linear fit to data with intercept at origin

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Lindsay Anderson
Lindsay Anderson 2016-3-22
评论： Star Strider 2018-3-30
Hi, I am new to matlab and am working on a problem trying to fit a straight line through a series of data points, but I need to force the line to cross the origin.
I am using the linfit.m code and would like to modify it to calculate a(2) and the uncertainty in a(2), sa(2) (since a(1) will be 0). The code for linfit.m is attached.
Can anyone provide a suggestion on how to modify the code to force the line through the origin and to estimate the uncertainty in the slope (without using cramer's rule)? I know how to do this by hand and in excel (only 12 data points in the problem), but can't seem to sort it out in the linfit.m code, since it uses cramer's rule which from what I understand will produce a degenerate solution in estimating the uncertainty in the slope a(2) when I force a(1) to be zero.
Any help is greatly appreciated.
Thanks!
  1 个评论
John D'Errico
John D'Errico 2016-3-22
Ye gawds! That is terribly poor code. Why not use something better like regress from the stats toolbox? Or you can download my own polyfitn from the FEX.
Is your goal to learn to solve the problem yourself? It is rarely a good idea to write code to solve a problem when there is high quality code written by professionals out there. The result is often much like what I saw in linfit. Bad code, using poor algorithms. It may look impressive to the person who knows no better, but linfit is simply poor code.

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回答（1 个）

Star Strider
Star Strider 2016-3-22
You can do that wit the mldivide,\ function (or operator), and by not including a vector of ones in the ‘x’ matrix (that would create an intercept term):
x = 0:10;
y = randi(99, 11, 1);
B = x(:)\y(:); % Estimate Parameter: x*B = y
y_est = x*B;
figure(1)
plot(x, y, 'bp')
hold on
plot(x, y_est, '-r')
hold off
grid
axis([0 10 ylim])
  2 个评论
Star Strider
Star Strider 2018-3-30
‘Hi, I am new to matlab and am working on a problem trying to fit a straight line through a series of data points, but I need to force the line to cross the origin.’
Yes it does!

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