I am trying to sort random coordinates on a 2D cartesian grid using MATLAB into "bins" defined by a grid.
For example if I have a 2D domain with X ranging from [-1,1] and Y from [-1,1] and I generate some random coordinates within the domain, how can I "count" how many coordinates fall into each quadrant?
I realize that for and if statements can be used to determine the if each coordinate is within the quadrants, but I would like to scale this to much larger square grids that have more than just 4 quadrants.
Any concise and efficient approach would be appreciated!
Below is an example adapted from the code I mentioned.
The resulting binned points are be stored the variable subs; Each row contains 2d subscript indices of the bin to which a point was assigned.
% 2D points, both coordinates in the range [-1,1]
XY = rand(1000,2)*2 - 1;
% define equal-sized bins that divide the [-1,1] grid into 10x10 quadrants
mn = [-1 -1]; mx = [1 1]; % mn = min(XY); mx = max(XY);
N = 10;
edges = linspace(mn(1), mx(1), N+1);
% map points to bins
% We fix HISTC handling of last edge, so the intervals become:
% [-1, -0.8), [-0.8, -0.6), ..., [0.6, 0.8), [0.8, 1]
% (note the last interval is closed on the right side)
[~,subs] = histc(XY, edges, 1);
subs(subs==N+1) = N;
% 2D histogram of bins count
H = accumarray(subs, 1, [N N]);
% plot histogram
imagesc(H.'); axis image xy
set(gca, 'TickDir','out')
colormap gray; colorbar
xlabel('X'); ylabel('Y')
% show bin intervals
ticks = (0:N)+0.5;
labels = strtrim(cellstr(num2str(edges(:),'%g')));
set(gca, 'XTick',ticks, 'XTickLabel',labels, ...
'YTick',ticks, 'YTickLabel',labels)
% plot 2D points on top, correctly scaled from [-1,1] to [0,N]+0.5
XY2 = bsxfun(#rdivide, bsxfun(#minus, XY, mn), mx-mn) * N + 0.5;
line(XY2(:,1), XY2(:,2), 'LineStyle','none', 'Marker','.', 'Color','r')
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.
Idea
Im trying to plot a plane delimited by two vectors, using cross in matlab
Code
NM= [1 3; 2 4] //matrix
figure
hold on;
z = zeros(size(NM, 1), 1); //to use quiver3
quiver3(z, z, z, NM(:,1), NM(:,2), z, 0); //vectors plotted
grid on
view(45, 45);
s=sum(NM);
p = 10*(rand(3,1) - 0.5); // generation of points
O1=[NM(1,:) 0] // new vectors of length 3 ,
O2=[NM(2,:) 0] // to be used by cross
v3 = cross(O1,O2) //cross product to find the norm
[ x , y ] = meshgrid( p(1)+(-5:5) , p(2)+(-5:5) ); // points inside the plane
z = p(3) - (v3(1)*(x-p(1)) + v3(2)*(y-p(2)))/v3(3); // plane equation
surf(x,y,z) //the plane itself
The output is
Issue
The plane must be delimited by the vectors or the vectors must be inside the plane not outside.
The vectors do not appear inside the plane because you are choosing (0,0,0) as the starting point of the vectors while you are making the plane pass by the randomly chosen point p.
You either make the plane pass by (0,0,0) or use p as the starting point for the vectors when plotted with quiver3().
Here is a solution where I have chosen the second option:
vplane = [1 3 0; 2 4 0]'; % (column) vectors defining the plane
vnormal = cross(vplane(:,1), vplane(:,2)); % normal defining the orientation of the plane
figure; hold on; grid on; view(45, 45);
rng(1313) % seed for reproducible output
p = 10*(rand(3,1) - 0.5); % a point defining the position of the plane we want to plot with the given normal vector
P = repmat(p, 1, 2); % matrix with the given point repeated for easier use of quiver3()
quiver3(P(1,:), P(2,:), P(3,:), vplane(1,:), vplane(2,:), vplane(3,:), 0); % arrows representing the vectors defining the plane
[x,y] = meshgrid( p(1)+(-5:5), p(2)+(-5:5) ); % set of equally spaced points inside the plane
z = p(3) - (vnormal(1)*(x-p(1)) + vnormal(2)*(y-p(2))) / vnormal(3); % plane equation
surf(x,y,z) % plot the plane
and the result is the following:
I am using MATLAB to print my simulation results. The results concerns a UAV's trajectory and waypoints that the UAV has to visit. The UAV is supposed to be equipped with a camera, whose range view is 10x10. Right now, the diagram shows the UAV's trajectory as a line visiting the waypoints. Is it possible, to show the camera's footprint, instead of the actual trajectory? I would like it to plot the rectangular camera's view to show the exhaustive coverage of the area. There is the option to plot the points as square, or cross, or cyrcles, but is it possible to set the boundaries of those?
