How would i display one by one dots that are in a 3x3 matrix such as in the code below?
I would like to have dot1 appears in position [x1,y1] of the grid for a time t1, then dot2 to appears in position [x2,y2] of the grid for a time t2. Only one dot is being shown at each time.
Thanks for help
%grid
dim = 1
[x, y] = meshgrid(-dim:1:dim, -dim:1:dim);
pixelScale = screenYpixels / (dim * 2 + 2);
x = x .* pixelScale;
y = y .* pixelScale;
% Calculate the number of dots
numDots = numel(x);
% Make the matrix of positions for the dots.
dotPositionMatrix = [reshape(x, 1, numDots); reshape(y, 1, numDots)];
% We can define a center for the dot coordinates to be relaitive to.
dotCenter = [xCenter yCenter];
dotColors = [1 0 0];
dotSizes = 20;
Screen('DrawDots', window, dotPositionMatrix,...
dotSizes, dotColors, dotCenter, 2);
I think you want something like this?
%positions of each successive dots:
x_vec = [1,2,3,1,2,3,1,2,3];
y_vec = [1,1,1,2,2,2,3,3,3];
%wait times in sec for each dot:
wait_times = [1,1,2,1,1,2,1,1,2]
dotColor = [1 0 0];
dotSize = 400;
num_dots = length(x_vec);
for i = 1:num_dots
scatter(x_vec(i),y_vec(i),dotSize,dotColor,'filled');
xlim([0,max(x_vec)])
ylim([0,max(y_vec)])
pause(wait_times(i));
end
Related
I have a line that I have drawn between two points
origin = [1 0 2];
point = [1 -2.8 2.8 ];
I want to draw a line between them at a certain distance. I have tried the following, which gives me wrong results... Is there a better way to do this?
distance = 0.5;
point3 = origin + (point-origin) * distance;
MY SOLUTION:
which assumes that I will start at the same origin, just the last point changes:
cff = 0.5; % Coefficient that determines the length of the line
point2(1)= origin(1) + cff* (point(1) - origin(1));
point2(2)= origin(2) + cff* (point(2)- origin(2));
point2(3)= origin(3) + cff* (point(3) - origin(3));
For a length of a certain distance from origin:
d = 3; % Distance
unit_v = point - origin;
u = unit_v / norm(unit_v);
point2(1)= origin(1) + d * u(1) ;
point2(2)= origin(2) + d * u(2) ;
point2(3)= origin(3) + d * u(3) ;
You can calculate the unit vector from start point to end point and move the start point:
A = [1 0 2];
B = [1 -2.8 2.8 ];
d = 0.5;
V = B-A; % A->B vector
l = norm(V); % length of V
U = V/l; % unit vector
C = A + d*U;
D = A + (l-d)*U;
hold on
plot3([A(1) B(1)], [A(2) B(2)],[A(3) B(3)], '--o')
plot3([C(1) D(1)], [C(2) D(2)],[C(3) D(3)], '-x', 'linewidth', 2)
I need to create a binary mask. I have some coordinates and I make those coordinates and inside that region equal to 1, and background equal to zero.
Here is what I have done, but the problem is ROI located not in the correct position and located on the right bottom of the image. I appreciate if someone could point me to the right direction.
function [X,Y, BW] = Create_mask(X,Y,im)
X = round(X);
Y = round(Y);
X ( X < 1 ) = 1;
Y ( Y < 1 ) = 1;
BW = im > 255;
for p = 1:length(X)
BW(Y(p),X(p)) = 1;
end
for n = 0:(1/round(sqrt((X(end)-X(1))^2 + (Y(end)-Y(1))^2 ))):1
xn = round(X(1) +(X(end) - X(1))*n);
yn = round(Y(1) +(Y(end) - Y(1))*n);
BW(yn,xn) = 1;
end
se = strel('disk',10);
BW = imclose(BW,se);
BW = imdilate(BW,se);
BW = imfill(BW,'holes');
im( im < 255 ) = 0;
im = imclose(im,se);
BW = BW * 255;
BW = im2uint8(BW);
% BW = imresize(BW, [256 256],'nearest');
figure;
imshow(BW);
% close all;
end
Here is the output the function:
I was expecting to be similar to this image. This is not the exact solution but it shows my expectation.
