How to distribute the points to be like Fig.A
This matlab code for Fig. B :
N = 30; % number of points
r = 0.5; % r = radius
d = 50; % dimension
C_point = 0; % center point
figure, clf
C = ones(1, d) * C_point;
C_rep = repmat( C,N,1);
X = randn(N,d);
s2 = sum(X.^2,2) ;
radius = r * (rand(N,1).^(1/d));
X = X.*repmat(radius./sqrt(s2),1,d) + C_rep;
%% Plot 2D
t = linspace(0, 2*pi, 100);
x = r*cos(t) + C(1);
y = r*sin(t) + C(2);
plot(x,y,'b')
hold on
plot(C(1),C(2),'b.', 'MarkerSize', 10) % center point
hold on
plot(X(:,1), X(:,2),'r.','markersize',10);
axis equal;rotate3d off; rotate3d on;drawnow;shg;
hold on
ax = axis;
Source of the code
What I should change to be like fig. A
The OP's code computes points uniformly distributed within a d-dimensional box, projects those onto a d-dimensional sphere, then samples the radius to move them inside the d-dimensional ball. This is perfect except that the points inside the box, when projected onto the sphere, do not form a uniform distribution on that sphere. If instead you find random points distributed in a Gaussian distribution, you are guaranteed uniform angle distribution.
First compute points with a Gaussian distribution in d dimensions (I do all here with minimal changes to the OP's code):
N = 1000; % number of points
r = 0.5; % r = radius
d = 3; % dimension
C_point = 0; % center point
C = ones(1,d) * C_point;
C_rep = repmat(C,N,1);
X = randn(N,d);
Note that I use randn, not rand. randn creates a Gaussian distribution.
Next we normalize the vectors so the points move to the sphere:
nX = sqrt(sum(X.^2,2));
X = X./repmat(nX,1,d);
These points are uniformly distributed, which you can verify by scatter3(X(:,1),X(:,2),X(:,3)); axis equal and turning the display around (a 2D rendering doesn't do it justice). This is the reason I set d=3 above, and N=1000. I wanted to be able to plot the points and see lots of them.
Next we compute, as you already did, a random distance to the origin, and correct it for the dimensionality:
radius = r * (rand(N,1).^(1/d));
X = X.*repmat(radius,1,d) + C_rep;
X now is distributed uniformly in the ball. Again, scatter3(X(:,1),X(:,2),X(:,3)); axis equal shows this.
However, if you set d=50 and then plot only two dimensions of your data, you will not see the data filling the circle. And you will not see a uniform distribution either. This is because you are projecting a 50-D ball onto 2 dimensions, this simply does not work. You either have to trust the math, or you have to slice the data:
figure, hold on
t = linspace(0, 2*pi, 100);
x = r*cos(t) + C(1);
y = r*sin(t) + C(2);
plot(x,y,'b')
plot(C(1),C(2),'b.', 'MarkerSize', 10) % center point
axis equal
I = all(abs(X(:,3:d))<0.1,2);
plot(X(I,1), X(I,2),'r.','markersize',10);
The I there indexes points that are close to the origin in dimensions perpendicular to the first two shown. Again, with d=50 you will have very few points there, so you will need to set N very large! To see the same density of points as in the case above, for every dimension you add, you need to multiply N by 10. So for d=5 you'd have N=1000*10*10=1e5, and for d=50 you'd need N=1e50. That is totally impossible to compute, of course.
Related
I try to solve the following 2D elliptic PDE electrostatic problem by fixing the Parallel plate Capacitors code. But I have problem to plot the circle region. How can I plot a circle region rather than the square?
