Does anyone know how to obtain a mean curve having a matrix with the correspondent x,y points from the original plot? I mean, I pretend a medium single curve.
Any code or just ideas would be very very helpful for me since I am new with matlab.
Thank you very much!
Well, one thing you can do is fit a parametric curve. Here's an example on how to do this for a figure-8 with noise on it:
function findParamFit
clc, clf, hold on
%# some sample data
noise = #(t) 0.25*rand(size(t))-0.125;
x = #(t) cos(t) + noise(t);
y = #(t) sin(2*t) + noise(t);
t = linspace(-100*rand, +100*rand, 1e4);
%# initial data
plot(x(t), y(t), 'b.')
%# find fits
options = optimset(...
'tolfun', 1e-12,...
'tolx', 1e-12);
a = lsqcurvefit(#myFun_x, [1 1], t, x(t), -10,10, options);
b = lsqcurvefit(#myFun_y, [1 2], t, y(t), -10,10, options);
%# fitted curve
xx = myFun_x(a,t);
yy = myFun_y(b,t);
plot(xx, yy, 'r.')
end
function F = myFun_x(a, tt)
F = a(1)*cos(a(2)*tt);
end
function F = myFun_y(b, tt)
F = b(1)*sin(b(2)*tt);
end
Note that this is a particularly bad way to fit parametric curves, as is apparent here by the extreme sensitivity of the solution to the quality of the initial values to lsqcurvefit. Nevertheless, fitting a parametric curve will be the way to go.
There's your google query :)
Related
The stairstep-look of the graph is unintentional. When I plot the same-sized vectors Arb and V (plot (Arb, V, 'r')) I get the following graph:
enter image description here
To make it smoother, I tried using 1-D data interpolation, interp1, as follows:
xq = 0:0.001:max(Arb);
Vq = interp1 (Arb, V, xq);
plot (xq, Vq);
However, I get this error message:
Error using interp1>reshapeAndSortXandV (line 416)
X must be a vector.
Error in interp1 (line 92)
[X,V,orig_size_v] = reshapeAndSortXandV(varargin{1},varargin{2})
interp1 will use linear interpolation on your data so it won't help you much. One thing you can try is to use a cubic spline with interp1 but again given the nature of the data I don't think it will help much. e.g.
Alternative 1. Cubic Spline
xq = 0:0.001:max(Arb);
Vq = interp1 (Arb, V, xq, 'spline');
plot (xq, Vq);
Alternative 2. Polynomial fit
Another alternative you can try is, polynomial interpolation using polyfit. e.g.
p = polifit(Arb, V, 2); % I think a 2nd order polynomial should do
xq = 0:0.001:max(Arb);
Vq = polyval(p, xq);
plot (xq, Vq);
Alternative 3. Alpha-beta filter
Last but not least, another thing to try is to use an smoothing alpha-beta filter on your V data. One possible implementation can be:
% x is the input data
% a is the "smoothing" factor from [0, 1]
% a = 0 full smoothing
% a = 1 no smoothing
function xflt = alphaBeta(x, a)
xflt = zeros(1,length(x));
xflt(1) = x(1);
a = max(min(a, 1), 0); % Bound coefficient between 0 and 1;
for n = 2:1:length(x);
xflt(n) = a * x(n) + (1 - a) * xflt(n-1);
end
end
Usage:
plot (Arb, alphaBeta(V, 0.65), 'r')
I have a set of 3d data (300 points) that create a surface which looks like two cones or ellipsoids connected to each other. I want a way to find the equation of a best fit ellipsoid or cone to this dataset. The regression method is not important, the easier it is the better. I basically need a way, a code or a matlab function to calculate the constants of the elliptic equation for these data.
You can also try with fminsearch, but to avoid falling on local minima you will need a good starting point given the amount of coefficients (try to eliminate some of them).
Here is an example with a 2D ellipse:
% implicit equation
fxyc = #(x, y, c_) c_(1)*x.^2 + c_(2).*y.^2 + c_(3)*x.*y + c_(4)*x + c_(5).*y - 1; % free term locked to -1
% solution (ellipse)
c_ = [1, 2, 1, 0, 0]; % x^2, y^2, x*y, x, y (free term is locked to -1)
[X,Y] = meshgrid(-2:0.01:2);
figure(1);
fxy = #(x, y) fxyc(x, y, c_);
c = contour(X, Y, fxy(X, Y), [0, 0], 'b');
axis equal;
grid on;
xlabel('x');
ylabel('y');
title('solution');
% we sample the solution to have some data to fit
N = 100; % samples
sample = unique(2 + floor((length(c) - 2)*rand(1, N)));
x = c(1, sample).';
y = c(2, sample).';
x = x + 5e-2*rand(size(x)); % add some noise
y = y + 5e-2*rand(size(y));
fc = #(c_) fxyc(x, y, c_); % function in terms of the coefficients
e = #(c) fc(c).' * fc(c); % squared error function
% we start with a circle
c0 = [1, 1, 0, 0, 0];
copt = fminsearch(e, c0)
figure(2);
plot(x, y, 'rx');
hold on
fxy = #(x, y) fxyc(x, y, copt);
contour(X, Y, fxy(X, Y), [0, 0], 'b');
hold off;
axis equal;
grid on;
legend('data', 'fit');
xlabel('x'); %# Add an x label
ylabel('y');
title('fitted solution');
The matlab function fit can take arbitrary fit expressions. It takes a bit of figuring out the parameters but it can be done.
You would first create a fittype object that has a string representing your expected form. You'll need to work out the expression yourself that best fits what you're expecting, I'm going to take a cone expression from the Mathworld site for an example and rearrange it for z
ft = fittype('sqrt((x^2 + y^2)/c^2) + z_0', ...
