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I'm trying to find the line which best fits to the data. I use the following code below but now I want to have the data placed into an array sorted so it has the data which is closest to the line first how can I do this? Also is polyfit the correct function to use for this?
x=[1,2,2.5,4,5];
y=[1,-1,-.9,-2,1.5];
n=1;
p = polyfit(x,y,n)
f = polyval(p,x);
plot(x,y,'o',x,f,'-')
PS: I'm using Octave 4.0 which is similar to Matlab
You can first compute the error between the real value y and the predicted value f
err = abs(y-f);
Then sort the error vector
[val, idx] = sort(err);
And use the sorted indexes to have your y values sorted
y2 = y(idx);
Now y2 has the same values as y but the ones closer to the fitting value first.
Do the same for x to compute x2 so you have a correspondence between x2 and y2
x2 = x(idx);
Sembei Norimaki did a good job of explaining your primary question, so I will look at your secondary question = is polyfit the right function?
The best fit line is defined as the line that has a mean error of zero.
If it must be a "line" we could use polyfit, which will fit a polynomial. Of course, a "line" can be defined as first degree polynomial, but first degree polynomials have some properties that make it easy to deal with. The first order polynomial (or linear) equation you are looking for should come in this form:
y = mx + b
where y is your dependent variable and X is your independent variable. So the challenge is this: find the m and b such that the modeled y is as close to the actual y as possible. As it turns out, the error associated with a linear fit is convex, meaning it has one minimum value. In order to calculate this minimum value, it is simplest to combine the bias and the x vectors as follows:
Xcombined = [x.' ones(length(x),1)];
then utilized the normal equation, derived from the minimization of error
beta = inv(Xcombined.'*Xcombined)*(Xcombined.')*(y.')
great, now our line is defined as Y = Xcombined*beta. to draw a line, simply sample from some range of x and add the b term
Xplot = [[0:.1:5].' ones(length([0:.1:5].'),1)];
Yplot = Xplot*beta;
plot(Xplot, Yplot);
So why does polyfit work so poorly? well, I cant say for sure, but my hypothesis is that you need to transpose your x and y matrixies. I would guess that that would give you a much more reasonable line.
x = x.';
y = y.';
then try
p = polyfit(x,y,n)
I hope this helps. A wise man once told me (and as I learn every day), don't trust an algorithm you do not understand!
Here's some test code that may help someone else dealing with linear regression and least squares
%https://youtu.be/m8FDX1nALSE matlab code
%https://youtu.be/1C3olrs1CUw good video to work out by hand if you want to test
function [a0 a1] = rtlinreg(x,y)
x=x(:);
y=y(:);
n=length(x);
a1 = (n*sum(x.*y) - sum(x)*sum(y))/(n*sum(x.^2) - (sum(x))^2); %a1 this is the slope of linear model
a0 = mean(y) - a1*mean(x); %a0 is the y-intercept
end
x=[65,65,62,67,69,65,61,67]'
y=[105,125,110,120,140,135,95,130]'
[a0 a1] = rtlinreg(x,y); %a1 is the slope of linear model, a0 is the y-intercept
x_model =min(x):.001:max(x);
y_model = a0 + a1.*x_model; %y=-186.47 +4.70x
plot(x,y,'x',x_model,y_model)
I want to plot relations like y^2=x^2(x+3) in MATLAB without using ezplot or doing algebra to find each branch of the function.
Does anyone know how I can do this? I usually create a linspace and then create a function over the linspace. For example
x=linspace(-pi,pi,1001);
f=sin(x);
plot(x,f)
Can I do something similar for the relation I have provided?
What you could do is use solve and allow MATLAB's symbolic solver to symbolically solve for an expression of y in terms of x. Once you do this, you can use subs to substitute values of x into the expression found from solve and plot all of these together. Bear in mind that you will need to cast the result of subs with double because you want the numerical result of the substitution. Not doing this will still leave the answer in MATLAB's symbolic format, and it is incompatible for use when you want to plot the final points on your graph.