% I use following two lines to label the 50V and 100V squares
% (but it should be two circles)
V(pp1-r_circle_small:pp1+r_circle_small,pp1-r_circle_small:pp1+r_circle_small) = 50;
V(pp2-r_circle_big:pp2+r_circle_big,pp2-r_circle_big:pp2+r_circle_big) = 100;
% Contour Display for electric potential
figure(1)
contour_range_V = -101:0.5:101;
contour(x,y,V,contour_range_V,'linewidth',0.5);
axis([min(x) max(x) min(y) max(y)]);
colorbar('location','eastoutside','fontsize',10);
xlabel('x-axis in meters','fontsize',10);
ylabel('y-axis in meters','fontsize',10);
title('Electric Potential distribution, V(x,y) in volts','fontsize',14);
h1=gca;
set(h1,'fontsize',10);
fh1 = figure(1);
set(fh1, 'color', 'white')
% Contour Display for electric field
figure(2)
contour_range_E = -20:0.05:20;
contour(x,y,E,contour_range_E,'linewidth',0.5);
axis([min(x) max(x) min(y) max(y)]);
colorbar('location','eastoutside','fontsize',10);
xlabel('x-axis in meters','fontsize',10);
ylabel('y-axis in meters','fontsize',10);
title('Electric field distribution, E (x,y) in V/m','fontsize',14);
h2=gca;
set(h2,'fontsize',10);
fh2 = figure(2);
set(fh2, 'color', 'white')
You're creating a square due to the way you're indexing (see this post on indexing). You've specified the rows to run from pp1-r_circle_small to pp1+r_circle_small and similar for the columns. Given that Swiss cheese is not an option, you're creating a complete square.
From geometry we know that all points within distance sqrt((X-X0)^2 - (Y-Y0)^2) < R from the centre of the circle at (X0,Y0) with radius R are within the circle, and the rest outside. This means that you can simply build a mask:
% Set up your grid
Xsize = 30; % Your case: 1
Ysize = 30; % Your case: 1
step = 1; % Amount of gridpoints; use 0.001 or something
% Build indexing grid for circle search, adapt as necessary
X = 0:step:Xsize;
Y = 0:step:Ysize;
[XX,YY] = meshgrid(X, Y);
V = zeros(numel(X), numel(Y));
% Repeat the below for both circles
R = 10; % Radius of your circle; your case 0.1 and 0.15
X0 = 11; % X coordinate of the circle's origin; your case 0.3 and 0.7
Y0 = 15; % Y coordinate of the circle's origin; your case 0.3 and 0.7
% Logical check to see whether a point is inside or outside
mask = sqrt( (XX - X0).^2 + (YY - Y0).^2) < R;
V(mask) = 50; % Give your circle the desired value
imagesc(V) % Just to show you the result
axis equal % Use equal axis to have no image distortion
mask is a logical matrix containing 1 where points are within your circle and 0 where points are outside. You can then use this mask to logically index your potential grid V to set it to the desired value.
Note: This will, obviously, not create a perfect circle, given you cannot plot a perfect circle on a square grid. The finer the grid, the more circle-like your "circle" will be. This shows the result with step = 0.01
Note 2: You'll need to tweek your definition of X, Y, X0, Y0 and R to match your values.
I'm having some problems in generating heatmaps given a defined figure.
I'm studying the temperature at the point of contact between two surfaces, a disk and a cylinder that could be modelled as one-dimensional, compared to the disk.
I have 3 sets of data, 1 for the radius(r) of a disk, another one for the angle (Theta) of the contact point and a last one for the temperature of the contacting point where the friction occurs.
So far I am able to create the disk and the differents points in a simulation obtained via another program, that gives me the previous sets of data.
Where I have the trouble is when I want to link the temperature obtained to its point and giving it a colour scale based on its temperature. I don't know how to establish this relationship.
As I say, this is what I have arrived to, which is only the definition of the points given by the results of the simulation.