X and Y coordinates are attached here, The first col is X and the second Y.
you can do this by calling function inpolygon, try this
function mask=createmask(x,y, cmin, cmax, dx)
if(nargin<3)
cmin=min([x(:) y(:)]);
end
if(nargin<4)
cmax=max([x(:) y(:)]);
end
if(nargin<5)
dx=(cmax-cmin)/100;
end
if(length(dx)==1)
dx=[dx dx];
end
[xi,yi]=meshgrid(cmin(1):dx(1):cmax(1),cmin(2):dx(2):cmax(2));
mask=reshape(inpolygon(xi(:),yi(:),x(:),y(:)), size(xi));
to test
xv = [0 3 3 0 0 NaN 1 1 2 2 1];
yv = [0 0 3 3 0 NaN 1 2 2 1 1];
mask=createmask(xv,yv, [-1 -1], [4 4]);
imagesc(mask)
I want to find Orientation, MajorAxisLengthand MinorAxisLength of contour which is plotted with below code.
clear
[x1 , x2] = meshgrid(linspace(-10,10,100),linspace(-10,10,100));
mu = [1,3];
sigm = [2,0;0,2];
xx_size = length(mu);
tem_matrix = ones(size(x1));
x_mesh= cell(1,xx_size);
for i = 1 : xx_size
x_mesh{i} = tem_matrix * mu(i);
end
x_mesh= {x1,x2};
temp_mesh = [];
for i = 1 : xx_size
temp_mesh = [temp_mesh x_mesh{i}(:)];
end
Z = mvnpdf(temp_mesh,mu,sigm);
z_plat = reshape(Z,size(x1));
figure;contour(x1, x2, z_plat,3, 'LineWidth', 2,'color','m');
% regionprops(z_plat,'Centroid','Orientation','MajorAxisLength','MinorAxisLength');
In my opinion, I may have to use regionprops command but I don't know how to do this. I want to find direction of axis of contour and plot something like this
How can I do this task? Thanks very much for your help
Rather than trying to process the graphical output of contour, I would instead recommend using contourc to compute the ContourMatrix and then use the x/y points to estimate the major and minor axes lengths as well as the orientation (for this I used this file exchange submission)
That would look something like the following. Note that I have modified the inputs to contourc as the first two inputs should be the vector form and not the output of meshgrid.
% Compute the three contours for your data
contourmatrix = contourc(linspace(-10,10,100), linspace(-10,10,100), z_plat, 3);
% Create a "pointer" to keep track of where we are in the output
start = 1;
count = 1;
% Now loop through each contour
while start < size(contourmatrix, 2)
value = contourmatrix(1, start);
nPoints = contourmatrix(2, start);
contour_points = contourmatrix(:, start + (1:nPoints));
% Now fit an ellipse using the file exchange
ellipsedata(count) = fit_ellipse(contour_points(1,:), contour_points(2,:));
% Increment the start pointer
start = start + nPoints + 1;
count = count + 1;
end
orientations = [ellipsedata.phi];
% 0 0 0
major_length = [ellipsedata.long_axis];
% 4.7175 3.3380 2.1539
minor_length = [ellipsedata.short_axis];
% 4.7172 3.3378 2.1532
As you can see, the contours are actually basically circles and therefore the orientation is zero and the major and minor axis lengths are almost equal. The reason that they look like ellipses in your post is because your x and y axes are scaled differently. To fix this, you can call axis equal
figure;contour(x1, x2, z_plat,3, 'LineWidth', 2,'color','m');
axis equal
Thank you #Suever. It help me to do my idea.
I add some line to code:
clear
[X1 , X2] = meshgrid(linspace(-10,10,100),linspace(-10,10,100));
mu = [-1,0];
a = [3,2;1,4];
a = a * a';
sigm = a;
xx_size = length(mu);
tem_matrix = ones(size(X1));
x_mesh= cell(1,xx_size);
for i = 1 : xx_size
x_mesh{i} = tem_matrix * mu(i);
end
x_mesh= {X1,X2};
temp_mesh = [];
for i = 1 : xx_size
temp_mesh = [temp_mesh x_mesh{i}(:)];
end
Z = mvnpdf(temp_mesh,mu,sigm);
z_plat = reshape(Z,size(X1));
figure;contour(X1, X2, z_plat,3, 'LineWidth', 2,'color','m');
hold on;
% Compute the three contours for your data
contourmatrix = contourc(linspace(-10,10,100), linspace(-10,10,100), z_plat, 3);
% Create a "pointer" to keep track of where we are in the output
start = 1;
count = 1;
% Now loop through each contour
while start < size(contourmatrix, 2)
value = contourmatrix(1, start);
nPoints = contourmatrix(2, start);
contour_points = contourmatrix(:, start + (1:nPoints));
% Now fit an ellipse using the file exchange
ellipsedata(count) = fit_ellipse(contour_points(1,:), contour_points(2,:));
% Increment the start pointer
start = start + nPoints + 1;
count = count + 1;
end
orientations = [ellipsedata.phi];
major_length = [ellipsedata.long_axis];
minor_length = [ellipsedata.short_axis];
tet = orientations(1);
x1 = mu(1);
y1 = mu(2);
a = sin(tet) * sqrt(major_length(1));
b = cos(tet) * sqrt(major_length(1));
x2 = x1 + a;
y2 = y1 + b;
line([x1, x2], [y1, y2],'linewidth',2);
tet = ( pi/2 + orientations(1) );
a = sin(tet) * sqrt(minor_length(1));
b = cos(tet) * sqrt(minor_length(1));
x2 = x1 + a;
y2 = y1 + b;
line([x1, x2], [y1, y2],'linewidth',2);
I would like to plot a plane using a vector that I calculated from 3 points where:
pointA = [0,0,0];
pointB = [-10,-20,10];
pointC = [10,20,10];
plane1 = cross(pointA-pointB, pointA-pointC)
How do I plot 'plane1' in 3D?