Also, what you'll need to do is that given equations like what you have posted above, you may have to loop over each solution, substitute your values of x into each, then add them to the plot.
Something like the following. Here, you also have control over the domain as you have desired:
syms x y;
eqn = solve('y^2 == x^2*(x+3)', 'y'); %// Solve for y, as an expression of x
xval = linspace(-1, 1, 1000);
%// Spawn a blank figure and remember stuff as we throw it in
figure;
hold on;
%// For as many solutions as we have...
for idx = 1 : numel(eqn)
%// Substitute our values of x into each solution
yval = double(subs(eqn(idx), xval));
%// Plot the points
plot(xval, yval);
end
%// Add a grid
grid;
Take special care of how I used solve. I specified y because I want to solve for y, which will give me an expression in terms of x. x is our independent variable, and so this is important. I then specify a grid of x points from -1 to 1 - exactly 1000 points actually. I spawn a blank figure, then for as many solutions to the equation that we have, we determine the output y values for each solution we have given the x values that I made earlier. I then plot these on a graph of these points. Note that I used hold on to add more points with each invocation to plot. If I didn't do this, the figure would refresh itself and only remember the most recent call to plot. You want to put all of the points on here generated from all of the solution. For some neatness, I threw a grid in.
This is what I get:
Ok I was about to write my answer and I just saw that #rayryeng proposed a similar idea (Good job Ray!) but here it goes. The idea is also to use solve to get an expression for y, then convert the symbolic function to an anonymous function and then plot it. The code is general for any number of solutions you get from solve:
clear
clc
close all
syms x y
FunXY = y^2 == x^2*(x+3);
%//Use solve to solve for y.
Y = solve(FunXY,y);
%// Create anonymous functions, stored in a cell array.
NumSol = numel(Y); %// Number of solutions.
G = cell(1,NumSol);
for k = 1:NumSol
G{k} = matlabFunction(Y(k))
end
%// Plot the functions...
figure
hold on
for PlotCounter = 1:NumSol
fplot(G{PlotCounter},[-pi,pi])
end
hold off
The result is the following:
n = 1000;
[x y] = meshgrid(linspace(-3,3,n),linspace(-3,3,n));
z = nan(n,n);
z = (y .^ 2 <= x .^2 .* (x + 3) + .1);
z = z & (y .^ 2 >= x .^2 .* (x + 3) - .1);
contour(x,y,z)
It's probably not what you want, but I it's pretty cool!
my question is quite trivial, but I'm looking for the vectorized form of it.
My code is:
HubHt = 110; % Hub Height
GridWidth = 150; % Grid length along Y axis
GridHeight = 150; % Grid length along Z axis
RotorDiameter = min(GridWidth,GridHeight); % Turbine Diameter
Ny = 31;
Nz = 45;
%% GRID DEFINITION
dy = GridWidth/(Ny-1);
dz = GridHeight/(Nz-1);
if isequal(mod(Ny,2),0)
iky = [(-Ny/2:-1) (1:Ny/2)];
else
iky = -floor(Ny/2):ceil(Ny/2-1);
end
if isequal(mod(Nz,2),0)
ikz = [(-Nz/2:-1) (1:Nz/2)];
else
ikz = -floor(Nz/2):ceil(Nz/2-1);
end
[Y Z] = ndgrid(iky*dy,ikz*dz + HubHt);
EDIT
Currently I am using this solution, which has reasonable performances:
coord(:,1) = reshape(Y,[numel(Y),1]);
coord(:,2) = reshape(Z,[numel(Z),1]);
dist_y = bsxfun(#minus,coord(:,1),coord(:,1)');
dist_z = bsxfun(#minus,coord(:,2),coord(:,2)');
dist = sqrt(dist_y.^2 + dist_z.^2);
I disagree with Dan and Tal.