'independent', {'x', 'y'}, 'coeff', {'c', 'z_0'});
If it's a simple form matlab can work out which are the variables and which the coefficients but with something more complex like this you'd want to give it a hand.
The 'fitoptions' object holds the configuration for the methods: depending on your dataset you might have to spend some time specifying upper and lower bounds, starting values etc.
fo = fitoptions('Upper', [one, for, each, of, your, coeffs, in, the, order, they, appear, in, the, string], ...
'Lower', [...], `StartPoint', [...]);
then get the output
[fitted, gof] = fit([xvals, yvals], zvals, ft, fo);
Caveat: I've done this plenty with 2D datasets and the docs state it works for three but I haven't done that myself so the above code might not work, check the docs to make sure you've got your syntax right.
It might be worth starting with a simple fit expression, something linear, so that you can get your code working. Then swap the expression out for the cone and play around until you get something that looks like what you're expecting.
After you've got your fit a good trick is that you can use the eval function on the string expression you used in your fit to evaluate the contents of the string as if it was a matlab expression. This means you need to have workspace variables with the same names as the variables and coefficients in your string expression.
I need to fit data e.g. x, y, CI (where CI is confidence index of y) in Matlab.
Now, I use this code:
pf = polyfit(x, y, 2);
x1 = min(x):.1:max(x);
y1 = polyval(pf, x1);
figure
hold on
errorbar(x, y, CI, 'ko');
plot(x1, y1, 'k');
hold off
Of course, the fit comes out of some errors bars, and it's correct.
I would like obtain a fit curve closer to the points with a low confidence index, and discard the points with a high confidence index.
Thank you and bye,
Giacomo
What you are looking for are Weighted Least Squares. You can compute them with the function lscov. There is a nice example in its help page, but I'll try to make it clearer.
Let us construct a simple parabola, with a corrupted point
x = (0:0.1:1)';
y = 0.5*x.^2;
y(5) = 3*y(5);
and give some weights
w = ones(size(y));
w(5) = 0.1;
Next build the Vandermonde matrix (see here for the code) and solve the system
%// V = [x.^2 x ones(size(x))];
V = bsxfun(#power, x, 2:-1:0);
coeff = lscov(V, y, w);
The estimated coefficients, with and without the weights, are
x^2 x 1
with weights [0.4797 0.0186 -0.0004]
no weights [0.3322 0.1533 -0.0034]
Note that in your case w will have to be inverted.
If you don't like to build the Vandermonde matrix, and you have a license for the Curve Fitting Toolbox, you can use the following code
ft = fittype('poly2');
opts = fitoptions('Method', 'LinearLeastSquares');
opts.Weights = w;
fitresult = fit(x, y, ft, opts);
and you'll obtain the same result.
I have three data points through which I have to fit a straight line of the form Y=m*X+C. I want the line to have pre-determined slope 'm' but the constant'C' can change to get the least error while fitting using matlab. Can someone help me out?
Just do the math:
C= mean(Y)-m*mean(X)
assuming Y is the vector containing the y coordinates, and X the x coordinates.
Reference: http://hotmath.com/hotmath_help/topics/line-of-best-fit.html
If you opt to use the Curve Fitting Toolbox the solution is as follows.
To start generate some data
m = 3;
x = (1:10).';
y = m*x + 2 + randn(size(x));
then select the model to fit and set the bounds for its coefficients
ft = fittype('poly1');
opts = fitoptions('Method', 'LinearLeastSquares');
opts.Lower = [m -Inf];
opts.Upper = [m Inf];
finally call the fitting routine
[fitresult, gof] = fit(x, y, ft, opts);
The intercept is stored in fitresult.p2.
I have been trying to fit a parabola to parts of data where y is positive. I am told, that P1(x1,y1) is the first data point, Pg(xg,yg) is the last, and that the top point is at x=(x1+xg)/2. I have written the following:
x=data(:,1);
y=data(:,2);
filter=y>0;
xp=x(filter);
yp=y(filter);
P1=[xp(1) yp(1)];
Pg=[xp(end) yp(end)];
xT=(xp(1)+xp(end))/2;
A=[1 xp(1) power(xp(1),2) %as the equation is y = a0 + x*a1 + x^2 * a2
1 xp(end) power(xp(end),2)
0 1 2*xT]; % as the top point satisfies dy/dx = a1 + 2*x*a2 = 0
b=[yg(1) yg(end) 0]'; %'
p=A\b;
x_fit=[xp(1):0.1:xp(end)];
y_fit=power(x_fit,2).*p(3)+x_fit.*p(2)+p(1);
figure
plot(xp,yp,'*')
hold on
plot(x_fit,y_fit,'r')
And then I get this parabola which is completely wrong. It doesn't fit the data at all! Can someone please tell me what's wrong with the code?
My parabola
Well, I think the primary problem is some mistake in your calculation. I think you should use three points on the parabola to obtain a system of linear equations. There is no need to calculate the derivative of your function as you do with
dy/dx = a1 + 2*x*a2 = 0
Instead of a point on the derivative you choose another point in your scatter plot, e.g. the maximum: PT = [xp_max yp_max]; and use it for your matrix A and b.
The equation dy/dx = a1 + 2*x*a2 = 0 does not fulfill the basic scheme of your system of linear equations: a0 + a1*x + a2*x^2 = y;
By the way: If you don't have to calculate your parabola necessarily in this way, you can maybe have a look at the Matlab/Octave-function polyfit() which calculates the least squares solution for your problem. This would result in a simple implementation:
p = polyfit(x, y, 2);
y2 = polyval(p, x);
figure(); plot(x, y, '*'); hold on;
plot(x, y2, 'or');