Theta = xlsread('Laiton1.xlsx',1,'G2:G3381'); % Parameter turning angle
r = xlsread('Laiton1.xlsx',1,'C2:C3381'); % Parameter radius
Tsurf_d = xlsread('Laiton1.xlsx',1,'E2:E3381'); % Temperature on the surface
x = r*cos(Theta'); % parametrical transformation of (r,Theta) for the X axis
y = r*sin(Theta'); % parametrical transformation of (r,Theta) for the Y axis
Theta1 = linspace(0,360,5000); % Angle to define the 2 circumferences of the disk
x1 = 0.0145*cos(Theta1); % X points for the inner circumerference
y1 = 0.0145*sin(Theta1); % Y points for the inner circumerference
x2 = 0.0475*cos(Theta1); % X points for the external circumerference
y2 = 0.0475*sin(Theta1); % Y points for the external circumerference
plot(X,Y,X1,Y1,'black',X2,Y2,'black')
I hope I understood your question right: You have three vectors of coordinates and measurements, and you wish to plot a heat map with these. The code below does this. Vary the parameter "resolution" in order to zoom in or out of the plot.
% Me simulating your data
numData = 100;
Theta = (0:2*pi/(numData-1):2*pi) + (rand(1,numData)-.5)/10;
r = 23 + (rand(1,numData)-.5);
Tsurf_d = rand(1,numData)*100;
% Creating a table on which you can gather your data for plotting
resolution = 1; % Smaller number -> bigger and fewer pixels
c = resolution*ceil(max(r)) + 1; % Which pixel will be your center coordinate
width = 2*c + 1; % The width and height of your table
tempSumMap = zeros(width);
numDataMap = zeros(width);
% Calculating corresponding positions of each data point
xCoords = round( resolution*r.*cos(Theta) );
yCoords = round( resolution*r.*sin(Theta) );
% Adding the data points. In situations where two data points want to add
% to the same pixel, they both add, and numDataMap remembers to later
% divide by 2
for dataNo = 1:numData
y = yCoords(dataNo) + c ;
x = xCoords(dataNo) + c ;
tempSumMap(y,x) = tempSumMap(y,x) + Tsurf_d(dataNo);
numDataMap(y,x) = numDataMap(y,x) + 1;
end
% Remember to divide by the number of times you added a certain temperature
% to a pixel
tempMap = tempSumMap./max(1,numDataMap);
% Display the result
imagesc(tempMap)
I have a piecewise function, where domain changes for each case. The function is as follows:
For
(x,y)greater than Divider v= f(x,y) (A1)
(x,y)less than Divider v = g(x,y) (A2)
The location of the divider changes with tilt angle of the rectangle given in figures 1 and 2.Figure 1 & 2 The divider will always be a bisector of the rectangle. For example, the divider makes an angle (alpha + 90) with the horizontal.
If the rectangle makes an angle 0, it's easy to implement above functions as I can create meshgrid from
x =B to C & y = A to D for A1
x =A to B & y = A to D for A2
However, when the angles for the rectangle are different, I can't figure out how to create the mesh to calculate the function v using the algorithm A1 and A2 above.
I was thinking of using some inequality and using the equation of the line (as I have the co-ordinates for the center of the rectangle and the angle of tilt). But, I can't seem to think of a way to do it for all angles (for example , slope of pi/2 as in the first figure, yields infinity). Even if I do create some kind of inequality, I can't create a mesh.
