I have three matrices x, y, z which are plotted via scatter3 in matlab. However I also need vertical lines dropping from every point in the graph for better visualization.
Using matlab 2017a, implemented 3D scatter plot in matlab.
enter code here
clc;
figure
x = [0,0,0,0,0,10,10,10,10,10];
y = [0,10,20,30,40,-10,0,10,20,30];
z = [46,52,51,59,53,85,56,87,86,88];
scatter3(x, y, z, 30, 'filled')
You could also use the built in function stem, which is doing exactly that.
The minor trick is that you cannot pass the z coordinates in the shorthand form stem(x,y,z), but the graphic object still accept z data, you just have to send them as additional parameter.
The nice part of it is you do not need a loop ;-)
x = [0,0,0,0,0,10,10,10,10,10];
y = [0,10,20,30,40,-10,0,10,20,30];
z = [46,52,51,59,53,85,56,87,86,88];
hp = stem(x,y,'filled','ZData',z) ;
Or as Gnovice nicely pointed out, even easier to use the stem3 function which accept z data directly:
hp = stem3(x,y,z,'filled') ;
Both example above will produce:
As #SardarUsama pointed out, plot3 should do the trick. Code could be more compact but kept it as is for clarity.
% MATLAB R2017a
x = [0,0,0,0,0,10,10,10,10,10];
y = [0,10,20,30,40,-10,0,10,20,30];
z = [46,52,51,59,53,85,56,87,86,88];
figure
scatter3(x, y, z, 30, 'filled') % scatter plot (3D)
zRng = zlim;
hold on
for k = 1:length(x)
xL = [x(k) x(k)];
yL = [y(k) y(k)];
zL = [zRng(1) z(k)];
plot3(xL,yL,zL,'r-') % plot vertical line (3D)
end
Related
I am stuck with an apparently simple problem. I have to revolve of 360° a 2D curve around an axis, to obtain a 3D plot. Say, I want to do it with this sine function:
z = sin(r);
theta = 0:pi/20:2*pi;
xx = bsxfun(#times,r',cos(theta));
yy = bsxfun(#times,r',sin(theta));
zz = repmat(z',1,length(theta));
surf(xx,yy,zz)
axis equal
I now want to visualize the numerical values of the Z plane, stored in a matrix. I would normally do it this way:
ch=get(gca,'children')
X=get(ch,'Xdata')
Y=get(ch,'Ydata')
Z=get(ch,'Zdata')
If I visualize Z with
imagesc(Z)
I don't obtain the actual values of Z of the plot, but the "un-revolved" projection. I suspect that this is related to the way I generate the curve, and from the fact I don't have a function of the type
zz = f(xx,yy)
Is there any way I can obtain the grid values of xx and yy, as well as the values of zz at each gridpoint?
Thank you for your help.
Instead of bsxfun you can use meshgrid:
% The two parameters needed for the parametric equation
h = linspace(0,2) ;
th = 0:pi/20:2*pi ;
[R,T] = meshgrid(h,th) ;
% The parametric equation
% f(x) Rotation along Z
% ↓ ↓
X = sin(R) .* cos(T) ;
Y = sin(R) .* sin(T) ;
% Z = h
Z = R ;
surf(X,Y,Z,'EdgeColor',"none")
xlabel('X')
ylabel('Y')
zlabel('Z')
Which produce:
And if you want to extract the contour on the X plane (X = 0) you can use contour:
contour(Y,Z,X,[0,0])
Which produce:
I'm trying to create a gradient fill inside a circular area according to a given function. I hope the plot below explains it at best
I'm not sure how to approach this, as in the simulation I'm working on the direction of the gradient changes (not always in the x direction as below, but free to be along all the defined angles), so I'm looking for a solution that will be flexible in that manner as well.
