I have three variables x, y and z. I have inequalities of the form
x >= a, y>= b, z>=c, x+y>=d, y+z>=e, x+z>=f, x+y+z>=g
where a to g are positive numbers. On a 3D plot with axes x, y and z, this is an open volume. I would like to fill the open side (i.e. away from 0) shape with color and show it in a plot. What is the way to do this on MATLAB?
I attempted to use fill3 and a mesh but the result was not very good
[x,y,z] = meshgrid(0:0.01:2,0:0.01:2,0:0.01:2);
ineq = (x>=1)& (y>0.5)&(z>=0.25)&(x+y>1.25)&(y+z>0.6)&(x+z>1.1)&(x+y+z>1.6);
fill3(x(:),y(:),z(:), 'r')
box on
grid on
Using plot3 also was not very good. Is there any other way to generate a nice 3D figure on MATLAB?
Mathematica does this using RegionPlot3D. I was hoping for a similar resultant image.
First of all, be careful when using 3D meshes, the one you defined contains 8M+ points.
Assuming your shape is convex, you can use convhull and trisurf:
Not that the option 'Simplify' is set as true to reduce the number of elements accounted for in the convex hull.
[x,y,z] = meshgrid(0:0.1:2,0:0.1:2,0:0.1:2);
ineq = (x>=1)& (y>0.5)&(z>=0.25)&(x+y>1.25)&(y+z>0.6)&(x+z>1.1)&(x+y+z>1.6);
figure;
x_ineq = x(ineq);
y_ineq = y(ineq);
z_ineq = z(ineq);
id_cvhl = convhull(x_ineq,y_ineq,z_ineq,'Simplify',true);
trisurf(id_cvhl,x_ineq,y_ineq,z_ineq,'FaceColor','cyan','edgecolor','none')
xlim([0 2])
ylim([0 2])
zlim([0 2])
In case you want the result to look a bit more than RegionPlot3D, don't use Simplify, and plot the edges (Be careful not too have a mesh with too many points!).
id_cvhl = convhull(x_ineq,y_ineq,z_ineq);
trisurf(id_cvhl,x_ineq,y_ineq,z_ineq,'Facecolor','yellow')
Related
I have a set of n=8000 cartesian coordinates X,Y and Z as vectors and also a vector V of same size which I want to use as values to create a heatmap on a sphere.
I saw this link (visualization of scattered data over a sphere surface MATLAB), but I don't understand how I convert this set of data into a meshgrid for plotting using surf.
Almost every example I saw uses meshgrids.
Right now, I am doing by plotting a sphere and then use scatter3 to plot my points as big balls and try to smooth them later. I looks like this:
I would like to get the figure as the plotting of the example in that link, where he uses:
k = 5;
n = 2^k-1;
[x,y,z] = sphere(n);
c = hadamard(2^k);
surf(x,y,z,c);
colormap([1 1 0; 0 1 1])
axis equal
EDIT:
(Sorry for taking so long to reply, the corona crises kept away from work)
What I am actually doing is:
for i=1:numel(pop0n)
ori(i,:)=ori(i,:)/norm(ori(i,:));
end
x = ori(:,1);
y = ori(:,2);
z = ori(:,3);
%// plot
m=100;
[aa,bb,cc] = sphere(m);
surf(aa,bb,cc,ones(m+1,m+1)*min(pop0n))
hold on
colormap jet;
scatter3(x,y,z,400,pop0n/norm(pop0n),'filled');
colorbar
shading interp
The array 'ori' is 8000x3, and contains the x, y and z coordinates of the points I want to plot and pop0n is a 8000 sized vector with the intensities of each coordinate.
My main question is how do I transform my x, y, z and pop0n, that are vectors, into 2D arrays (meshgrid) to use surf?
Because I cannot simply do surf(x,y,z,pop0n) if they are vectors.
Thanks in advance
As David suggested, griddata does the job.
