indepvar and depvar are of inconsistent sizes with polyfitn

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.

[Walter Roberson]

polyfitn appears to be https://www.mathworks.com/matlabcentral/fileexchange/34765-polyfitn

Answer 1

[Bjorn Gustavsson]

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

[mohamed gryaa]

thanks for helping me

[LOGAN]

ok how does that help him

[Steven Lord]

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.'].