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I have a discrete dataset, X=[x1,x2,..,x12] & Y=[y1,y2,...,y12]. X ranges from [-25, 0] and Y ranges from [1e-6, 1e0]. X does not increase uniformly - as x approaches a value of 0, data sampling density increases from increments of 2.5 to increments of 1. Each x value is units of cm. I cannot get a good fit to the data from fitting a function (I've tried quite a few). I'm left with the discrete data. My need is to sweep the X, Y data completly around the Z axis and put the resulting swept data values into a matrix Z of size (51, 51). I've tried using the cylinder function, [u,v,w] = cylinder(Y) thinking I could extract the data or create a matrix Z from [u, v, w]. I can't seem to sort that out. surf(u,v,w) plots almost correctly - the scaling on the (u, v) axes ranges from [-1, 1] instead of [-25, 25]. This is, I assume, because I'm using cylinder(Y). When I try [u,v,w] = cylinder(X,Y) I get, error: linspace: N must be a scalar. It seems like there should be a better way then my approach of using cylinder to take the X & Y data, interpolate between points (to fill Z where data isn't), rotate it, and put the result into a matrix Z. Any suggestions are welcome. I'm using Octave 6.3.0. Thank you in advance.
Create a matrix R containing distance from origin values.
Use for loops and single value interpolation to cover the R space.
Place the interpolated values into the matrix Z.
% Original data X = [-25,-22.5,...,0]; size(X) = 1 12
% Original data Y = [1e-6, 1.3e-6,...,1] size(Y) = 1 12
u = [-range(X):1:range(X)]; v = [-range(X):1:range(X)]';
R = -sqrt(u.^2.+v.^2);
Z = zeros( 2 .* range(X) + 1);
for i = 1:size(R,1)
for j = 1:size(R,2)
if R(i,j) < min(X); Z(i,j) = 0; endif
if R(i,j) >= min(X); Z(i,j) = interp1(X,Y,R(i,j)); endif
endfor
endfor
Given a matrix X of size (5,3), and a vector of y size (1,3), I need to calculate the Euclidean distance of vector y to all vectors in X and return to the minimal.
For example
X =
0.1338 0.0346 0.2961
0.5320 0.4681 0.6784
0.4484 0.5954 0.2847
0.1437 0.5310 0.3946
0.2854 0.0793 0.8621
y = 0.4484 0.5954 0.2847
So in that case, the minimal Euclidean distance of y is with row 3 in matrix Xsince it's the same.
I have made the code as below :
X = rand(5,3) %The matrix x
y = rand(1,3) %The vector y
[~, size_y] = size(y) %size of y
[size_x, ~] = size(X) %size of matrix x
min_distance = zeros(size_x,size_y); %Initialize the minimal distance
%% Calculate minimum distance square of vector y to every vector in x
for i = 1 : size_y
min_distance(:,i) = sum(abs(repmat(y(:,i),1,size_x) - X).^2,2);
end
min_distance_1 = min_distance;
[index, ~] = (min(min_distance_1,[],1));
results = index - 1;
The results of that code is mismatch error, however the index of row in matrix X which has the minimal Euclidean distance to vector y should be shown!!
Is there any mistake in code ? or how can I do that ?
I think there are a few things missing in your code. For the Euclidean distance there should be a root somewhere. Also index is the second return value, not the first and there is no reason for -1 and an index in matlab.
There is a build in function to calculate the euclidean distance called norm(). For your case where you want to calculate the distance to each vector there is a special vecnorm() thats fits perfectly.
differences = X-y
%you dont need repmat but keep it if it helps your understanding of the code
distances = vecnorm(differences,2,2)
%the first 2 is for 2-norm, which is Euclidean distance
%the second 2 for row-wise calculation
[~,index]= min(distances)
Example:
Formula:
The difference between the two points gets the difference of the value of the received points of the matrix. If possible, could you suggest me a formula.
Minimum Distance between two points which are n-dimensional vectors in R^n can be found as below ( e.g: x= [ 1 3 5], y= [4 -4 7] 3-D points)
function result = minDistance (x, y) {
if ~isvector(x)
error('Input must be a vector')
result = sqrt(sum((x-y).^2))
end
if any additional information needed please explain the formula then solution can be fullfilled.
Based on your clarification, I think you only need a loop to do the job. The loop will go through each pixel, or point, in the picture, compare it with all other pixels and calculate the displacement vectors.
So for example, if you have a picture of 10x10, this means you have 100 pixels in the picture. For each pixel, you can calculate 99 displacement vectors. For the entire picture, you will have 99*100 displacement vectors.