What I did was:
for i=1:numel(pop0n)
ori(i,:)=ori(i,:)/norm(ori(i,:));
end
x = ori(:,1);
y = ori(:,2);
z = ori(:,3);
%// plot
m=100;
[aa,bb,cc] = sphere(m);
v = griddata(x,y,z,pop0n,aa,bb,cc,'nearest');
surf(aa,bb,cc,v)
colormap jet;
colorbar
shading interp
I want to represent data with 2 variables in 2D format. The value is represented by color and the 2 variables as the 2 axis. I am using the contourf function to plot my data:
clc; clear;
load('dataM.mat')
cMap=jet(256); %set the colomap using the "jet" scale
F2=figure(1);
[c,h]=contourf(xrow,ycol,BDmatrix,50);
set(h, 'edgecolor','none');
xlim([0.0352 0.3872]);
ylim([0.0352 0.3872]);
colormap(cMap);
cb=colorbar;
caxis([0.7 0.96]);
% box on;
hold on;
Both xrow and ycol are 6x6 matrices representing the coordinates. BDmatrix is the 6x6 matrix representing the corresponding data. However, what I get is this:
The following is the xrow and yrow matices:
The following is the BDmatrix matices:
Would it be possible for the contour color to vary smoothly rather than appearing as straight lines joining the data points? The problem of this figure is the coarse-granularity which is not appealing. I have tried to replace contourf with imagec but it seems not working. I am using MATLAB R2015b.
You can interpolate your data.
newpoints = 100;
[xq,yq] = meshgrid(...
linspace(min(min(xrow,[],2)),max(max(xrow,[],2)),newpoints ),...
linspace(min(min(ycol,[],1)),max(max(ycol,[],1)),newpoints )...
);
BDmatrixq = interp2(xrow,ycol,BDmatrix,xq,yq,'cubic');
[c,h]=contourf(xq,yq,BDmatrixq);
Choose the "smoothness" of the new plot via the parameter newpoints.
To reduce the Color edges, you can increase the number of value-steps. By default this is 10. The following code increases the number of value-steps to 50:
[c,h]=contourf(xq,yq,BDmatrixq,50);
A 3D-surf plot would be more suitable for very smooth color-shading. Just rotate it to a top-down view. The surf plot is also much faster than the contour plot with a lot of value-steps.
f = figure;
ax = axes('Parent',f);
h = surf(xq,yq,BDmatrixq,'Parent',ax);
set(h, 'edgecolor','none');
view(ax,[0,90]);
colormap(Jet);
colorbar;
Note 1: Cubic interpolation is not shape-preserving. That means, the interpolated shape can have maxima which are greater than the maximum values of the original BDmatrix (and minima which are less). If BDmatrix has noisy values, the interpolation might be bad.
Note 2: If you generated xrow and yrow by yourself (and know the limits), than you do not need that min-max-extraction what I did.
Note 3: After adding screenshots of your data matrices to your original posting, one can see, that xrow and ycol come from an ndgrid generator. So we also must use this here in order to be consistent. Since interp2 needs meshgrid we have to switch to griddedInterpolant:
[xq,yq] = ndgrid(...
linspace(min(min(xrow,[],1)),max(max(xrow,[],1)),newpoints ),...
linspace(min(min(ycol,[],2)),max(max(ycol,[],2)),newpoints )...
);
F = griddedInterpolant(xrow,ycol,BDmatrix,'cubic');
BDmatrixq = F(xq,yq);
I have a 3D matrix C=51x51x11 dimensions, obtained from a function in a separate script, the x,y,z represent length, depth and height, and the value represent a concentration per x,y,z point. I want to create a slice crossing x and another crossing y showing the difference in concentration by color. I have tried using ngrid and meshgrid but didn't work. may i have some help with this please?
Use slice()
C = randi(1,[51,51,11]);
x= 25; y = 25; z = 5;
sl = slice(C,x,y,z);
Using slice inside a function to make it easy to view in 3d:
function eslice(V,sx,sy,sz)
slice(V,sx,sy,sz)
shading interp
axis equal
axis vis3d
end
This is from my personal library, enjoy.
MATLAB's surf command allows you to pass it optional X and Y data that specify non-cartesian x-y components. (they essentially change the basis vectors). I desire to pass similar arguments to a function that will draw a line.
How do I plot a line using a non-cartesian coordinate system?
My apologies if my terminology is a little off. This still might technically be a cartesian space but it wouldn't be square in the sense that one unit in the x-direction is orthogonal to one unit in the y-direction. If you can correct my terminology, I would really appreciate it!
