Troubleshooting Best Fit Line and Distance Calculation in MATLAB

In summary: The line is used to calculate the equation of the best fit line by finding the slope (m) and intercept (y1) using the first and last data points (x1 and y1). The resulting equation represents the line that best fits the given data points.
  • #1
Wrichik Basu
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I have some experimental data, and I would like to plot a graph in MATLAB and also find the best fit line. I also want to find the distance between each point and the best fit line, and if the distance is greater than 1 unit, an error will be shown.

For this, I have written a function. Here it is:

Matlab:
function [] = plot2d(X,Y);
 
    %The aim of this function is to read
    % X and Y from the user, and plot the points.
    % The best fit line shall be shown.
 
    hold off;
 
    %this plots the given data points
    plot(X,Y,'.k', 'MarkerSize', 20);
 
    hold on;
 
    %here, we are calculating the best fit line
    p = polyfit(X,Y,1);
    f = polyval(p, X);
    plot(X, f, '-r');
 
    %necessary plot settings
    zoom on;
    set(gca, 'xminorgrid', 'on', 'yminorgrid', 'on');
 
    syms x1 y1 x2 y2 x y;
 
    %finding the equation of the best fit line
    x1 = X(1);
    x2 = X(end);
    y1 = f(1);
    y2 = f(end);
    m = (y2 - y1) / (x2 - x1);
 
    eqn = y - y1 == m * (x - x1);
 
    %simplifying the equation
    eqn = simplify(eqn);
 
    %writing the equation in Ax + By + C = 0 form and removing the rhs to make it an algebraic expression
    eqn2 = lhs(eqn) - rhs(eqn);
 
    coefx = coeffs(eqn2,x);
 
    %coefficient of x
    cx = coefx(end);
 
    coefy = coeffs(eqn2,y);
 
    %coefficient of y
    cy = coefy(end);
 
    %finding number of data points given
    sz = numel(X);
 
    for i = 1:1:sz
     
        x = X(i);
        y = f(i);
     
        %evaluating the expression by putting in values
        p = subs(eqn2);
     
        %calculating final distance
        dist = p / sqrt(cx^2 + cy^2);
     
        if dist > 1
            disp("Error for point (" + x + "," + y + ")");
        end
         
             
    end
 
    hold off;
 
end
Matlab gives me a good graph:

Figure 2018-08-01 20_19_01.png


The data is:

Screenshot_20180801-202541.png


But all distances of points are coming to be 0. Because the subs(eqn2) is yielding a 0. From the graph, it is evident that all points do not lie on the line. But I'm still getting a 0. I know this because I printed p from the function. This is what I got:

Screenshot_20180801-202755.png


Screenshot_20180801-202801.png


Any help is appreciated. Where am I going wrong?

Any ideas?

I use MATLAB mobile, version R2018a.
 

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Last edited:
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  • #2
My suggestion is to make some test data and a small program that illustrates the problem. Focus on the subs and at the Matlab api description of subs to see if you’re using it correctly.

Usually there are examples of the function that you can try too.
 
  • #3
jedishrfu said:
My suggestion is to make some test data and a small program that illustrates the problem. Focus on the subs and at the Matlab api description of subs to see if you’re using it correctly.

Usually there are examples of the function that you can try too.
The problem is that, I tried by putting values in eqn2 myself. I did this by printing eqn2, and then manually putting in values. The answers are truly coming to 0. But from the graph, the distance cannot be 0. That is the bottleneck. The points do not lie on the best fit line, but the equation of the best fit line shows that all the points lie on it. I think the problem is coming from the polyfit or polyval functions.
 
  • #4
What is this line used for?

Matlab:
    eqn = y - y1 == m * (x - x1);
 
  • #5
jedishrfu said:
What is this line used for?

Matlab:
    eqn = y - y1 == m * (x - x1);
That line finds the equation of the best fit line using the formula $$y - y_1 = m(x-x_1)$$
 
  • #6
Is your "distance" the Y error or is it the perpondicular distance to the line? If it is the former, there are easier ways to get the Y error. If it is the latter, consider principle components.
 
