Linear Fitting starting from a value

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Nicola De Noni
Nicola De Noni am 10 Okt. 2022
Bearbeitet: Torsten am 10 Okt. 2022
Hi everyone!
I need to fit these data [100 90 80 70 50 45 40 43 42 40 35] whom are distanced by 10 seconds.
The problem is that I would fit these data starting from the first value (100) in order to understand the slope of the line.
Is it possible?
Thanks!

Akzeptierte Antwort

Torsten
Torsten am 10 Okt. 2022
Bearbeitet: Torsten am 10 Okt. 2022
Sure. Use
l(x) = a*x + 100
as model function and fit the parameter a.
xdata = ...; % your xdata as column vector
ydata = ...; % your ydata as column vector
a = xdata\(ydata-100)
But starting from a certain x-value, the slope seems to change. You should separate the fit into two parts:
l1(x) = a*x + 100 0 <= x<= x1
l2(x) = b*x + [(a-b)*x1 + 100] x1 <= x <= x(end)

Weitere Antworten (1)

Hiro Yoshino
Hiro Yoshino am 10 Okt. 2022
Bearbeitet: Hiro Yoshino am 10 Okt. 2022
refline is the easiest way to achieve that.
y = [100 90 80 70 50 45 40 43 42 40 35];
x = 0:10:10*(length(y)-1);
plot(x,y,'o');
refline
if you want to know the coefficients:
p = polyfit(x,y,1)
p = 1×2
-0.6391 89.6818
polyfit for reference.
  1 Kommentar
Nicola De Noni
Nicola De Noni am 10 Okt. 2022
Thanks @Hiro but i would like that the fitting line passe through the first point 100,0

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