I have two vectors A(1,512), B(1,8) , and one matrix C(8,512). I am trying to plot contour by using contour(X,Y,Z). I do not know to do it.
A vector represents distance, B vector is frequency, and C matrix is velocity.
This can be done by using function contourf(X,Y,Z)
C1=C';
contourf(B,A,C1);
Transposed of C used because length(B) must equal size(C,2) and length(A) must equal size(C,1)
I quote from the documentation:
contourf(X,Y,Z), contourf(X,Y,Z,n), and contourf(X,Y,Z,v) draw filled contour plots of Z using X and Y to determine the x and y values.
How would I plot in 3D space (plot3() I presume) from a matrix containing N rows of coordinates where column 1 is x, column 2 is y and column 3 is z?
Given that your matrix is in X, it's as simple as:
plot3(X(:,1), X(:,2), X(:,3), 'b.');
plot3 takes in three arguments as a base. The first argument are the x coordinates, the second are the y coordinates and the third are the z coordinates. Because you have all three coordinates conveniently in a matrix and each are in separate columns, you just have to pluck out each coordinate and put that into plot3. I'm also assuming that the points are discrete and you don't want to join any of the points together, so the fourth argument denotes both the colour of the points as well as the style of the points. Here, I've made them blue and single dots.
Another option would be scatter3.
X = rand(30,3);
scatter3(X(:,1), X(:,2), X(:,3), 'b.');
I want to draw a contour plot for 3D data.
I have a force in x,y,z directions I want to plot the contour3 for that
the dimensions of the Fx = 21x21X21 same for Fy and Fz
I am finding force = f*vector(x,y,z)
Then
Fx(x,y,z) = force(1)
Fy(x,y,z) = force(2)
Fz(x,y,z) = force(3)
I did the following but it is not working with me ?? why and how can I plot that
FS = sqrt(Fx.^2 + Fy.^2 + Fz.^2);
x = -10:1:10;
[X,Y] = meshgrid(x);
for i=1:length(FS)
for j = 1:length(FS)
for k=1:length(FS)
contour3(X,Y,FS(i,j,k),10)
hold on
end
end
end
This is the error I am getting
Error using contour3 (line 129)
When Z is a vector, X and Y must also be vectors.
Your problem is that FS is not the same shape as X and Y.
Lets illustrate with a simple example:
X=[1 1 1
2 2 2
3 3 3];
Y=[1 2 3
1 2 3
1 2 3];
Z=[ 2 4 5 1 2 5 5 1 2];
Your data is probably something like this. How does Matlab knows which Z entry corresponds to which X,Y position? He doesnt, and thats why he tells you When Z is a vector, X and Y must also be vectors.
You could solve this by doing reshape(FS,size(X,1),size(X,2)) and will probably work in your case, but you need to be careful. In your example, X and Y don't seem programatically related to FS in any way. To have a meaningful contour plot, you need to make sure that FS(ii,jj,k)[ 1 ] corresponds to X(ii,jj), else your contour plot would not make sense.
Generally you'd want to plot the result of FS against the variables your are using to compute it, such as ii, jj or k, however, I dont know how these look like so I will stop my explanation here.
[ 1 ]: DO NOT CALL VARIABLES i and j IN MATLAB!
I'm not sure if this solution is what you want.
Your problem is that contour and contour3 are plots to represent scalar field in 2D objects. Note that ball is 2D object - every single point is defined by angles theta and phi - even it is an object in "space" not in "plane".
For representation of vector fields there is quiver, quiver3, streamslice and streamline functions.
If you want to use contour plot, you have to transform your data from vector field to scalar field. So your data in form F = f(x,y,z) must be transformed to form of H = f(x,y). In that case H is MxN matrix, x and y are Mx1 and Nx1 vectors, respectively. Then contour3(x,y,H) will work resulting in so-called 3D graph.
If you rely on vector field You have to specify 6 vectors/matrices of the same size of corresponding x, y, z coordinates and Fx, Fy, Fz vector values.
In that case quiver3(x,y,z,Fx,Fy,Fz) will work resulting in 6D graph. Use it wisely!
As I comment the Ander's answer, you can use colourspace to get more dimensions, so You can create 5D or, theoretically, 6D, because you have x, y, z coordinates for position and R, G, B coordinates for the values. I'd recommend using static (x,y,R,G,B) for 5D graph and animated (x,y,t,R,G,B) for 6D. Use it wisely!
In the example I show all approaches mentioned above. i chose gravity field and calculate the plane 0.25 units below the centre of gravity.
Assume a force field defined in polar coordinates as F=-r/r^3; F=1/r^2.
Here both x and yare in range of -1;1 and same size N.
F is the MxMx3 matrix where F(ii,jj) is force vector corresponding to x(ii) and y(jj).
Matrix H(ii,jj) is the norm of F(ii,jj) and X, Y and Z are matrices of coordinates.