1Please help me with this problem. I have wasted a lot of time on this. It seems to be out of my reach
%% Constants
Angle1=0;
Angle1=Angle1.*pi./180;
rect_center=0; % in m
rect_length=5; % in m
rect_width=1; % in m
rect_strength=1.8401e-06;
Angle2=0;
Angle2 =Angle2.*pi./180;
%% This code calculates the outer coordinates of the rectangle by using the central point
% the following code calculates the vertices
vertexA=rect_center+(-rect_width./2.*exp(1i.*1.5708)-rect_length./2).*exp(1i.*Angle2);
vertexA=[vertexA,vertexA+2.*(rect_width./2.*exp(1i.*1.5708)).*exp(1i.*Angle2)];
vertexB=rect_center+(-rect_width./2.*exp(1i.*1.5708)+rect_length./2).*exp(1i.*Angle2);
vertexB=[vertexB,vertexB+2.*(rect_width./2.*exp(1i.*1.5708)).*exp(1i.*Angle2)];
za1=vertexA(1:numel(vertexA)/2);
za2=vertexA(1+numel(vertexA)/2:numel(vertexA));
zb1=vertexB(1:numel(vertexB)/2);
zb2=vertexB(1+numel(vertexB)/2:numel(vertexB));
arg1=exp(-1i.*Angle2);
%% This Section makes the two equations necessary for making the graphs
syms var_z
% Equation 1
Eqn1(var_z)=1.5844e-07.*exp(-1i.*Angle1).*var_z./9.8692e-13;
% subparts of the Equation 2
A = 1.0133e+12.*(-1i.*rect_strength.*exp(-1i*Angle2)./(2*pi.*rect_length.*rect_width*0.2));
ZA1 = var_z+za1-2*rect_center;
ZA2 = var_z+za2-2*rect_center;
ZB1 = var_z+zb1-2*rect_center;
ZB2 = var_z+zb2-2*rect_center;
ZAA2 = log(abs(ZA2)) + 1i*mod(angle(ZA2),2*pi);
ZAA1 = log(abs(ZA1)) + 1i*mod(angle(ZA1),2*pi);
ZBB1 = log(abs(ZB1)) + 1i*mod(angle(ZB1),2*pi);
ZBB2 = log(abs(ZB2)) + 1i*mod(angle(ZB2),2*pi);
%Equation 2 ; this is used for the left side of the center
Eqn2= A*(ZA2*(log(ZA2)-1)-(ZA1*(log(ZA1)-1))+(ZB1*(log(ZB1)-1))-(ZB2*(log(ZB2)-1)));
%Equation 3 ; this is used for the right side of the center
Eqn3 = A.*(ZA2*(ZAA2-1)-(ZA1*(ZAA1-1))+(ZB1*(ZBB1-1))-(ZB2*(ZBB2-1)));
%Equation 4 :Add Equation 2 and Equation 1; this is used for the left side of the center
Eqn4 = matlabFunction(Eqn1+Eqn2,'vars',var_z);
%Equation 5: Add Equation 3 and Equation 1; this is used for the right side of the center
Eqn5 = matlabFunction(Eqn1+Eqn3,'vars',var_z);
%% Prepare for making the plots
minx=-10; %min x coordinate
maxx=10; %max x coordinate
nr_x=1000; %nr x points
miny=-10; %min y coordinate
maxy=10; %max y coordinate
nr_y=1000; %nr y points
%This vector starts from left corner (minx) to the middle of the plot surface,
%The middle of the plot surface lies at the center of the rectange
%created earlier
xvec1=minx:(rect_center-minx)/(0.5*nr_x-1):rect_center;
%This vector starts from middle to the right corner (maxx) of the plot surface,
%The middle of the plot surface lies at the center of the rectange
%created earlier
xvec2=rect_center:(maxx-rect_center)/(0.5*nr_x-1):maxx;
%the y vectors start from miny to maxy
yvec1=miny:(maxy-miny)/(nr_y-1):maxy;
yvec2=miny:(maxy-miny)/(nr_y-1):maxy;
% create mesh from above vectors
[x1,y1]=meshgrid(xvec1,yvec1);
[x2,y2]=meshgrid(xvec2,yvec2);
z1=x1+1i*y1;
z2=x2+1i*y2;
% Calculate the above function using equation 4 and equation 5 using the mesh created above
r1 = -real(Eqn5(z1));
r2 = -real(Eqn4(z2));
%Combine the calculated functions
Result = [r1 r2];
%Combine the grids
x = [x1 x2];
y = [y1 y2];
% plot contours
[c,h]=contourf(x,y,Result(:,:,1),50,'LineWidth',1);
% plot the outerboundary of the rectangle
line_x=real([vertexA;vertexB]);
line_y=imag([vertexA;vertexB]);
line(line_x,line_y,'color','r','linestyle',':','linewidth',5)
The final Figure is supposed to look like this.Final Expected Figure.
