# mean surface distance and residual mean square distance of 2 borders

14 vues (au cours des 30 derniers jours)
talayeh ghodsi le 26 Juin 2020
Commenté : Image Analyst le 1 Juil 2020
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?
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### Réponse acceptée

Image Analyst le 26 Juin 2020
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.
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Image Analyst le 30 Juin 2020
You didn't get the boundaries correctly. The boundaries are not an edge image, you need to call bwboundaries().
Image Analyst le 1 Juil 2020
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 = imbinarize(image1);
% Fill holes
image1 = imfill(image1, 'holes');
subplot(1, 2, 1);
imshow(image1, []);
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.

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