Fitting lines and polynomialon on edge after edge extraction and finding a certain distance
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I got the following edge after extracting it from a image which has been provided by me in attachmnent. I need to fit a red line along the lower edge and green horizontal line as shown in 2nd figure and find the distance.
What approach should I use???
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/336919/image.png)
The desired result I seek??
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/336922/image.png)
1 件のコメント
Matt J
2020 年 7 月 25 日
I have attached the matlab code for extracting the edge.
It would be more convenient for us if you simply attach a .mat file containing h.
採用された回答
Image Analyst
2020 年 7 月 25 日
編集済み: Image Analyst
2020 年 7 月 25 日
Really easy. Call bwconvhull(bw, 'union'). Then scan column-by-column using find() to get the last/lowest white point. Then fit those bottom rows to a quadratic with polyfit(), if you want - it's optional because the curve is not really needed to find the height of the last column. Then you simple need to find the last column where there is some white and subtract it from the max value. Something like (untested)
mask = bwconvhull(mask, 'union');
[rows, columns] = size(mask);
bottomRows = zeros(1, columns);
for col = 1 : columns
thisCol = mask(:, col);
r = find(thisCol, 1, 'last');
if ~isempty(r)
bottomRows(col) = r;
end
end
lastColumn = find(bottomRows > 0, 1, 'last');
bottomLine = max(bottomRows);
yline(bottomLine, 'Color', 'g', 'LineWidth', 2)
distance = bottomLine - bottomRows(lastColumn)
% Optional fitting
goodIndexes = bottomRows > 0;
c = 1:columns;
x = c(goodIndexes);
y = bottomRows(goodIndexes);
coefficients = polyfit(x, y, 2);
xFit = c;
yFit = polyval(coefficients, xFit);
hold on;
plot(xFit, yFit, 'r-', 'LineWidth', 2)
yLast = yFit(lastColumn);
plot(lastColumn, yLast, 'r+', 'MarkerSize', 30)
fittedDistance = bottomLine - yLast
% Will be different than other distance because the other one was actual
% and this one is from the regression.
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