Thank you in advance
The problem with using the marker size to indicate the range view is that there is no direct relation between the data units of your waypoints and the marker size. In other words, a value of 10 for the marker size doesn't necessarily mean that a side of a square marker is going to be 10 data units long (as defined by the scaling and limits of the axes).
An alternative is to plot square patches at each of your waypoints where the patch is aligned with the trajectory of the UAV. Here's how you can do this:
% Generate some sample data:
N = 20; % Number of waypoints
x = cumsum(5.*rand(1, N)); % X coordinates of UAV
y = cumsum(5.*rand(1, N)); % Y coordinates of UAV
% Compute vectors parallel and perpendicular to the trajectory at each point:
v = [diff(x); diff(y); zeros(1, N-1)]; % Vectors (1 per column)
v = bsxfun(#rdivide, v, sqrt(sum(v.^2, 1))); % Normalize each column to a unit vector
v = v(:, [1 1:end]); % Replicate a vector for starting point
vCross = cross(v, [zeros(2, N); ones(1, N)]); % Perpendicular vector
% Generate patch coordinates:
R = 10; % Range view
xPatch = [x+(R/2).*(v(1, :)+vCross(1, :)); ...
x+(R/2).*(v(1, :)-vCross(1, :)); ...
x-(R/2).*(v(1, :)+vCross(1, :)); ...
x-(R/2).*(v(1, :)-vCross(1, :))];
yPatch = [y+(R/2).*(v(2, :)+vCross(2, :)); ...
y+(R/2).*(v(2, :)-vCross(2, :)); ...
y-(R/2).*(v(2, :)+vCross(2, :)); ...
y-(R/2).*(v(2, :)-vCross(2, :))];
% Plot the patches and trajectory:
patch(xPatch, yPatch, [0 0.3 0], 'FaceAlpha', 0.25, 'EdgeColor', 'none');
hold on;
plot(x, y, '-', 'Color', [0.8 0 0], 'Marker', '.', 'MarkerSize', 12);
axis equal;
And here's a sample plot:
As a first attempt you can specify marker shape as square and set constant marker size, e.g.
plot(x,y,'s','markersize',10)
Here x and y are the vectors, holding the UAV coordinates. The letter 's' sets marker shape as square, and size is set to 10.
In reality, UAV trajectory is defined in a 3d space, where varying height above the ground corresponds to varying footprint size and shape. Taking this into account would require a bit more effort.
Also this assumes that the points are spaced closely enough otherwise there would be empty areas between markers.
I have polar coordinates, radius 0.05 <= r <= 1 and 0 ≤ θ ≤ 2π. The radius r is 50 values between 0.05 to 1, and polar angle θ is 24 values between 0 to 2π.
How do I interpolate r = 0.075 and theta = pi/8?