Here's an easy way to plot the plane using fill3:
points=[pointA' pointB' pointC']; % using the data given in the question
fill3(points(1,:),points(2,:),points(3,:),'r')
grid on
alpha(0.3)
You have already calculated the normal vector. Now you should decide what are the limits of your plane in x and z and create a rectangular patch.
An explanation : Each plane can be characterized by its normal vector (A,B,C) and another coefficient D. The equation of the plane is AX+BY+CZ+D=0. Cross product between two differences between points, cross(P3-P1,P2-P1) allows finding (A,B,C). In order to find D, simply put any point into the equation mentioned above:
D = -Ax-By-Cz;
Once you have the equation of the plane, you can take 4 points that lie on this plane, and draw the patch between them.
normal = cross(pointA-pointB, pointA-pointC); %# Calculate plane normal
%# Transform points to x,y,z
x = [pointA(1) pointB(1) pointC(1)];
y = [pointA(2) pointB(2) pointC(2)];
z = [pointA(3) pointB(3) pointC(3)];
%Find all coefficients of plane equation
A = normal(1); B = normal(2); C = normal(3);
D = -dot(normal,pointA);
%Decide on a suitable showing range
xLim = [min(x) max(x)];
zLim = [min(z) max(z)];
[X,Z] = meshgrid(xLim,zLim);
Y = (A * X + C * Z + D)/ (-B);
reOrder = [1 2 4 3];
figure();patch(X(reOrder),Y(reOrder),Z(reOrder),'b');
grid on;
alpha(0.3);
Here's what I came up with:
function [x, y, z] = plane_surf(normal, dist, size)
normal = normal / norm(normal);
center = normal * dist;
tangents = null(normal') * size;
res(1,1,:) = center + tangents * [-1;-1];
res(1,2,:) = center + tangents * [-1;1];
res(2,2,:) = center + tangents * [1;1];
res(2,1,:) = center + tangents * [1;-1];
x = squeeze(res(:,:,1));
y = squeeze(res(:,:,2));
z = squeeze(res(:,:,3));
end
Which you would use as:
normal = cross(pointA-pointB, pointA-pointC);
dist = dot(normal, pointA)
[x, y, z] = plane_surf(normal, dist, 30);
surf(x, y, z);
Which plots a square of side length 60 on the plane in question
I want to add to the answer given by Andrey Rubshtein, his code works perfectly well except at B=0. Here is the edited version of his code
Below Code works when A is not 0
normal = cross(pointA-pointB, pointA-pointC);
x = [pointA(1) pointB(1) pointC(1)];
y = [pointA(2) pointB(2) pointC(2)];
z = [pointA(3) pointB(3) pointC(3)];
A = normal(1); B = normal(2); C = normal(3);
D = -dot(normal,pointA);
zLim = [min(z) max(z)];
yLim = [min(y) max(y)];
[Y,Z] = meshgrid(yLim,zLim);
X = (C * Z + B * Y + D)/ (-A);
reOrder = [1 2 4 3];
figure();patch(X(reOrder),Y(reOrder),Z(reOrder),'r');
grid on;
alpha(0.3);
Below Code works when C is not 0
normal = cross(pointA-pointB, pointA-pointC);
x = [pointA(1) pointB(1) pointC(1)];
y = [pointA(2) pointB(2) pointC(2)];
z = [pointA(3) pointB(3) pointC(3)];
A = normal(1); B = normal(2); C = normal(3);
D = -dot(normal,pointA);
xLim = [min(x) max(x)];
yLim = [min(y) max(y)];
[Y,X] = meshgrid(yLim,xLim);
Z = (A * X + B * Y + D)/ (-C);
reOrder = [1 2 4 3];
figure();patch(X(reOrder),Y(reOrder),Z(reOrder),'r');
grid on;
alpha(0.3);
I am trying to implement least squares circle fitting following this paper (sorry I can't publish it). The paper states, that we could fit a circle, by calculating the geometric error as the euclidean distance (Xi'') between a specific point (Xi) and the corresponding point on the circle (Xi'). We have three parametres: Xc (a vector of coordinates the center of circle), and R (radius).