The code I have is below
clear t
N=10;
for i=0:N
t(i+1) = 0+(2*i*pi) / N;
end
F = exp(-cos(t))./(2.*pi*besseli(1,1));
figure(1)
subplot(1,3,1)
plot(t*180/pi,F,'-ob')
xlim([0 360])
xlabel('angle')
subplot(1,3,2)
hold on
plot(cos([t 2*pi]), sin([t 2*pi]),'-k','linewidth',2);
plot(cos([t 2*pi]), sin([t 2*pi]),'ob');
plot(cos(t).*F,sin(t).*F,'b','linewidth',2);
subplot(1,3,3)
hold on
plot(cos([t 2*pi]), sin([t 2*pi]),'-k','linewidth',2);
plot(cos([t 2*pi]), sin([t 2*pi]),'ob');
To fill surface, you need to use the patch command.
t = linspace(0, 2*pi, 100);
x = cos(t);
y = sin(t);
c = x; % colored following x value and current colormap
figure
patch(x,y,c)
hold on
scatter(x,y)
hold off
colorbar
Resulting graph:
Colors are defined in c per point, and are interpolated inside the shape, so I'm sure that you should have all freedom to color as you want!
For example, the rotated version:
t = linspace(0, 2*pi, 100);
x = cos(t);
y = sin(t);
c = cos(t+pi/4)
figure
patch(x,y,c)
colorbar
To understand how it is going on, just think that every point has a color, and matlab interpolate inside. So here I just rotated the intensity per point by pi /4.
For this to work you need to have a filled shape, and you may need to customize the color (c) parameter so that it matches your need. For example, if your gradient direction is encoded in a vector, you want to project all your point onto that vector to get the value along the gradient for all points.
For example:
% v controls the direction of the gradient
v = [0.1, 1];
t = linspace(0, 2*pi, 100);
F = exp(-cos(t))./(2.*pi*besseli(1,1));
% reconstructing point coordinate all around the surface
% this closes the path so with enough points so that interpolation works correctly
pts = [[t', F']; [t(end:-1:1)', ones(size(t'))*min(F)]];
% projecting all points on the vector to get the color
c = pts * (v');
clf
patch(pts(:,1),pts(:,2),c)
hold on
scatter(t, F)
hold off
I'm trying to graph a solution obtained through the quadratic formula in Matlab. Since it's obtained by the quadratic formula, there are two parts: plus and minus. The graph should be a hyperbola. How can I place the upper part and the bottom part on the same graph?
There are different ways. Let's say you want to plot the solution of y^2 = x, that is y = ±sqrt(x):
You can plot the two parts with the same color using a plot once…
x = 0:0.1:10;
plot(x, sqrt(x), 'k', x, -sqrt(x), 'k')
…or twice:
x = 0:0.1:10;
plot(x, sqrt(x), 'k')
hold on
plot(x, -sqrt(x), 'k')
hold off
Or you can plot everything in one go like you might draw it with a pen:
x = [10:-0.1:0 0.1:0.1:10];
y = [-sqrt(10:-0.1:0) sqrt(0.1:0.1:10)];
plot(x, y)
I have a problem and maybe you will be able to help me. Like in the title i have cross section data of a symmetric lens - coordinates s=-100:1:100 and height y - and I would like to create 3D plot the whole lens (x,y,z). Is there any build in function that helps with that? Thanks for help in advance!
If I'm understanding correctly, you have a 1-D array that you'd effectively like to 'sweep' around a circle to produce a 3-D plot. Here is an example of how to do that
% Some dummy data
Npts = 100;
z = sin(linspace(0, pi, Npts));
Nreps = 100; % How many times to repeat around circle
% Create polar meshgrid and convert to Cartesian
[r, theta] = meshgrid( ...
linspace(-length(z)/2, length(z)/2, Npts), ...
linspace(0, pi, Nreps));
[X, Y] = pol2cart(theta, r);
% Copy data Nreps times
Z = repmat(z, Nreps, 1);
% Plot!
surf(X, Y, Z)
Without more specs (such as if your y is a 2D matrix or a 1D array), it's not possible to give you the exact answer. However here is how you draw a surface in Matlab:
% create a meshgrid used as the independent variables of your surface
[sx, sy] = meshgrid(-100:100);
% if you have your own 2D matrix, ignore this line.