I dunno what you have tried, but interp2 works just as well on polar data as it does on Cartesian. Here is some evidence:
% Coordinates
r = linspace(0.05, 1, 50);
t = linspace(0, 2*pi, 24);
% Some synthetic data
z = sort(rand(50, 24));
% Values of interest
ri = 0.075;
ti = pi/8;
% Manually interpolate
rp = find(ri <= r, 1, 'first');
rm = find(ri >= r, 1, 'last');
tp = find(ti <= t, 1, 'first');
tm = find(ti >= t, 1, 'last');
drdt = (r(rp) - r(rm)) * (t(tp) - t(tm));
dr = [r(rp)-ri ri-r(rm)];
dt = [t(tp)-ti ti-t(tm)];
fZ = [z(rm, tm) z(rm, tp)
z(rp, tm) z(rp, tp)];
ZI_manual = (dr * fZ * dt.') / drdt
% Interpolate with MATLAB
ZI_MATLAB = interp2(r, t, z', ri, ti, 'linear')
Result:
ZI_manual =
2.737907208525297e-002
ZI_MATLAB =
2.737907208525298e-002
Based on comments you have the following information
%the test point
ri=0.53224;
ti = pi/8;
%formula fo generation of Z
g=9.81
z0=#(r)0.01*(g^2)*((2*pi)^-4)*(r.^-5).*exp(-1.25*(r/0.3).^-4);
D=#(t)(2/pi)*cos(t).^2;
z2=#(r,t)z0(r).*D(t) ;
%range of vlaues of r and theta
r=[0.05,0.071175,0.10132,0.14422,0.2053, 0.29225,0.41602,0.5922,0.84299,1.2];
t=[0,0.62832,1.2566,1.885, 2.5133,3.1416,3.7699,4.3982,5.0265,5.6549,6.2832];
and you want interplation of the test point.
When you sample some data to use them for interpolation you should consider how to sample data according to your requirements.
So when you are sampling a regular grid of polar coordinates ,those coordinates when converted to rectangular will form a circular shape that
most of the points are concentrated in the center of the cricle and when we move from the center to outer regions distance between the points increased.
%regular grid generated for r and t
[THETA R] = meshgrid(t ,r);
% Z for polar grid
Z=z2(R,THETA);
%convert coordinate from polar to cartesian(rectangular):
[X, Y] = pol2cart (THETA, R);
%plot points
plot(X, Y, 'k.');
axis equal
So when you use those point for interpolation the accuracy of the interpolation is greater in the center and lower in the outer regions where the distance between points increased.
In the other word with this sampling method you place more importance on the center region related to outer ones.
To increase accuracy density of grid points (r and theta) should be increased so if length of r and theta is 11 you can create r and theta with size 20 to increase accuracy.
In the other hand if you create a regular grid in rectangular coordinates an equal importance is given to each region . So accuracy of the interpolation will be the same in all regions.
For it first you create a regular grid in the polar coordinates then convert the grid to rectangular coordinates so you can calculate the extents (min max) of the sampling points in the rectangular coordinates. Based on this extents you can create a regular grid in the rectangular coordinates
Regular grid of rectangular coordinates then converted to polar coordinated to get z for grid points using z2 formula.
%get the extent of points
extentX = [min(X(:)) max(X(:))];
extentY = [min(Y(:)) max(Y(:))];
%sample 100 points(or more or less) inside a region specified be the extents
X_samples = linspace(extentX(1),extentX(2),100);
Y_samples = linspace(extentY(1),extentY(2),100);
%create regular grid in rectangular coordinates
[XX YY] = meshgrid(X_samples, Y_samples);
[TT RR] = cart2pol(XX,YY);
Z_rect = z2(RR,TT);
For interpolation of a test point say [ri ti] first it converted to rectangular then using XX ,YY value of z is interpolated
[xi yi] = pol2cart (ti, ri);
z=interp2(XX,YY,Z_rect,xi,yi);
If you have no choice to change how you sample the data and only have a grid of polar points as discussed with #RodyOldenhuis you can do the following:
Interpolate polar coordinates with interp2 (interpolation for gridded data)
this approach is straightforward but has the shortcoming that r and theta are not of the same scale and this may affect the accuracy of the interpolation.
z = interp2(THETA, R, Z, ti, ri)
convert polar coordinates to rectangular and then apply an interpolation method that is for scattered data.
this approach requires more computations but result of it is more reliable.
MATLAB has griddata function that given scattered points first generates a triangulation of points and then creates a regular grid on top of the triangles and interpolates values of grid points.
So if you want to interpolate value of point [ri ti] you should then apply a second interpolation to get value of the point from the interpolated grid.
