mean surface distance and residual mean square distance of 2 borders

Question

talayeh ghodsi 2020-6-26

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/555034-mean-surface-distance-and-residual-mean-square-distance-of-2-borders

评论： Image Analyst 2020-7-1

hi every body. I have 2 borders of 2 surfaces called S1 and S2. I need to compute the surface distance and after that the mean surface distance and residual mean square distance from that.

A complete description of the concept and the code in python is presented in https://mlnotebook.github.io/post/surface-distance-function/

Is there any matlab code for that?

Thanks in advance

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Image Analyst 2020-6-26

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/555034-mean-surface-distance-and-residual-mean-square-distance-of-2-borders#answer_457636

在 MATLAB Online 中打开

You can simply use sqrt() and min():

x1 = S1(:, 1);
y1 = S1(:, 2);
x2 = S2(:, 1);
y2 = S2(:, 2);
for k = 1 : length(S1)
    % Get the distance from the kth point of S1 to all points in S2.
    distances = sqrt((x1(k) - x2) .^ 2 + (y1(k) - y2) .^ 2);
    % Find the min of those distances to find how how close point k came to curve 2
    minDistance(k) = min(distances);
end

If you have 3-D (x,y,z) coordinates, the extension to 3-D should be obvious.

6 个评论
显示 4更早的评论隐藏 4更早的评论

Image Analyst 2020-6-29

编辑：Image Analyst 2020-6-29

在 MATLAB Online 中打开

It's pretty simple so you probably already have it by now. Did you just simply try two loops with the reference changed in each? You probably did. So did you get this:

x1 = S1(:, 1);
y1 = S1(:, 2);
x2 = S2(:, 1);
y2 = S2(:, 2);
for k = 1 : length(S1)
    % Get the distance from the kth point of S1 to all points in S2.
    distances = sqrt((x1(k) - x2) .^ 2 + (y1(k) - y2) .^ 2);
    % Find the min of those distances to find how how close point k came to curve 2
    minDistance1(k) = min(distances);
end
for k = 1 : length(S2)
    % Get the distance from the kth point of S2 to all points in S1.
    distances = sqrt((x2(k) - x1) .^ 2 + (y2(k) - y1) .^ 2);
    % Find the min of those distances to find how how close point k came to curve 1
    minDistance2(k) = min(distances);
end
numPairs = numel(x1) + numel(x2)
MSD = (sum(minDistance1) + sum(minDistance2)) / numPairs
RMS = sqrt((sum(minDistance1.^2) + sum(minDistance2.^2)) / numPairs)

Does that agree with your code?

talayeh ghodsi 2020-6-30

编辑：talayeh ghodsi 2020-6-30

在 MATLAB Online 中打开

I have used the Sobel filter in order to extract the border of my image as attached below, but the contents in workspace are all zero. I also attach my images

S1=imread('ECHO-mask.jpg');
S2=rgb2gray(imread('CTBWslice.jpg'));
S2=imresize(S2,[512 512]);
S1 = edge(S1,'sobel');
S2 = edge(S2,'sobel');
% BW1=imcomplement(BW1);
% BW2=imcomplement(BW2);
C = rgb2gray(imfuse(S1,S2));
imshow(C)
%%%% S1 and S2 are two contours
x1 = S1(:, 1);
y1 = S1(:, 2);
x2 = S2(:, 1);
y2 = S2(:, 2);
for k = 1 : length(S1)
    % Get the distance from the kth point of S1 to all points in S2.
    distances = sqrt((x1(k) - x2) .^ 2 + (y1(k) - y2) .^ 2);
    % Find the min of those distances to find how how close point k came to curve 2
    minDistance1(k) = min(distances);
end
for k = 1 : length(S2)
    % Get the distance from the kth point of S2 to all points in S1.
    distances = sqrt((x2(k) - x1) .^ 2 + (y2(k) - y1) .^ 2);
    % Find the min of those distances to find how how close point k came to curve 1
    minDistance2(k) = min(distances);
end
numPairs = numel(x1) + numel(x2)
MSD = (sum(minDistance1) + sum(minDistance2)) / numPairs
RMS = sqrt((sum(minDistance1.^2) + sum(minDistance2.^2)) / numPairs)

Image Analyst 2020-6-30

You didn't get the boundaries correctly. The boundaries are not an edge image, you need to call bwboundaries().

