Issue in plotting binary image boundaries with real axis values
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Turbulence Analysis
2021 年 6 月 29 日
コメント済み: Turbulence Analysis
2021 年 6 月 30 日
Hi,
I have extracted the image boundaries from binary image and mm). I am able to diplay the binary image with real axis values i.e. x1, y1, (which is in mm), However, while plotting the extracted boundaries, the axes values are displayed in the pixels. (as shown in the attachment of binary image boundary.bmp).. Have anyone encountered simiar problem before ??
A = imgaussfilt(grayImage,5);
A1 = imbinarize (A,1900); %% binary image conversion
ax = gca;
A2 = imagesc (x1, y1,A1); %% Display binary image
set(gca, 'YDir','normal')
[b,~] = bwboundaries(A1, 8); b1 = b{1}; %% boundary extraction from binary image
b2 = smoothdata (b1(:,2),'sgolay',25);
b3 = smoothdata(b1(:,1),'sgolay',25);
plot(b2,b3,'b'); axis equal %% Plot the extrcated boundary
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Steve Eddins
2021 年 6 月 29 日
The function bwboundaries returns boundary values in pixel coordinates. You'll need to scale those to world coordinates yourself. For this purpose, imref2d and intrinsicToWorld might be useful to you.
R = imref2d(imageSize,xWorldLimits,yWorldLimits)
[xWorld, yWorld] = intrinsicToWorld(R,xIntrinsic,yIntrinsic)
3 件のコメント
Steve Eddins
2021 年 6 月 29 日
Look at the input to bwboundaries: it is just a MATLAB matrix. It does not have any information embedded in it about how matrix subscripts relate to some real-world coordinate system. Therefore, bwboundaries knows nothing about the world coordinate system and must return the output in what we call intrinsic coordinates. The syntax you used for imagesc suggests that, in your case, the intrinsic coordinates are related to the world coordinates by some combination of a multiplicative scale factor and an additive offset, or shift. You'll need to do this computation yourself on the output of bwboundaries. The other functions that I mentioned, imref2d and intrinsicToWorld, are capable of performing this computation for you if you wish.
その他の回答 (1 件)
Joseph Cheng
2021 年 6 月 29 日
so you'll need to find the conversion between pixel to linear xy (mm). sort of the mm/px scale factor + offset to 0;
from the looks of it half of the image (row or column) is 0,
xscale = (max(x1)-min(x1))/size(A,2); %mm/pixel
newb2 = (b2-size(A,2)/2)*xscale;
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