I'm not sure which angle defines the dividing line so I assume it's Angle1. It looks like logical indexing is the way to go here. Instead of creating two separate mesh grids we simply create the entire mesh grid then partition it into two sets and operate on each independently.
%% Prepare for making the plots
minx=-10; %min x coordinate
maxx=10; %max x coordinate
nr_x=1000; %nr x points
miny=-10; %min y coordinate
maxy=10; %max y coordinate
nr_y=1000; %nr y points
% create full mesh grid
xvec=linspace(minx,maxx,nr_x);
yvec=linspace(miny,maxy,nr_y);
[x,y]=meshgrid(xvec,yvec);
% Partition mesh based on divider line
% Assumes the line passes through (ox,oy) with normal vector defined by Angle1
ox = rect_center;
oy = rect_center;
a = cos(Angle1);
b = sin(Angle1);
c = -(a*ox + b*oy);
% use logical indexing to opperate on the appropriate parts of the mesh
idx1 = a*x + b*y + c < 0;
idx2 = ~idx1;
z = zeros(size(x));
z(idx1) = x(idx1) + 1i*y(idx1);
z(idx2) = x(idx2) + 1i*y(idx2);
% Calculate the above function using equation 4 and equation 5
% using the mesh created above
Result = zeros(size(z));
Result(idx1) = -real(Eqn5(z(idx1)));
Result(idx2) = -real(Eqn4(z(idx2)));
For example with Angle1 = 45 and Angle2 = 45 we get the following indexing
>> contourf(x,y,idx1);
>> line(line_x,line_y,'color','r','linestyle',':','linewidth',5);
where the yellow region uses Eqn5 and the blue region uses Eqn4. This agrees with the example you posted but I don't know what the resulting contour map for other cases is supposed to look like.
Hope this helps.
I have a set of 2D points (not ordered) forming a closed contour, and I would like to resample them to 14 equally spaced points. It is a contour of a kidney on an image. Any ideas?
One intuitive approach (IMO) is to create an independent variable for both x and y. Base it on arc length, and interpolate on it.
% close the contour, temporarily
xc = [x(:); x(1)];
yc = [y(:); y(1)];
% current spacing may not be equally spaced
dx = diff(xc);
dy = diff(yc);
% distances between consecutive coordiates
dS = sqrt(dx.^2+dy.^2);
dS = [0; dS]; % including start point
% arc length, going along (around) snake
d = cumsum(dS); % here is your independent variable
perim = d(end);
Now you have an independent variable and you can interpolate to create N segments:
N = 14;
ds = perim / N;
dSi = ds*(0:N).'; %' your NEW independent variable, equally spaced
dSi(end) = dSi(end)-.005; % appease interp1
xi = interp1(d,xc,dSi);
yi = interp1(d,yc,dSi);
xi(end)=[]; yi(end)=[];
Try it using imfreehand:
figure, imshow('cameraman.tif');
h = imfreehand(gca);
xy = h.getPosition; x = xy(:,1); y = xy(:,2);
% run the above solution ...
Say your contour is defined by independent vector x and dependent vector y.
You can get your resampled x vector using linspace:
new_x = linspace(min(x),max(x),14); %14 to get 14 equally spaced points
Then use interp1 to get new_y values at each new_x point:
new_y = interp1(x,y,new_x);
There are a few interpolation methods to choose from - default is linear. See interp1 help for more info.
I have a 3D density function q(x,y,z) that I am trying to plot in Matlab 8.3.0.532 (R2014a).