% read your picture
im = imread('myImage.jpg');
imageSize = size(im);
xsize = imageSize(1);
ysize = imageSize(2);
% construct pixel matrix
pixelPosXM = repmat(1:xsize, ysize, 1);
pixelPosYM = repmat(1:ysize, 1, xsize);
% convert into array
pixelPosX = pixelPosXM(:);
pixelPosY = pixelPosYM(:);
nsize = length(pixelPosX);
for i = 1:nsize
% construct matrix for calculation
px = repmat([pixelPosX(i), pixelPosY(i)], nsize - 1, 1);
otherPixels = [pixelPosX, pixelPosY]; otherPixels(i,:) = [];
% calculate displacement vectors
displacementVectors(:,:,i) = [px - otherPixels];
end
If you are not familiar with 3-dimentional matrix, you can replace
displacementVectors(:,:,i) = [px - otherPixels];
with
displacementVectors{i} = [px - otherPixels];
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 would like to reproduce the following figure in MATLAB:
There are two classes of points with X and Y coordinates. I'd like to surround each class with an ellipse with one parameter of standard deviation, which determine how far the ellipse will go along the axis.
The figure was created with another software and I don't exactly understand how it calculates the ellipse.
Here is the data I'm using for this figure. The 1st column is class, 2nd - X, 3rd - Y. I can use gscatter to draw the points itself.
A = [
0 0.89287 1.54987
0 0.69933 1.81970
0 0.84022 1.28598
0 0.79523 1.16012
0 0.61266 1.12835
0 0.39950 0.37942
0 0.54807 1.66173
0 0.50882 1.43175
0 0.68840 1.58589
0 0.59572 1.29311
1 1.00787 1.09905
1 1.23724 0.98834
1 1.02175 0.67245
1 0.88458 0.36003
1 0.66582 1.22097
1 1.24408 0.59735
1 1.03421 0.88595
1 1.66279 0.84183
];
gscatter(A(:,2),A(:,3),A(:,1))
FYI, here is the SO question on how to draw ellipse. So, we just need to know all the parameters to draw it.
Update:
I agree that the center can be calculated as the means of X and Y coordinates. Probably I have to use principal component analysis (PRINCOMP) for each class to determine the angle and shape. Still thinking...
Consider the code:
%# generate data
num = 50;
X = [ mvnrnd([0.5 1.5], [0.025 0.03 ; 0.03 0.16], num) ; ...
mvnrnd([1 1], [0.09 -0.01 ; -0.01 0.08], num) ];
G = [1*ones(num,1) ; 2*ones(num,1)];
gscatter(X(:,1), X(:,2), G)
axis equal, hold on
for k=1:2
%# indices of points in this group
idx = ( G == k );
%# substract mean
Mu = mean( X(idx,:) );
X0 = bsxfun(#minus, X(idx,:), Mu);
%# eigen decomposition [sorted by eigen values]
[V D] = eig( X0'*X0 ./ (sum(idx)-1) ); %#' cov(X0)
[D order] = sort(diag(D), 'descend');
D = diag(D);
V = V(:, order);
t = linspace(0,2*pi,100);
e = [cos(t) ; sin(t)]; %# unit circle
VV = V*sqrt(D); %# scale eigenvectors
e = bsxfun(#plus, VV*e, Mu'); %#' project circle back to orig space
%# plot cov and major/minor axes
plot(e(1,:), e(2,:), 'Color','k');
%#quiver(Mu(1),Mu(2), VV(1,1),VV(2,1), 'Color','k')
%#quiver(Mu(1),Mu(2), VV(1,2),VV(2,2), 'Color','k')
end
EDIT
If you want the ellipse to represent a specific level of standard deviation, the correct way of doing is by scaling the covariance matrix:
STD = 2; %# 2 standard deviations
conf = 2*normcdf(STD)-1; %# covers around 95% of population
scale = chi2inv(conf,2); %# inverse chi-squared with dof=#dimensions
Cov = cov(X0) * scale;
[V D] = eig(Cov);
I'd try the following approach:
Calculate the x-y centroid for the center of the ellipse (x,y in the linked question)
Calculate the linear regression fit line to get the orientation of the ellipse's major axis (angle)
Calculate the standard deviation in the x and y axes
Translate the x-y standard deviations so they're orthogonal to the fit line (a,b)
I'll assume there is only one set of points given in a single matrix, e.g.
B = A(1:10,2:3);
you can reproduce this procedure for each data set.
Compute the center of the ellipsoid, which is the mean of the points. Matlab function: mean
Center your data. Matlab function bsxfun
Compute the principal axis of the ellipsoid and their respective magnitude. Matlab function: eig
The successive steps are illustrated below:
Center = mean(B,1);
Centered_data = bsxfun(#minus,B,Center);
[AX,MAG] = eig(Centered_data' * Centered_data);
The columns of AX contain the vectors describing the principal axis of the ellipsoid while the diagonal of MAG contains information on their magnitude.
To plot the ellipsoid, scale each principal axis with the square root of its magnitude.