polyfitn throws an error with seemingly sufficient input data
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I have two data points that lie on a straight line:
x = [1;2];
y = [3;4];
Matlab's built-in polyfit function nicely fits the data to a linear equation:
P = polyfit(x,y,1);
which can be used to get an equation for the line y = x + 2. As I understand it, polyfitn should work the same way:
Pn = polyfitn(x,y,1);
but it gives
Error using betainc
X must be in the interval [0,1].
Error in polyfitn (line 233)
polymodel.p = betainc(polymodel.DoF./(t.^2 + polymodel.DoF),polymodel.DoF/2,1/2);
I can trick polyfitn by over-defining the problem:
Pn = polyfitn([x;x],[y;y],1);
but is that legal? It gives the answer I expect, but I want to make sure I'm not missing something. I'd like to understand why polyfitn only works when I give it superfluous data.
1 个评论
snoop58
2016-3-4
I get the same error with polyfitn. Looks like polyfitn thinks there is not enough data points; however, the builtin polyfit is good with it:
x = [-1;0;1]
y = [-0.1;0;0.1]
polyfit(x, y, 1)
ans = 0.1 0
polyfitn(x, y, 1)
Error using betainc
X must be in the interval [0,1].
Error in polyfitn (line 233)
polymodel.p = betainc(polymodel.DoF./(t.^2 + polymodel.DoF),polymodel.DoF/2,1/2);
回答（1 个）
Walter Roberson
2015-5-12
polyfitn is a File Exchange Contribution, not part of MATLAB. You can read the source there.
If you trace through the code, your degrees of freedom is coming out as 0, and t is coming out as 0, so t^2+DoF is coming out as 0 for the denominator. 0/0 is NaN and so is not in range for betainc
The t is coming out as 0 because the variances are being set to infinity because
% we cannot form variance or standard error estimates
% unless there are at least as many data points as
% parameters to estimate.
The particular test involves "n > nt" when n (2 points) = nt (number of terms generated for middleterms). Since the comment says "at least as many" and there appear to be exactly as many there just might be a bug there. But I'll let John D'Errico the author comment on that.
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2015-5-11
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