With the help of some information from spatialanalysisonline and Wikipedia linear interpolation based on triangulation calculated this way (tested in Octave. In newer versions of MATLAB use of triangulation and pointLocation recommended instead of delaunay and tsearch ):
ri=0.53224;
ti = pi/8;
[THETA R] = meshgrid(t ,r);
[X, Y] = pol2cart (THETA, R);
[xi yi] = pol2cart (ti, ri);
%generate triangulation
tri = delaunay (X, Y);
%find the triangle that contains the test point
idx = tsearch (X, Y, tri, xi, yi);
pts= tri(idx,:);
%create a matrix that repesents equation of a plane (triangle) given its 3 points
m=[X(pts);Y(pts);Z(pts);ones(1,3)].';
%calculate z based on det(m)=0;
z= (-xi*det(m(:,2:end)) + yi*det([m(:,1) m(:,3:end)]) + det(m(:,1:end-1)))/det([m(:,1:2) m(:,end)]);
More refinement:
Since it is known that the search point is surrounded by 4 points we can use only those point for triangulation. these points form a trapezoid. Each diagonal of trapezoid forms two triangles so using vertices of the trapezoid we can form 4 triangles, also a point inside a trapezoid can lie in at least 2 triangles.
the previous method based on triangulation only uses information from one triangle but here z of the test point can be interpolated two times from data of two triangles and the calculated z values can be averaged to get a better approximation.
%find 4 points surrounding the test point
ft= find(t<=ti,1,'last');
fr= find(cos(abs(diff(t(ft+(0:1))))/2) .* r < ri,1,'last');
[T4 R4] = meshgrid(t(ft+(0:1)), r(fr+(0:1)));
[X4, Y4] = pol2cart (T4, R4);
Z4 = Z(fr+(0:1),ft+(0:1));
%form 4 triangles
tri2= nchoosek(1:4,3);
%empty vector of z values that will be interpolated from 4 triangles
zv = NaN(4,1);
for h = 1:4
pts = tri2(h,:);
% test if the point lies in the triangle
if ~isnan(tsearch(X4(:),Y4(:),pts,xi,yi))
m=[X4(pts) ;Y4(pts) ;Z4(pts); [1 1 1]].';
zv(h)= (-xi*det(m(:,2:end)) + yi*det([m(:,1) m(:,3:end)]) + det(m(:,1:end-1)))/det([m(:,1:2) m(:,end)]);
end
end
z= mean(zv(~isnan(zv)))
Result:
True z:
(0.0069246)
Linear Interpolation of (Gridded) Polar Coordinates :
(0.0085741)
Linear Interpolation with Triangulation of Rectangular Coordinates:
(0.0073774 or 0.0060992) based on triangulation
Linear Interpolation with Triangulation of Rectangular Coordinates(average):
(0.0067383)
Conclusion:
Result of interpolation related to structure of original data and the sampling method. If the sampling method matches pattern of original data result of interpolation is more accurate, so in cases that grid points of polar coordinates follow pattern of data result of interpolation of regular polar coordinate can be more reliable. But if regular polar coordinates do not match the structure of data or structure of data is such as an irregular terrain, method of interpolation based on triangulation can better represent the data.
please check this example, i used two for loops, inside for loop i used condition statement, if u comment this condition statement and run the program, u'll get correct answer, after u uncomment this condition statement and run the program, u'll get wrong answer. please check it.
% Coordinates
r = linspace(0.05, 1, 10);
t = linspace(0, 2*pi, 8);
% Some synthetic data
%z = sort(rand(50, 24));
z=zeros();
for i=1:10
for j=1:8
if r(i)<0.5||r(i)>1
z(i,j)=0;
else
z(i,j) = r(i).^3'*cos(t(j)/2);
end
end
end
% Values of interest
ri = 0.55;
ti = pi/8;
% Manually interpolate
rp = find(ri <= r, 1, 'first');
rm = find(ri >= r, 1, 'last');
tp = find(ti <= t, 1, 'first');
tm = find(ti >= t, 1, 'last');
drdt = (r(rp) - r(rm)) * (t(tp) - t(tm));
dr = [r(rp)-ri ri-r(rm)];
dt = [t(tp)-ti ti-t(tm)];
fZ = [z(rm, tm) z(rm, tp)
z(rp, tm) z(rp, tp)];
ZI_manual = (dr * fZ * dt.') / drdt
% Interpolate with MATLAB
ZI_MATLAB = interp2(r, t, z', ri, ti, 'linear')
Result:
z1 =
0.1632
ZI_manual =
0.1543
ZI_MATLAB =
0.1582
I have points in 3D space and their corresponding 2D image points. How can I make a mesh out of the 3D points, then texture the triangle faces formed by the mesh?