EDIT:
Below better demonstrates what I mean:
The commands:
datA=1:10;
datB=1:10;
X=cosd(8*datA)'*datB;
Y=datA'*log10(datB*3);
Z=ones(size(datA'))*cosd(datB);
XX=X./(1+Z);
YY=Y./(1+Z);
surf(XX,YY,eye(10)); view([0 0 1])
produces the following graph:
Here, the X and Y dimensions are not orthogonal nor equi-spaced. One unit in x could correspond to 5 cm in the x direction but the next one unit in x could correspond to 2 cm in the x direction + 1 cm in the y direction. I desire to replicate this functionality but drawing a line instead of a surf For instance, I'm looking for a function where:
straightLine=[(1:10)' (1:10)'];
my_line(XX,YY,straightLine(:,1),straightLine(:,2))
would produce a line that traced the red squares on the surf graph.
I'm still not certain of what your input data are about, and what you want to plot. However, from how you want to plot it, I can help.
When you call
surf(XX,YY,eye(10)); view([0 0 1]);
and want to get only the "red parts", i.e. the maxima of the function, you are essentially selecting a subset of the XX, YY matrices using the diagonal matrix as indicator. So you could select those points manually, and use plot to plot them as a line:
Xplot = diag(XX);
Yplot = diag(YY);
plot(Xplot,Yplot,'r.-');
The call to diag(XX) will take the diagonal elements of the matrix XX, which is exactly where you'll get the red patches when you use surf with the z data according to eye().
Result:
Also, if you're just trying to do what your example states, then there's no need to use matrices just to take out the diagonal eventually. Here's the same result, using elementwise operations on your input vectors:
datA = 1:10;
datB = 1:10;
X2 = cosd(8*datA).*datB;
Y2 = datA.*log10(datB*3);
Z2 = cosd(datB);
XX2 = X2./(1+Z2);
YY2 = Y2./(1+Z2);
plot(Xplot,Yplot,'rs-',XX2,YY2,'bo--','linewidth',2,'markersize',10);
legend('original','vector')
Result:
Matlab has many built-in function to assist you.
In 2D the easiest way to do this is polar that allows you to make a graph using theta and rho vectors:
theta = linspace(0,2*pi,100);
r = sin(2*theta);
figure(1)
polar(theta, r), grid on
So, you would get this.
There also is pol2cart function that would convert your data into x and y format:
[x,y] = pol2cart(theta,r);
figure(2)
plot(x, y), grid on
This would look slightly different
Then, if we extend this to 3D, you are only left with plot3. So, If you have data like:
theta = linspace(0,10*pi,500);
r = ones(size(theta));
z = linspace(-10,10,500);
you need to use pol2cart with 3 arguments to produce this:
[x,y,z] = pol2cart(theta,r,z);
figure(3)
plot3(x,y,z),grid on
Finally, if you have spherical data, you have sph2cart:
theta = linspace(0,2*pi,100);
phi = linspace(-pi/2,pi/2,100);
rho = sin(2*theta - phi);
[x,y,z] = sph2cart(theta, phi, rho);
figure(4)
plot3(x,y,z),grid on
view([-150 70])
That would look this way
I want to plot a surface with a texture map on it, but the conditions are not the "ideal" ones.
first lets explain what I have.
I have a set of points (~7000) that are image coordinates, in a grid. This points do NOT define perfect squares. IT IS NOT A MESHGRID. For the sake of the question, lets assume that we have 9 points. Lets ilustrate what we have with an image:
X=[310,270,330,430,410,400,480,500,520]
Y=[300,400,500,300,400,500,300,400,500]
Lets say we can get the "structure" of the grid, so
size1=3;
size2=3;
points=zeros(size1,size2,2)
X=[310,270,330;
430,410,400;
480,500,520]
Y=[300,400,500;
300,400,500;
300,400,500]
points(:,:,1)=X;
points(:,:,2)=Y;
And now lets say we have a 3rd dimension, Z.
EDIT: Forgot to add a piece if info. I triangulate the points in the image and get a 3D correspondence, so when displayed in a surface they don't have the X and Y coords of the image, for a simplification of the given data lets say X=X/2 Y=Y/3
And we have:
points=zeros(size1,size2,3)
Z=[300,330,340;
300,310,330;
290,300,300]
surf(points(:,:,1)/2,points(:,:,2)/3,points(:,:,3))
What I want is to plot the surface in 3D with the image texture. Each element should have the texture piece that have in the first image.