I came up with the following MATLAB code (note that I am trying to fit circles, not spheres as it is indicated on the images):
function [ circle ] = fit_circle( X )
% Kör paraméterstruktúra inicializálása
% R - kör sugara
% Xc - kör középpontja
circle.R = NaN;
circle.Xc = [ NaN; NaN ];
% Kezdeti illesztés
% A köz középpontja legyen a súlypont
% A sugara legyen az átlagos négyzetes távolság a középponttól
circle.Xc = mean( X );
d = bsxfun(#minus, X, circle.Xc);
circle.R = mean(bsxfun(#hypot, d(:,1), d(:,2)));
circle.Xc = circle.Xc(1:2)+random('norm', 0, 1, size(circle.Xc));
% Optimalizáció
options = optimset('Jacobian', 'on');
out = lsqnonlin(#ort_error, [circle.Xc(1), circle.Xc(2), circle.R], [], [], options, X);
end
%% Cost function
function [ error, J ] = ort_error( P, X )
%% Calculate error
R = P(3);
a = P(1);
b = P(2);
d = bsxfun(#minus, X, P(1:2)); % X - Xc
n = bsxfun(#hypot, d(:,1), d(:,2)); % || X - Xc ||
res = d - R * bsxfun(#times,d,1./n);
error = zeros(2*size(X,1), 1);
error(1:2:2*size(X,1)) = res(:,1);
error(2:2:2*size(X,1)) = res(:,2);
%% Jacobian
xdR = d(:,1)./n;
ydR = d(:,2)./n;
xdx = bsxfun(#plus,-R./n+(d(:,1).^2*R)./n.^3,1);
ydy = bsxfun(#plus,-R./n+(d(:,2).^2*R)./n.^3,1);
xdy = (d(:,1).*d(:,2)*R)./n.^3;
ydx = xdy;
J = zeros(2*size(X,1), 3);
J(1:2:2*size(X,1),:) = [ xdR, xdx, xdy ];
J(2:2:2*size(X,1),:) = [ ydR, ydx, ydy ];
end
The fitting however is not too good: if I start with the good parameter vector the algorithm terminates at the first step (so there is a local minima where it should be), but if I perturb the starting point (with a noiseless circle) the fitting stops with very large errors. I am sure that I've overlooked something in my implementation.
For what it's worth, I implemented these methods in MATLAB a while ago. However, I did it apparently before I knew about lsqnonlin etc, as it uses a hand-implemented regression. This is probably slow, but may help to compare with your code.
function [x, y, r, sq_error] = circFit ( P )
%# CIRCFIT fits a circle to a set of points using least sqaures
%# P is a 2 x n matrix of points to be fitted
per_error = 0.1/100; % i.e. 0.1%
%# initial estimates
X = mean(P, 2)';
r = sqrt(mean(sum((repmat(X', [1, length(P)]) - P).^2)));
v_cen2points = zeros(size(P));
niter = 0;
%# looping until convergence
while niter < 1 || per_diff > per_error
%# vector from centre to each point
v_cen2points(1, :) = P(1, :) - X(1);
v_cen2points(2, :) = P(2, :) - X(2);
%# distacnes from centre to each point
centre2points = sqrt(sum(v_cen2points.^2));
%# distances from edge of circle to each point
d = centre2points - r;
%# computing 3x3 jacobean matrix J, and solvign matrix eqn.
R = (v_cen2points ./ [centre2points; centre2points])';
J = [ -ones(length(R), 1), -R ];
D_rXY = -J\d';
%# updating centre and radius
r_old = r; X_old = X;
r = r + D_rXY(1);
X = X + D_rXY(2:3)';
%# calculating maximum percentage change in values
per_diff = max(abs( [(r_old - r) / r, (X_old - X) ./ X ])) * 100;
%# prevent endless looping
niter = niter + 1;
if niter > 1000
error('Convergence not met in 1000 iterations!')
end
end
x = X(1);
y = X(2);
sq_error = sum(d.^2);
This is then run with:
X = [1 2 5 7 9 3];
Y = [7 6 8 7 5 7];
[x_centre, y_centre, r] = circFit( [X; Y] )
And plotted with:
[X, Y] = cylinder(r, 100);
scatter(X, Y, 60, '+r'); axis equal
hold on
plot(X(1, :) + x_centre, Y(1, :) + y_centre, '-b', 'LineWidth', 1);
Giving: