fitlm() with 0 intercept function returns different values for R2 betwen MATLAB 2020a and 2020b?

I am trying to find the coefficient of determination. I ran the same fitlm function in version 2020a and 2020b, but calling the 'Rsquared.Ordinary' variable gives me drastically different results.

2020a, R2 = 0.5432

2020b, R2 = -8.0637

Why are they different?

Here is the code I am using:

data =[1.8286, 2.0125; 2.5571, 2.3886; 2.0747, 2.3476; 1.6216, 2.0296;
    4.6010, 2.4861;  2.6089, 2.4834; 2.0093,  2.5387;  2.4757, 2.5660;
    2.5475, 2.6642; 2.4854, 2.6650];

mdl = fitlm(data(:,1), data(:,2),'Intercept',false);
R2 = mdl.Rsquared.Ordinary;

Answer 1

[the cyclist]

Interesting. The 2020b result seems to be correct, based on the formula using sums of squares. (See, e.g. the definition on wikipedia page.)

% Original data
data =[1.8286, 2.0125; 2.5571, 2.3886; 2.0747, 2.3476; 1.6216, 2.0296;
    4.6010, 2.4861;  2.6089, 2.4834; 2.0093,  2.5387;  2.4757, 2.5660;
    2.5475, 2.6642; 2.4854, 2.6650];

% Define x,y for convenience
x = data(:,1);
y = data(:,2);

mdl = fitlm(x,y,'Intercept',false);
R2 = mdl.Rsquared.Ordinary

R2 = -8.0618

% Predicted y value, at x data
y_pred = predict(mdl,data(:,1));

% Calculate R^2 via sums of squares of residuals and total
R_squared_via_sum_of_squares = 1 - sum((y - y_pred).^2)/sum((y - mean(y)).^2)

R_squared_via_sum_of_squares = -8.0618

Based on some playing around, but not quite well formulated enough to post here, I think 2020a might have been using a formula based on the correlation of y data vs. y predicted that is only valid when an intercept term is estimated, and therefore not accurate in your case.