I believe you should use pdist rather than pdist2.
D = pdist( [Y(:) Z(:)] ); % a compact form
D = squareform( D ); % square m*n x m*n distances.
I agree with Tal Darom, pdist2 is exactly the function you need. It finds the distance for each pair of coordinates specified in two vectors and NOT the distance between two matrices.
So I'm pretty sure in your case you want this:
pdist2([Y(:), Z(:)], [Y(:), Z(:)])
The matrix [Y(:), Z(:)] is a list of every possible coordinate combination over the 2D space defined by Y-Z. If you want a matrix containing the distance from each point to each other point then you must call pdist2 on this matrix with itself. The result is a 2D matrix with dimensions numel(Y) x numel(Y) and although you haven't defined it I'm pretty sure that both Y and Z are n*m matrices meaning numel(Y) == n*m
EDIT:
A more correct solution suggested by #Shai is just to use pdist since we are comparing points within the same matrix:
pdist([Y(:), Z(:)])
You can use the matlab function pdist2 (I think it is in the statistics toolbox) or you can search online for open source good implementations of this function.
Also,
look at this unswer: pdist2 equivalent in MATLAB version 7
I'm going to start off by stating that, yes, this is homework (my first homework question on stackoverflow!). But I don't want you to solve it for me, I just want some guidance!
The equation in question is this:
I'm told to take N = 50, phi1 = 300, phi2 = 400, 0<=x<=1, and 0<=y<=1, and to let x and y be vectors of 100 equally spaced points, including the end points.
So the first thing I did was set those variables, and used x = linspace(0,1) and y = linspace(0,1) to make the correct vectors.
The question is Write a MATLAB script file called potential.m which calculates phi(x,y) and makes a filled contour plot versus x and y using the built-in function contourf (see the help command in MATLAB for examples). Make sure the figure is labeled properly. (Hint: the top and bottom portions of your domain should be hotter at about 400 degrees versus the left and right sides which should be at 300 degrees).
However, previously, I've calculated phi using either x or y as a constant. How am I supposed to calculate it where both are variables? Do I hold x steady, while running through every number in the vector of y, assigning that to a matrix, incrementing x to the next number in its vector after running through every value of y again and again? And then doing the same process, but slowly incrementing y instead?
If so, I've been using a loop that increments to the next row every time it loops through all 100 values. If I did it that way, I would end up with a massive matrix that has 200 rows and 100 columns. How would I use that in the linspace function?
If that's correct, this is how I'm finding my matrix:
clear
clc
format compact
x = linspace(0,1);
y = linspace(0,1);
N = 50;
phi1 = 300;
phi2 = 400;
phi = 0;
sum = 0;
for j = 1:100
for i = 1:100
for n = 1:N
sum = sum + ((2/(n*pi))*(((phi2-phi1)*(cos(n*pi)-1))/((exp(n*pi))-(exp(-n*pi))))*((1-(exp(-n*pi)))*(exp(n*pi*y(i)))+((exp(n*pi))-1)*(exp(-n*pi*y(i))))*sin(n*pi*x(j)));
end
phi(j,i) = phi1 - sum;
end
end
for j = 1:100
for i = 1:100
for n = 1:N
sum = sum + ((2/(n*pi))*(((phi2-phi1)*(cos(n*pi)-1))/((exp(n*pi))-(exp(-n*pi))))*((1-(exp(-n*pi)))*(exp(n*pi*y(j)))+((exp(n*pi))-1)*(exp(-n*pi*y(j))))*sin(n*pi*x(i)));
end
phi(j+100,i) = phi1 - sum;
end
end
This is the definition of contourf. I think I have to use contourf(X,Y,Z):
contourf(X,Y,Z), contourf(X,Y,Z,n), and contourf(X,Y,Z,v) draw filled contour plots of Z using X and Y to determine the x- and y-axis limits. When X and Y are matrices, they must be the same size as Z and must be monotonically increasing.
