Correlation regression lines between two parameters

Hello dear Researchers,

I have a query need your expertise to resolve.

I want to add correlation regression for two paramters for comparison purpose.

I have attached sub-plots which are scatter plots among two orientations.

Now purpose is to add correlation regression line i.e., one plot I added from a reference.

Answer 1

[yanqi liu]

sir, may be upload some data, use lsqcurvefit、polyfit and so on to compute the coef

please check the follow code to get some information

for more information,you can read the book 《计算机视觉与深度学习实战》，Thank you！

clc; clear all; close all;
xdata = linspace(0,3);
ydata = 1.3*xdata + 0.05*randn(size(xdata));
lb = [];
ub = [];
fun = @(x,xdata)x(1)*xdata+x(2);
x0 = [0,0];
x = lsqcurvefit(fun,x0,xdata,ydata,lb,ub)

Local minimum found.

Optimization completed because the size of the gradient is less than
the value of the optimality tolerance.

x = 1×2

1.3048 -0.0023

plot(xdata,ydata,'ko',xdata,fun(x,xdata),'r-','LineWidth',2)
legend('Data','Fitted exponential')
title('Data and Fitted Curve')

[yanqi liu]

yes，sir，may be use the color list

clc; clear all; close all;
xdata = linspace(0,3,2e2);
ydata = 1.3*xdata + 0.05*randn(size(xdata));
lb = [];
ub = [];
fun = @(x,xdata)x(1)*xdata+x(2);
x0 = [0,0];
x = lsqcurvefit(fun,x0,xdata,ydata,lb,ub)

Local minimum found.

Optimization completed because the size of the gradient is less than
the value of the optimality tolerance.

x = 1×2

1.2999 -0.0017

%scatter(xdata,ydata,10,randn(size(xdata)));
scatter(xdata,ydata,10,ydata);
hold on
plot(xdata,fun(x,xdata),'r-','LineWidth',2)
legend('Data','Fitted exponential')
title('Data and Fitted Curve')

[yanqi liu]

yes，sir，you are welcome，加油！

[Adam Danz]

Using lsqcurvefit requires setting an initial guess which is not necessary with linear fits. This function is better for nonlinear data.

[Amjad Iqbal]

Point to be noted it seems non linear curve fitting. I am working with your methods, till I succeed with any outcomes. Yeah, you are right as curve line appears isolatedly

[Adam Danz]

The objective function in yanqi Liu's answer is linear. Your scatter plots show a linear relationship. And the regression line you plotted his linear. I don't see anything that's nonlinear.

[Amjad Iqbal]

Like in the reference plot in my question, regression line is not isolated from scatter. In this case, line is seems isolated. Although relationship is linear.

[Adam Danz]

What does it mean for a line to be isolated? Are you describing the fit of the line?

[Adam Danz]

Are you describing the color of the line? Line properties can be changed in any method you use.

In the first image of your previous comment, the legend was moved and now I see some points in the lower left corner where the legend was that do not follow the linear trend (around (3,1)). But they don't appear to be affecting the linear fit too much if the blue line is the linear fit.

[Amjad Iqbal]

I understand the fact linear fit. It seems in a good agreement by using both methods recommended by you and yanqi liu. Many many thanks once agian all, also @Image Analyst for your expertised suggestions. It really helps me. Final result is now up to the mark.

[Adam Danz]

Yes, lsqcurvefit will provide the same results as polyfit or fitlm but the latter two are designed for linear models and do not require making initial guesses to the parameter values. I'm not trying to convince anyone to change their approach (or their selected answer). I'm arguing that lsqcurvefit is not the best tool for linear regression.

polyfit is much more efficient than lsqcurvefit (95x faster):

x = rand(1,100);
y = 4.8*x+2.1 +rand(1,100);

n = 5000;

tic
for i = 1:n
    opts = struct('Display','off');
    fun = @(x,xdata)x(1)*xdata+x(2);
    p = lsqcurvefit(fun, [0,0],x,y,[],[],opts);
end
T1 = toc % time in seconds

T1 = 13.2634

p

p = 1×2

4.9170 2.4408

tic
for i = 1:n
    p2 = polyfit(x,y,1);
end
T2 = toc % time in seconds

T2 = 0.1386

p2 % same results

p2 = 1×2

4.9170 2.4408

% Difference in time
T1/T2

ans = 95.7249

Answer 2

[Adam Danz]

Common methods of adding a simple linear regression line

1. Use lsline which will add a regression line for each set of data in the plot.

2. Use polyfit (degree 1) & refline to compute the regression coefficients and plot the line.

3. Use fitlm & refline to compute the regression coefficients and plot the line.

Examples of #1 & #2

https://www.mathworks.com/matlabcentral/answers/421072-add-trend-line-to-a-scatter-plot#answer_338738

Example of #3

https://www.mathworks.com/matlabcentral/answers/277495-how-to-fit-linear-regression-models-to-a-scatter-graph#answer_492643

If you want to compute correlation, see corr or corrcoeff.

[Image Analyst]

Just to build on Adam's #3, this is how you'd do it:

% First get your model for fitted y values.
% Determine and say how well we did with our predictions, numerically, using several metrics like RMSE and MAE.
% Fit a linear model between predicted and true so we can get the R squared.
mdl = fitlm(actualy, fittedy)
rSquared = mdl.Rsquared.Ordinary;
rmse = mdl.RMSE;
mdlMSE = mdl.MSE;
residuals = abs(y - x);
% Find the mean and median absolute deviations of the elements in X.
meandev = mad(residuals,0,'all');
mediandev = mad(residuals,1,'all');
aveResidual = mean(residuals, 'omitnan');
maxResidual = max(residuals);
fprintf('The RMSE value is %.5f.\n', rmse);
fprintf('The  R^2 value is %.5f.\n', rSquared);
fprintf('The  MSE value is %.5f.\n', mdlMSE);
fprintf('The fitted y is different from the actual y by an average of %.2f units.\n', aveResidual);
fprintf('The fitted y is different from the actual y by an average of %.2f units.\n', meandev);
fprintf('The fitted y is different from the actual y by a median of %.2f units.\n', mediandev);
fprintf('The max residual from the average human grade is %.2f PSU.\n', maxResidual);

[Amjad Iqbal]

Thanks a lot dear Adam Danz and Image Analyst

That's seems a proper way to add several numerical comparisons. It helps a lot.