Note that the function trisurf that you were originally trying to use returns a handle to a patch object. If you look at the 'FaceColor' property for patch objects, you can see that there is no 'texturemap' option. That option is only valid for the 'FaceColor' property of surface objects. You will therefore have to find a way to plot your triangular surface as a surface object instead of a patch object. Here are two ways to approach this:
If your data is in a uniform grid...
If the coordinates of your surface data represent a uniform grid such that z is a rectangular set of points that span from xmin to xmax in the x-axis and ymin to ymax in the y-axis, you can plot it using surf instead of trisurf:
Z = ... % N-by-M matrix of data
x = linspace(xmin, xmax, size(Z, 2)); % x-coordinates for columns of Z
y = linspace(ymin, ymax, size(Z, 1)); % y-coordinates for rows of Z
[X, Y] = meshgrid(x, y); % Create meshes for x and y
C = imread('image1.jpg'); % Load RGB image
h = surf(X, Y, Z, flipdim(C, 1), ... % Plot surface (flips rows of C, if needed)
'FaceColor', 'texturemap', ...
'EdgeColor', 'none');
axis equal
In order to illustrate the results of the above code, I initialized the data as Z = peaks;, used the built-in sample image 'peppers.png', and set the x and y values to span from 1 to 16. This resulted in the following texture-mapped surface:
If your data is non-uniformly spaced...
If your data are not regularly spaced, you can create a set of regularly-spaced X and Y coordinates (as I did above using meshgrid) and then use one of the functions griddata or TriScatteredInterp to interpolate a regular grid of Z values from your irregular set of z values. I discuss how to use these two functions in my answer to another SO question. Here's a refined version of the code you posted using TriScatteredInterp (Note: as of R2013a scatteredInterpolant is the recommended alternative):
x = ... % Scattered x data
y = ... % Scattered y data
z = ... % Scattered z data
xmin = min(x);
xmax = max(x);
ymin = min(y);
ymax = max(y);
F = TriScatteredInterp(x(:), y(:), z(:)); % Create interpolant
N = 50; % Number of y values in uniform grid
M = 50; % Number of x values in uniform grid
xu = linspace(xmin, xmax, M); % Uniform x-coordinates
yu = linspace(ymin, ymax, N); % Uniform y-coordinates
[X, Y] = meshgrid(xu, yu); % Create meshes for xu and yu
Z = F(X, Y); % Evaluate interpolant (N-by-M matrix)
C = imread('image1.jpg'); % Load RGB image
h = surf(X, Y, Z, flipdim(C, 1), ... % Plot surface
'FaceColor', 'texturemap', ...
'EdgeColor', 'none');
axis equal
In this case, you have to first choose the values of N and M for the size of your matrix Z. In order to illustrate the results of the above code, I initialized the data for x, y, and z as follows and used the built-in sample image 'peppers.png':
x = rand(1, 100)-0.5; % 100 random values in the range -0.5 to 0.5
y = rand(1, 100)-0.5; % 100 random values in the range -0.5 to 0.5
z = exp(-(x.^2+y.^2)./0.125); % Values from a 2-D Gaussian distribution
This resulted in the following texture-mapped surface:
Notice that there are jagged edges near the corners of the surface. These are places where there were too few points for TriScatteredInterp to adequately fit an interpolated surface. The Z values at these points are therefore nan, resulting in the surface point not being plotted.
If your texture is already in the proper geometry you can just use regular old texture mapping.
The link to the MathWorks documentation of texture mapping:
http://www.mathworks.com/access/helpdesk/help/techdoc/visualize/f0-18164.html#f0-9250
Re-EDIT: Updated the code a little:
Try this approach (I just got it to work).
a=imread('image.jpg');
b=double(a)/255;
[x,y,z]=peaks(30); %# This is a surface maker that you do have
%# The matrix [x,y,z] is the representation of the surface.
surf(x,y,z,b,'FaceColor','texturemap') %# Try this with any image and you
%# should see a pretty explanatory
%# result. (Just copy and paste) ;)
So [x,y,z] is the 'surface' or rather a matrix containing a number of points in the form (x,y,z) that are on the surface. Notice that the image is stretched to fit the surface.