Last command ensures that F values are in (-1;1) range. The F./2+0.5 moves values of F so they fit into RGB range. The colour meaning will be:
black for (-1,-1,-1),
red for (1,-1,-1),
grey for (0,0,0)
Un-comment the type of plot You want to see. For quiver use resolution of 0.1, for other cases use 0.01.
clear all,close all
% Definition of coordinates
resolution=0.1;
x=-1:resolution:1;
y=x;
z=-.25;
%definition of matrices
F=zeros([max(size(x))*[1 1],3]); % matrix of the force
X=zeros(max(size(x))*[1 1]); % X coordinates for quiver3
Y=X; % Y coordinates for quiver3
Z=X+z; % Z coordinates for quiver3
% Force F in polar coordinates
% F=-1/r^2
% spherical -> cartesian transformation
for ii=1:max(size(x))
for jj=1:max(size(y))
% temporary variables for transformations
xyz=sqrt(x(ii)^2+y(jj)^2+z^2);
xy= sqrt(x(ii)^2+y(jj)^2);
sinarc=sin(acos(z/xyz));
%filling the quiver3 matrices
X(ii,jj)=x(ii);
Y(ii,jj)=y(jj);
F(ii,jj,3)=-z/xyz^2;
if xy~=0 % 0/0 error for x=y=0
F(ii,jj,2)=-y(jj)/xyz/xy*sinarc;
F(ii,jj,1)=-x(ii)/xyz/xy*sinarc;
end
H(ii,jj)=sqrt(F(ii,jj,1)^2+F(ii,jj,2)^2+F(ii,jj,3)^2);
end
end
F=F./max(max(max(F)));
% quiver3(X,Y,Z,F(:,:,1),F(:,:,2),F(:,:,3));
% image(x,y,F./2+0.5),set(gca,'ydir','normal');
% surf(x,y,Z,F./2+.5,'linestyle','none')
% surf(x,y,H,'linestyle','none')
surfc(x,y,H,'linestyle','none')
% contour3(x,y,H,15)
I have 3 dimensional (X-Y-Z) data which includes-
11 Y vectors (100*1 each (range 0:0.01:1)) each is unique and correspond to the same X vector (100*1 (range 0:0.01:1)). Thus, Y can be treated as matrix with dimension 100*11.
Different Y vectors correspond to different Z values on Z axis (11 values of Z ranging from 0 to 1).
e.g. => Every value of Z on Z axis i.e. Z1=0.05 ... Z11=0.95 has different Y vector for the same X vector. Thus it is like: X-Y1-Z1, X-Y2-Z2, X-Y3-Z3...X-Y11-Z11 (Z1 to Z11 are fixed values like 0.05, 0.1 ,..., 0.95 which are repeated equal to the length of X and Y).
I have already plotted this data as 11 discrete contour lines stacked one above other in 3D space using plot3 hold on. But, I want to create surface out of it. I think I can use interpolation using griddata but I didn't understand how to approach this problem. Can anybody help?
I am trying to fit a B-spline to a set of ordered discrete data points which represent pixels of a contour extracted from an image.
While the below code works fine for some simple shapes, but not for others (please see attached image for examples). Why does this happen, and what would be a better way to approach this problem?
I am quite new to differential geometry, appreciate any insights or inputs. Thanks.
% data contains two columns representing x,y coordinates of pixels
x = data(:, 1);
y = data(:, 2);
plot(x, y, 'bo');
fittedmodel = fit(x, y, 'cubicinterp');
plot(fittedmodel, 'r-');
What went wrong?
You have two sets of numbers x and y with the same number of elements in both sets.
You assume:
a. That there is a function f such that f( x_i ) = y_i for all pairs x_i,y_i in your sets.
b. That the points in the set are ordered: that is, if you follow the curve of f(x) then x_i "comes before" x_{i+1}.
While these assumptions hold for the "correct fit" example you have. They are no longer valid for the "incorrect fit" example.
As you can see for yourself, the input contour on the top cannot be expressed as y = f(x) since there are values of x for which there are two possible corresponding values of y (see the definition of mathematical function). The fit you got is the closest thing to a mathematical function y = f(x) that can be given the pairs x,y you gave (the red curve has the property of each x having only one y value).
What can you do?
In most cases when you try and fit a 2D curve you search for a parametric curve: that is, you introduce an auxilary parameter t such that each point along the curve can be represented as [x(t), y(t)] for some 0<=t<=1.
Now, if assumption b holds (and by looking at your examples, I'm not certain it is), what you can do is
t = linspace( 0, 1, numel(x) ); % define the parameter t
fitX = fit( t, x, 'cubicinterp'); % fit x as a function of parameter t
fitY = fit( t, y, 'cubicinterp'); % fit y as a function of parameter t
plot( fitX, fitY, 'r-' ); % plot the parametric curve