This needs to work for huge datasheets. I don't specially need it to be fast.
related post (but I has a meshgrid as initial set of points) : Texture map for a 2D grid
PD: I can post original images + real data if needed, just posted this because i think it is easier with small data.
You can use the texturemap property of surf which works with rectangular meshes as well as with non-rectangular ones.
Creating non-rectangular data points
% creating non-rectangular data points
[X, Y] = meshgrid(1:100, 1:100);
X = X+rand(size(X))*5;
Y = Y+rand(size(X))*5;
which results in the following data points:
Generating height data:
Z = sin(X/max(X(:))*2*pi).*sin(Y/max(Y(:))*2*pi);
Loading picture:
[imageTest]=imread('peppers.png');
and mapping it as texture to the mesh:
surf(X,Y,Z, imageTest, ...
'edgecolor', 'none','FaceColor','texturemap')
Note that, for the sake of demonstration, this non-rectangular grid is quite sparsely populated which results in a rather jagged texture. With more points, the result gets much better, irrespective of the distortion of the grid points.
Note also that the number of grid points does not have to match the number of pixels in the texture image.
~edit~
If X and Y coordinates are only available for parts of the image, you can adjust the texture accordingly by
minX = round(min(X(:)));
maxX = round(max(X(:)));
minY = round(min(Y(:)));
maxY = round(max(Y(:)));
surf(X,Y,Z, imageTest(minX:maxX, minY:maxY, :), ...
'edgecolor', 'none','FaceColor','texturemap')
I don't think you can do what you want with Matlab's built in commands and features. But using the technique from my other answer with a high-res version of the grid can do it for you.
By "high-res", I mean an interpolated version of the non-uniform grid with denser data points. That is used to sample the texture at denser data points so it can be drawn using the texturemap feature of surf. You can't use a normal 2D interpolation, however, because you need to preserve the non-uniform grid shape. This is what I came up with:
function g = nonUniformGridInterp2(g, sx, sy)
[a,b] = size(g);
g = interp1(linspace(0,1,a), g, linspace(0,1,sy)); % interp columns
g = interp1(linspace(0,1,b), g', linspace(0,1,sx))'; % interp rows
Note that you have to call this twice to interpolate the X and Y points independently. Here's an example of the original grid and an interpolated version with 10 points in each direction.
Here's how to use that high-res grid with interp2 and texturemap.
function nonUniformTextureMap
% define the non-uniform surface grid
X = [310,270,330; 430,410,400; 480,500,520];
Y = [300,400,500; 300,400,500; 300,400,500];
Z = [300,330,340; 300,310,330; 290,300,300];
% get texture data
load penny % loads data in variable P
% define texture grid based on image size
% note: using 250-550 so that a,b covers the range used by X,Y
[m,n] = size(P);
[a,b] = meshgrid(linspace(250,550,n), linspace(250,550,m));
% get a high-res version of the non-uniform grid
s = 200; % number of samples in each direction
X2 = nonUniformGridInterp2(X, s, s);
Y2 = nonUniformGridInterp2(Y, s, s);
% sample (map) the texture on the non-uniform grid
C = interp2(a, b, P, X2, Y2);
% plot the original and high-res grid
figure
plot(X(:),Y(:),'o',X2(:),Y2(:),'.')
legend('original','high-res')
% plot the surface using sampled points for color
figure
surf(X, Y, Z, C, 'edgecolor', 'none', 'FaceColor','texturemap')
colormap gray
I'm not sure I understand your question, but I think that what you need to do is sample (map) the texture at your grid's X,Y points. Then you can simply plot the surface and use those samples as colors.
Here's an example using the data you gave in your question. It doesn't look like much, but using more X,Y,Z points should give the result you're after.
% define the non-uniform surface grid
X = [310,270,330; 430,410,400; 480,500,520];
Y = [300,400,500; 300,400,500; 300,400,500];
Z = [300,330,340; 300,310,330; 290,300,300];
% get texture data
load penny % loads data in variable P
% define texture grid based on image size
% note: using 600 so that a,b covers the range used by X,Y
[m,n] = size(P);
[a,b] = meshgrid(linspace(0,600,n), linspace(0,600,m));
% sample (map) the texture on the non-uniform grid
C = interp2(a, b, P, X, Y);
% plot the surface using sampled points for color
figure
surf(X, Y, Z, C)
colormap gray