This would be easier with your data. Lacking them, I created my own.
Try this:
x = 0:32;
y = [-0.5*(0:11)+10+randi([-1 1], 1, 12), 4.5*(12:14)-50+randi([-1 1], 1, 3), 0*(15:28)+13+randi([-1 1], 1, 14), -4*(29:32)+129+randi([-1 1], 1, 4)];
Lv = ischange(y, 'linear', 'Threshold',5);
Idx = [1 find(Lv), numel(y)];
for k = 1:numel(Idx)-1
P(:,k) = [x(Idx(k)) 1; x(Idx(k+1)) 1] \ y(Idx(k:k+1)).';
end
Slopes = P(1,:)
Intercepts = P(2,:)
figure
plot(x, y, '-o')
hold on
plot(x(Idx), y(Idx), 'p')
hold off
grid
axis([0 35 0 15])
In this run, that produced this plot:
and these data:
Slopes =
-0.6667 3.6667 0 -3.0000
Intercepts =
11.0000 -41.0000 14.0000 98.0000
You will likely need to adjust the 'Threshold' value (here 5) to work with your data. It should produce the appropriate slopes and intercepts for the regression lines. The ischange function was introduced in R2017b. A similar function, findchangepts in the Signal Processing Toolbox (with significantly different arguments and outputs) was introduced in R2016a.