The domain of my function starts at a and ends at b, with uniform spacing ds. I want to plot the density on a ternary surface plot, where each dimension in the plot represents the proportion of x,y,z at a given point. For example, if I have a unit of density on the domain at q(1,1,1) and another unit of density on the domain at q(17,17,17), in both cases there is equal proportions of x,y,z and I will therefore have two units of density on my ternary surface plot at coordinates (1/3,1/3,1/3). I have code that works using ternsurf. The problem is that the number of proportion points grows exponentially fast with the size of the domain. At the moment I can only plot a domain of size 10 (in each dimension) with unit spacing (ds = 1). However, I need a much larger domain than this (size 100 in each dimension) and much smaller than unit spacing (ideally as small as 0.1) - this would lead to 100^3 * (1/0.1)^3 points on the grid, which Matlab just cannot handle. Does anyone have any ideas about how to somehow bin the density function by the 3D proportions to reduce the number of points?
My working code with example:
a = 0; % start of domain
b = 10; % end of domain
ds = 1; % spacing
[x, y, z] = ndgrid((a:ds:b)); % generate 3D independent variables
n = size(x);
q = zeros(n); % generate 3D dependent variable with some norm distributed density
for i = 1:n(1)
for j = 1:n(2)
for k = 1:n(2)
q(i,j,k) = exp(-(((x(i,j,k) - 10)^2 + (y(i,j,k) - 10)^2 + (z(i,j,k) - 10)^2) / 20));
end
end
end
Total = x + y + z; % calculate the total of x,y,z at every point in the domain
x = x ./ Total; % find the proportion of x at every point in the domain
y = y ./ Total; % find the proportion of y at every point in the domain
z = z ./ Total; % find the proportion of z at every point in the domain
x(isnan(x)) = 0; % set coordinate (0,0,0) to 0
y(isnan(y)) = 0; % set coordinate (0,0,0) to 0
z(isnan(z)) = 0; % set coordinate (0,0,0) to 0
xP = reshape(x,[1, numel(x)]); % create a vector of the proportions of x
yP = reshape(y,[1, numel(y)]); % create a vector of the proportions of y
zP = reshape(z,[1, numel(z)]); % create a vector of the proportions of z
q = reshape(q,[1, numel(q)]); % create a vector of the dependent variable q
ternsurf(xP, yP, q); % plot the ternary surface of q against proportions
shading(gca, 'interp');
colorbar
view(2)
I believe you meant n(3) in your innermost loop. Here are a few tips:
1) Loose the loops:
q = exp(- ((x - 10).^2 + (y - 10).^2 + (z - 10).^2) / 20);
2) Loose the reshapes:
xP = x(:); yP = y(:); zP = z(:);
3) Check Total once, instead of doing three checks on x,y,z:
Total = x + y + z; % calculate the total of x,y,z at every point in the domain
Total( abs(Total) < eps ) = 1;
x = x ./ Total; % find the proportion of x at every point in the domain
y = y ./ Total; % find the proportion of y at every point in the domain
z = z ./ Total; % find the proportion of z at every point in the domain
PS: I just recognized your name.. it's Jonathan ;)
Discretization method probably depends on use of your plot, maybe it make sense to clarify your question from this point of view.
Overall, you probably struggling with an "Out of memory" error, a couple of relevant tricks are described here http://www.mathworks.nl/help/matlab/matlab_prog/resolving-out-of-memory-errors.html?s_tid=doc_12b?refresh=true#brh72ex-52 . Of course, they work only up to certain size of arrays.
A more generic solution is too save parts of arrays on hard drive, it makes processing slower but it'll work. E.g., you can define several q functions with the scale-specific ngrids (e.g. ngridOrder0=[0:10:100], ngridOrder10=[1:1:9], ngridOrder11=[11:1:19], etc... ), and write an accessor function which will load/save the relevant grid and q function depending on the part of the plot you're looking.