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
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 have two arrays
x = [0 9.8312 77.1256 117.9810 99.9979];
y = [0 2.7545 4.0433 5.3763 5.0504];
figure; plot(x, y)
I want to make more samples of x and y then I interpolated both arrays. I tried this code
xi =min(x):.1:max(x);
yi = interp1(x,y,xi);
figure; plot(xi, yi)
but the trajectory is not same as previous plot. Its because the xi is not fluctuating same as x. How should I interpolate both arrays with same trajectory as original one?
This is an issue because when interpolating, MATLAB is going to ignore the order that you feed in the points and instead just sort them based upon their x location.
Rather than interpolating in x/y coordinates, you can instead use a parameter which represents the cumulative arc length of the line segments and use that to interpolate both the x and y coordinates. This will provide you with an interpolant that respects the order and guarantees monotonicity even for multiple values at the same x coordinate.
% Compute the distance between all points.
distances = sqrt(diff(x).^2 + diff(y).^2);
% Compute the cumulative arclength
t = cumsum([0 distances]);
% Determine the arclengths to interpolate at
tt = linspace(t(1), t(end), 1000);
% Now interpolate x and y at these locations
xi = interp1(t, x, tt);
yi = interp1(t, y, tt);
I intend to use Matlab to plot the probability distribution from stochastic process on its state space. The state space can be represented by the lower triangle of a 150x150 matrix. Please see the figure (a surf plot without mesh) for a probability distribution at a certain time point.
As we can see, there is a high degree of symmetry in the graph, but because it is plotted as a square matrix, it looks kind of weird. It we could transform the rectangle the plot would look perfect. My question is, how can I use Matlab to plot/transform the lower triangle portion as/to an equal-lateral triangle?
This function should do the job for you. If not, please let me know.
function matrix_lower_tri_to_surf(A)
%Displays lower triangle portion of matrix as an equilateral triangle
%Martin Stålberg, Uppsala University, 2013-07-12
%mast4461 at gmail
siz = size(A);
N = siz(1);
if ~(ndims(A)==2) || ~(N == siz(2))
error('Matrix must be square');
end
zeds = #(N) zeros(N*(N+1)/2,1); %for initializing coordinate vectors
x = zeds(N); %x coordinates
y = zeds(N); %y coordinates
z = zeds(N); %z coordinates, will remain zero
r = zeds(N); %row indices
c = zeds(N); %column indices
l = 0; %base index
xt = 1:N; %temporary x coordinates
yt = 1; %temporary y coordinates
for k = N:-1:1
ind = (1:k)+l; %coordinate indices
l = l+k; %update base index
x(ind) = xt; %save temporary x coordinates
%next temporary x coordinates are the k-1 middle pairwise averages
%calculated by linear interpolation through convolution
xt = conv(xt,[.5,.5]);
xt = xt(2:end-1);
y(ind) = yt; %save temporary y coordinates
yt = yt+1; %update temporary y coordinates
r(ind) = N-k+1; %save row indices
c(ind) = 1:k; % save column indices
end
v = A(sub2ind(size(A),r,c)); %extract values from matrix A
tri = delaunay(x,y); %create triangular mesh
h = trisurf(tri,x,y,z,v,'edgecolor','none','facecolor','interp'); %plot surface
axis vis3d; view(2); %adjust axes projection and proportions
daspect([sqrt(3)*.5,1,1]); %adjust aspect ratio to display equilateral triangle
end %end of function
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)
I am trying to convert an image from cartesian to polar coordinates.
I know how to do it explicitly using for loops, but I am looking for something more compact.
I want to do something like:
[x y] = size(CartImage);
minr = floor(min(x,y)/2);
r = linspace(0,minr,minr);
phi = linspace(0,2*pi,minr);
[r, phi] = ndgrid(r,phi);
PolarImage = CartImage(floor(r.*cos(phi)) + minr, floor(r.sin(phi)) + minr);
But this obviously doesn't work.
Basically I want to be able to index the CartImage on a grid.
The polar image would then be defined on the grid.
given a matrix M (just a 2d Gaussian for this example), and a known origin point (X0,Y0) from which the polar transform takes place, we expect that iso-intensity circles will transform to iso-intensity lines:
M=fspecial('gaussian',256,32); % generate fake image
X0=size(M,1)/2; Y0=size(M,2)/2;
[Y X z]=find(M);
X=X-X0; Y=Y-Y0;
theta = atan2(Y,X);
rho = sqrt(X.^2+Y.^2);
% Determine the minimum and the maximum x and y values:
rmin = min(rho); tmin = min(theta);
rmax = max(rho); tmax = max(theta);
% Define the resolution of the grid:
rres=128; % # of grid points for R coordinate. (change to needed binning)
tres=128; % # of grid points for theta coordinate (change to needed binning)
F = TriScatteredInterp(rho,theta,z,'natural');
%Evaluate the interpolant at the locations (rhoi, thetai).
%The corresponding value at these locations is Zinterp:
[rhoi,thetai] = meshgrid(linspace(rmin,rmax,rres),linspace(tmin,tmax,tres));
Zinterp = F(rhoi,thetai);
subplot(1,2,1); imagesc(M) ; axis square
subplot(1,2,2); imagesc(Zinterp) ; axis square
getting the wrong (X0,Y0) will show up as deformations in the transform, so be careful and check that.
I notice that the answer from bla is from polar to cartesian coordinates.
However the question is in the opposite direction.
I=imread('output.png'); %read image
I1=flipud(I);
A=imresize(I1,[1024 1024]);
A1=double(A(:,:,1));
A2=double(A(:,:,2));
A3=double(A(:,:,3)); %rgb3 channel to double
[m n]=size(A1);
[t r]=meshgrid(linspace(-pi,pi,n),1:m); %Original coordinate
M=2*m;
N=2*n;
[NN MM]=meshgrid((1:N)-n-0.5,(1:M)-m-0.5);
T=atan2(NN,MM);
R=sqrt(MM.^2+NN.^2);
B1=interp2(t,r,A1,T,R,'linear',0);
B2=interp2(t,r,A2,T,R,'linear',0);
B3=interp2(t,r,A3,T,R,'linear',0); %rgb3 channel Interpolation
B=uint8(cat(3,B1,B2,B3));
subplot(211),imshow(I); %draw the Original Picture
subplot(212),imshow(B); %draw the result