Here is the new code:
N = 50;
phi1 = 300;
phi2 = 400;
[x, y, n] = meshgrid(linspace(0,1),linspace(0,1),1:N)
f = phi1-((2./(n.*pi)).*(((phi2-phi1).*(cos(n.*pi)-1))./((exp(n.*pi))-(exp(-n.*pi)))).*((1-(exp(-1.*n.*pi))).*(exp(n.*pi.*y))+((exp(n.*pi))-1).*(exp(-1.*n.*pi.*y))).*sin(n.*pi.*x));
g = sum(f,3);
[x1,y1] = meshgrid(linspace(0,1),linspace(0,1));
contourf(x1,y1,g)
Vectorize the code. For example you can write f(x,y,n) with:
[x y n] = meshgrid(-1:0.1:1,-1:0.1:1,1:10);
f=exp(x.^2-y.^2).*n ;
f is a 3D matrix now just sum over the right dimension...
g=sum(f,3);
in order to use contourf, we'll take only the 2D part of x,y:
[x1 y1] = meshgrid(-1:0.1:1,-1:0.1:1);
contourf(x1,y1,g)
The reason your code takes so long to calculate the phi matrix is that you didn't pre-allocate the array. The error about size happens because phi is not 100x100. But instead of fixing those things, there's an even better way...
MATLAB is a MATrix LABoratory so this type of equation is pretty easy to compute using matrix operations. Hints:
Instead of looping over the values, rows, or columns of x and y, construct matrices to represent all the possible input combinations. Check out meshgrid for this.
You're still going to need a loop to sum over n = 1:N. But for each value of n, you can evaluate your equation for all x's and y's at once (using the matrices from hint 1). The key to making this work is using element-by-element operators, such as .* and ./.
Using matrix operations like this is The Matlab Way. Learn it and love it. (And get frustrated when using most other languages that don't have them.)
Good luck with your homework!
I need to draw an elliptic curve over the finite field F17(in other words, I want to draw some specific dots on the curve), but somehow I don't get it right.
The curve is defined by the equation:
y^2 = x^3 +x + 1 (mod 17)
I tried the way below, but it can't work.
for x = 0:16, plot(x, mod(sqrt(x^3+x+1), 16),'r')', end
Can someone help ?
[Update]
According to Nathan and Bill's suggestions, here is a slightly modified version.
x = 0:18
plot(mod(x,16), mod(sqrt(x.^3+x+1), 16),'ro')
However, I feel the figure is WRONG , e.g.,y is not an integer when x=4 .
You have to test all points that fulfill the equation y^2 = x^3 +x + 1 (mod 17). Since it is a finite field, you cannot simply take the square root on the right side.
This is how I would go about it:
a=0:16 %all points of your finite field
left_side = mod(a.^2,17) %left side of the equation
right_side = mod(a.^3+a+1,17) %right side of the equation
points = [];
%testing if left and right side are the same
%(you could probably do something nicer here)
for i = 1:length(right_side)
I = find(left_side == right_side(i));
for j=1:length(I)
points = [points;a(i),a(I(j))];
end
end
plot(points(:,1),points(:,2),'ro')
set(gca,'XTick',0:1:16)
set(gca,'YTick',0:1:16)
grid on;
Matlab works with vectors natively.
your syntax was close, but needs to be vectorized:
x = 0:16
plot(x, mod(sqrt(x.^3+x+1), 16),'r')
Note the . in x.^3. This tells Matlab to cube each element of x individually, as opposed to raising the vector x to the 3rd power, which doesn't mean anything.
You can use this code if you want to plot on Real numbers:
syms x y;
v=y^2-x^3-x-1;
ezplot(v, [-1,3,-5,5]);
But, for plot in modulo, at first you can write below code;
X=[]; for x=[0:16], z=[x; mod(x^3+x+1,17)]; X=[X, z]; end, X,
Then, you can plot X with a coordinate matrix.