Image Analyst 2020-7-1

在 MATLAB Online 中打开

talayeh, try this:

clc;    % Clear the command window.
fprintf('Beginning to run %s.m.\n', mfilename);
close all;  % Close all figures (except those of imtool.)
clear;  % Erase all existing variables. Or clearvars if you want.
workspace;  % Make sure the workspace panel is showing.
format long g;
format compact;
fontSize = 20;
% Read in and display the images.
image1=imread('ECHO-mask.jpg');
image1 = imbinarize(image1);
% Fill holes
image1 = imfill(image1, 'holes');
subplot(1, 2, 1);
imshow(image1, []);
title('ECHO-mask.jpg');
image2=rgb2gray(imread('CTBWslice.jpg'));
image2 = imbinarize(image2);
% Fill holes
image2 = imfill(image2, 'holes');
subplot(1, 2, 2);
imshow(image2, []);
title('CTBWslice.jpg');
% Get the boundaries
boundaries1 = bwboundaries(image1);
boundaries2 = bwboundaries(image2);
for k1 = 1 : length(boundaries1)
	b1 = boundaries1{k1};
	S1 = fliplr(b1); % Flip so it's (x,y), not (row, column).
	for k2 = 1 : length(boundaries2)
		b2 = boundaries2{k2};
		S2 = fliplr(b2); % Flip so it's (x,y), not (row, column).
		
		%%%% S1 and S2 are two contours
		x1 = S1(:, 1);
		y1 = S1(:, 2);
		x2 = S2(:, 1);
		y2 = S2(:, 2);
		for k = 1 : length(S1)
			% Get the distance from the kth point of S1 to all points in S2.
			distances = sqrt((x1(k) - x2) .^ 2 + (y1(k) - y2) .^ 2);
			% Find the min of those distances to find how how close point k came to curve 2
			minDistance1(k) = min(distances);
		end
		for k = 1 : length(S2)
			% Get the distance from the kth point of S2 to all points in S1.
			distances = sqrt((x2(k) - x1) .^ 2 + (y2(k) - y1) .^ 2);
			% Find the min of those distances to find how how close point k came to curve 1
			minDistance2(k) = min(distances);
		end
		
		% Compute the metrics we're interested in.
		numPairs = numel(x1) + numel(x2);
		MSD = (sum(minDistance1) + sum(minDistance2)) / numPairs;
		RMS = sqrt((sum(minDistance1.^2) + sum(minDistance2.^2)) / numPairs);
		fprintf('Comparing image1 boundary #%d with image2 boundary #%d,\n  we get MSD = %f, and RMS = %f.\n',...
			k1, k2, MSD, RMS);
	end
end
fprintf('Done running %s.m.\n', mfilename);

In the command window, you'll see:

Comparing image1 boundary #1 with image2 boundary #1,

we get MSD = 5.545358, and RMS = 7.403608.

Comparing image1 boundary #1 with image2 boundary #2,

we get MSD = 60.859496, and RMS = 75.365100.

Comparing image1 boundary #2 with image2 boundary #1,

we get MSD = 57.429919, and RMS = 74.059752.

Comparing image1 boundary #2 with image2 boundary #2,

we get MSD = 3.204780, and RMS = 8.873175.

请先登录，再进行评论。

mean surface distance and residual mean square distance of 2 borders

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

采纳的回答

6 个评论
显示 4更早的评论隐藏 4更早的评论

更多回答（0 个）

另请参阅

类别

标签

Community Treasure Hunt

mean surface distance and residual mean square distance of 2 borders

0 个评论 显示 -2更早的评论隐藏 -2更早的评论

采纳的回答

6 个评论 显示 4更早的评论隐藏 4更早的评论

更多回答（0 个）

另请参阅

类别

标签

Community Treasure Hunt

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

6 个评论
显示 4更早的评论隐藏 4更早的评论