  • #7
FactChecker said:
Is your "distance" the Y error or is it the perpondicular distance to the line? If it is the former, there are easier ways to get the Y error. If it is the latter, consider principle components.
It is the perpendicular distance of each point from the best fit line.

Can you elaborate a bit on the principle components? For computation of distance, I have used the formula $$\dfrac{Ax+By+C}{\sqrt{A^2+B^2}}$$ where ##x## and ##y## are from the data.
 
  • #8
The MATLAB function polyfit is not doing what you want to do. It is minimizing the sum-squares of the y errors -- the vertical distance to the line.
To minimize the sum-squared perpendicular distances, the first principle component is what you want. From https://en.wikipedia.org/wiki/Principal_component_analysis#Further_considerations :

"Given a set of points in Euclidean space, the first principal component corresponds to a line that passes through the multidimensional mean and minimizes the sum of squares of the distances of the points from the line."
 
  • #9
FactChecker said:
The MATLAB function polyfit is not doing what you want to do. It is minimizing the sum-squares of the y errors -- the vertical distance to the line.
To minimize the sum-squared perpendicular distances, the first principle component is what you want. From https://en.wikipedia.org/wiki/Principal_component_analysis#Further_considerations :

"Given a set of points in Euclidean space, the first principal component corresponds to a line that passes through the multidimensional mean and minimizes the sum of squares of the distances of the points from the line."
Two questions:

1. What is the difference between vertical and perpendicular distances?
2. Is there any known function in MATLAB to make a graph from first PCA?
 
  • #11
FactChecker said:
1) See, for instance, https://benediktehinger.de/blog/sci...sion-lines-and-the-first-principal-component/
2) In MATLAB, use [COEFF,SCORE] = princomp(X). The first row of COEFF will give you the first principle component. I think that you will have to plot it yourself.
Thanks for the function. But princomp has been removed; it has now become pca in R2018a.

However, I will not be able to use it, because it requires me to buy the Statistics toolbox. In fact, my MATLAB is not licensed, and that's why I can use it only on mobile. :frown:
 
  • #12
A good, free statistics package is R. It is well respected and documented. In fact, it ranks fairly high on the list of most often used programming languages. But naturally there is a learning curve. If you will do a lot of statistics, it may be worth getting and learning.

Apparently, PCA can be done in Excel. I am not familiar with that. See https://www.quora.com/Which-is-the-...xcel-to-perform-principal-components-analysis
 
Last edited:
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  • #13
I am indeed heartbroken with matlab. Just now, I found that they have stopped supporting 32-bit pc. The last supporting version was R2015b, which is no longer sold now. Perhaps I am done with MATLAB for the moment, and I should shift to some other language.
 
  • #14
Or perhaps you should upgrade your computer.
 
  • #15
Image Analyst said:
Or perhaps you should upgrade your computer.
That's costly. What I have decided is, I will buy the software and additional toolboxes, but keep on using it in the mobile app. I don't need simulink at this moment, so I think I'm fine with whatever I can do on the app.
 

FAQ: Troubleshooting Best Fit Line and Distance Calculation in MATLAB

1. What is a best fit line for data points?

A best fit line for data points is a straight line that represents the trend or pattern in a set of data. It is used to summarize and visualize the relationship between two variables.

2. How is a best fit line calculated?

A best fit line is calculated using a statistical method called linear regression. This involves finding the line that minimizes the sum of the squared distances between the line and each data point.

3. What does the slope of a best fit line represent?

The slope of a best fit line represents the rate of change between the two variables. It indicates how much the dependent variable changes for every one unit change in the independent variable.

4. What does the y-intercept of a best fit line represent?

The y-intercept of a best fit line represents the predicted value of the dependent variable when the independent variable is equal to 0. It can also be interpreted as the initial value of the dependent variable.

5. How do you interpret the correlation coefficient of a best fit line?

The correlation coefficient of a best fit line is a measure of the strength and direction of the relationship between the two variables. It ranges from -1 to 1, with values closer to 1 indicating a strong positive correlation, values closer to -1 indicating a strong negative correlation, and values close to 0 indicating no correlation.

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