

Thank you for the reply.Actually the image need to be inverted first then have to scan the image from centroid column to image width and from centroid row to image height then have to find out the first black pixel row and it's distance from the centroid column. This process will be repeated for left hand side and top too. Then have to find out the max distance

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