I have three matrices x, y, z which are plotted via scatter3 in matlab. However I also need vertical lines dropping from every point in the graph for better visualization.
Using matlab 2017a, implemented 3D scatter plot in matlab.
enter code here
clc;
figure
x = [0,0,0,0,0,10,10,10,10,10];
y = [0,10,20,30,40,-10,0,10,20,30];
z = [46,52,51,59,53,85,56,87,86,88];
scatter3(x, y, z, 30, 'filled')
You could also use the built in function stem, which is doing exactly that.
The minor trick is that you cannot pass the z coordinates in the shorthand form stem(x,y,z), but the graphic object still accept z data, you just have to send them as additional parameter.
The nice part of it is you do not need a loop ;-)
x = [0,0,0,0,0,10,10,10,10,10];
y = [0,10,20,30,40,-10,0,10,20,30];
z = [46,52,51,59,53,85,56,87,86,88];
hp = stem(x,y,'filled','ZData',z) ;
Or as Gnovice nicely pointed out, even easier to use the stem3 function which accept z data directly:
hp = stem3(x,y,z,'filled') ;
Both example above will produce:
As #SardarUsama pointed out, plot3 should do the trick. Code could be more compact but kept it as is for clarity.
% MATLAB R2017a
x = [0,0,0,0,0,10,10,10,10,10];
y = [0,10,20,30,40,-10,0,10,20,30];
z = [46,52,51,59,53,85,56,87,86,88];
figure
scatter3(x, y, z, 30, 'filled') % scatter plot (3D)
zRng = zlim;
hold on
for k = 1:length(x)
xL = [x(k) x(k)];
yL = [y(k) y(k)];
zL = [zRng(1) z(k)];
plot3(xL,yL,zL,'r-') % plot vertical line (3D)
end
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');
Hi I have a set of data A with each element corresponding to an x and y combination. When I plot this dat using mesh I get a graph with many spikes on it. This is not unnexpected but I would like a way to smooth these out to get a smooth surface.
I've tried to use the smooth3 command but cannot figure out how to make the suitable input.
Any help would be appreciated. Thanks
This is how my data is generated.
function v = f(x,y) % Not actual function
return x*rand()+y*rand()
end
x = 0.05:0.01:0.95;
y = 0.05:0.01:0.95;
o = zeros(length(x),length(y));
A = zeros(length(x), length(y));
for k = 1:5
for i = 1:length(x)
for j = 1:length(y)
o(i,j) = f([x(i), y(j)]);
end
end
A= A+o;
end
A = A/5;
This is what produces the plot.
[X,Y] = meshgrid(x);
mesh(A)
my be you can try a convolution of your variable A with a filter(the following is an example of a Gaussian filter).
C = conv2(A,fspecial('gaussian', hsize, sigma));
check conv2 and fspecial in matlab help
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 :)