indepvar and depvar are of inconsistent sizes with polyfitn

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mohamed gryaa
mohamed gryaa 2019-9-4
评论： Walter Roberson 2024-12-15
Ran in:
when i use polyfitn with matlab say indepvar and depvar are of inconsistent sizes and i don't now where i am wrong can yo help me?
this is the data:
x1=[2.2 2.2 2.2 2.44 2.49 2.48 2.45 2.2 2.39 2.38];
x2=[29.4 30.9 32.9 26.9 28 28.8 30.1 29.4 28.8 30];
y=[7.5 7.02 6.94 7.91 7.96 7.91 7.78 7.42 7.86 7.47];
p=polyfitn([x1,x2],y,1)
Unrecognized function or variable 'polyfitn'.
Error using polyfitn (line 172)
indepvar and depvar are of inconsistent sizes.

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Bjorn Gustavsson
Bjorn Gustavsson 2019-9-4
Yes, you send in an array [x1,x2] of size [1 x 20] and an array y of size [ 1 x 10] into polyfitn. The polyfitn version I have (a matlab-central submission by John d'Errico from 2006) gives me a more sensible result after transposing your inputs:
p=polyfitn([x1;x2]',y',1)
p =
ModelTerms: [3x2 double]
Coefficients: [1.5589 -0.11979 7.4607]
ParameterVar: [0.1693 0.0010284 2.9687]
ParameterStd: [0.41146 0.032068 1.723]
R2: 0.91523
RMSE: 0.10284
VarNames: {}
This was my curse when starting to use matlab - I solved it by the very sofisticated technique:
If it doesn't work transpose, transpose again until all combinations have been tried.
You might do the same but a more intelligent aproach is to check the documentation and input variables.
HTH
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Steven Lord
Steven Lord 2024-12-15
Ran in:
Compare the sizes:
x1=[2.2 2.2 2.2 2.44 2.49 2.48 2.45 2.2 2.39 2.38];
x2=[29.4 30.9 32.9 26.9 28 28.8 30.1 29.4 28.8 30];
y=[7.5 7.02 6.94 7.91 7.96 7.91 7.78 7.42 7.86 7.47];
If the poster was using this File Exchange submission, what does its help text state about the requirements it places on its inputs?
% indepvar - (n x p) array of independent variables as columns
% n is the number of data points
% p is the dimension of the independent variable space
%
% IF n == 1, then I will assume there is only a
% single independent variable.
%
% depvar - (n x 1 or 1 x n) vector - dependent variable
% length(depvar) must be n.
If you passed [x1, x2] into polyfitn as the indepvar input argument, what are n and p?
[n, p] = size([x1, x2])
n = 1
p = 20
Is y (the depvar input argument) either an n x 1 or 1 x n vector? If we look at its size, it's clear that it is not.
sz = size(y)
sz = 1×2
1 10
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If we transposed the x and y data:
[n, p] = size([x1.', x2.'])
n = 10
p = 2
sz = size(y.')
sz = 1×2
10 1
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We see that y.' is an n x 1 vector. So polyfitn can operate on it and the combination [x1.', x2.'].

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