% if you have a formula, replace this line with your own formula
y = cos(sqrt(((sx/100).^2+(sy/100).^2)/2)*(pi/2));
% draw the surface
surf(sx, sy, y);
To have the opposite side as well, draw another surf on the same figure:
hold on;
surf(sx, sy, -y);
I have a dataset that describes a point cloud of a 3D cylinder (xx,yy,zz,C):
and I would like to make a surface plot from this dataset, similar to this
In order to do this I thought I could interpolate my scattered data using TriScatteredInterp onto a regular grid and then plot it using surf:
F = TriScatteredInterp(xx,yy,zz);
max_x = max(xx); min_x = min(xx);
max_y = max(yy); min_y = min(yy);
max_z = max(zz); min_z = min(zz);
xi = min_x:abs(stepSize):max_x;
yi = min_y:abs(stepSize):max_y;
zi = min_z:abs(stepSize):max_z;
[qx,qy] = meshgrid(xi,yi);
qz = F(qx,qy);
F = TriScatteredInterp(xx,yy,C);
qc = F(qx,qy);
figure
surf(qx,qy,qz,qc);
axis image
This works really well for convex and concave objects but ends in this for the cylinder:
Can anybody help me as to how to achieve a nicer plot?
Have you tried Delaunay triangulation?
http://www.mathworks.com/help/matlab/ref/delaunay.html
load seamount
tri = delaunay(x,y);
trisurf(tri,x,y,z);
There is also TriScatteredInterp
http://www.mathworks.com/help/matlab/ref/triscatteredinterp.html
ti = -2:.25:2;
[qx,qy] = meshgrid(ti,ti);
qz = F(qx,qy);
mesh(qx,qy,qz);
hold on;
plot3(x,y,z,'o');
I think what you are loking for is the Convex hull function. See its documentation.
K = convhull(X,Y,Z) returns the 3-D convex hull of the points (X,Y,Z),
where X, Y, and Z are column vectors. K is a triangulation
representing the boundary of the convex hull. K is of size mtri-by-3,
where mtri is the number of triangular facets. That is, each row of K
is a triangle defined in terms of the point indices.
Example in 2D
xx = -1:.05:1; yy = abs(sqrt(xx));
[x,y] = pol2cart(xx,yy);
k = convhull(x,y);
plot(x(k),y(k),'r-',x,y,'b+')
Use plot to plot the output of convhull in 2-D. Use trisurf or trimesh to plot the output of convhull in 3-D.
A cylinder is the collection of all points equidistant to a line. So you know that your xx, yy and zz data have one thing in common, and that is that they all should lie at an equal distance to the line of symmetry. You can use that to generate a new cylinder (line of symmetry taken to be z-axis in this example):
% best-fitting radius
% NOTE: only works if z-axis is cylinder's line of symmetry
R = mean( sqrt(xx.^2+yy.^2) );
% generate some cylinder
[x y z] = cylinder(ones(numel(xx),1));
% adjust z-range and set best-fitting radius
z = z * (max(zz(:))-min(zz(:))) + min(zz(:));
x=x*R;
y=y*R;
% plot cylinder
surf(x,y,z)
TriScatteredInterp is good for fitting 2D surfaces of the form z = f(x,y), where f is a single-valued function. It won't work to fit a point cloud like you have.
Since you're dealing with a cylinder, which is, in essence, a 2D surface, you can still use TriScatterdInterp if you convert to polar coordinates, and, say, fit radius as a function of angle and height--something like:
% convert to polar coordinates:
theta = atan2(yy,xx);
h = zz;
r = sqrt(xx.^2+yy.^2);
% fit radius as a function of theta and h
RFit = TriScatteredInterp(theta(:),h(:),r(:));
% define interpolation points
stepSize = 0.1;
ti = min(theta):abs(stepSize):max(theta);
hi = min(h):abs(stepSize):max(h);
[qx,qy] = meshgrid(ti,hi);
% find r values at points:
rfit = reshape(RFit(qx(:),qy(:)),size(qx));
% plot
surf(rfit.*cos(qx),rfit.*sin(qx),qy)