Using tabular data for Curve fitting of function z = f (x, y) with incomplete values of z. These blank z (i, j) automatically takes zero values and this is a problem

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Boyan
Boyan 2024-9-23
评论： Boyan 2024-9-29
Dear Community Members,
I have tabular data for data selection in the Curve fitter in the form of:.
I have to use curve fitter for z = f (x, y), but I don't have values for z(2, 4), z(3,3) and z(3,4) (red one).
In the workspase variable (matrix Z) these blank places automatically take values, which is zeros. Is there any instrument to ignore these zeros and the regression analyses to be made only with the available z values. In other words to ignore the three zero values in the down right corner? May be something with the weights instrument? If I simply use the matrix with zero values, without any additional actions, of course the regression method uses these z(2, 4) = 0, z(3,3) = 0 and z(3,4) = 0 and the results are unacceptable.
  2 个评论
Torsten
Torsten 2024-9-23
Supply z as a vector instead of a matrix and don't list these three data points in the vector. What's the problem ?
Boyan
Boyan 2024-9-25
Hi, Torsten,
The data is shown in picture below. x(i) is in green, y(j) is in blue. Z values simply form a matix, I think that I can't supply them as a vector.
I have no problem to obtain formula coefficients, if there are values in the right down corner.
In fact there is second issue, but this is asked in another question. The second issue is that there is no problem with regression analyses if the formula is z = a*x^n/y^m, but have problem when adding the "b" in the denominator and the formula is z = a*x^n/(y+b)^m.
So, to return to the first issue – how to make the regression method just to ignore missing values in the right down corner instead thinking for them like zeros.

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采纳的回答

Torsten
Torsten 2024-9-25
编辑：Torsten 2024-9-25
Ran in:
green = [5 5 5 5 5 4 4 4 4 1 1 1];
blue = [5 10 15 20 25 5 10 15 20 5 10 15];
yellow = [455 322 261 205 169 410 306 246 191 257 185 151];
F = @(p) fun(p,green,blue,yellow);
p0 = [1 1 1 1];
p = lsqnonlin(F,p0);
Local minimum possible. lsqnonlin stopped because the final change in the sum of squares relative to its initial value is less than the value of the function tolerance.
F(p)
ans = 1×12
1.7303 9.7596 9.9255 -10.0906 -21.7707 -12.4649 14.9799 11.9889 -9.4727 -5.7667 3.9900 5.4487
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
norm(F(p))
ans = 38.2707
a = p(1)
a = 674.6751
n = p(2)
n = -0.5377
b = p(3)
b = -0.2283
m = -p(4)/2
m = -0.0748
function res = fun(p,green,blue,yellow)
a = p(1);
n = p(2);
b = p(3);
m = p(4);
res = yellow - a*blue.^n.*((green+b).^2).^m;
end

更多回答（1 个）

Cris LaPierre
Cris LaPierre 2024-9-25
I'd suggest using the standardizeMissing function to convert the 0 values to nan. The Curve Fitter app will ignore NANs.
  1 个评论
Boyan
Boyan 2024-9-29
Thank you, @Cris LaPierre,
Your advice also works (like "the vector approach" of @Torsten). I implemented it with small change. I have simply import (with "import data" tool) Excel table (the matrix) and the the blank cells became NaN. Already with NaN it works. Before I simply input the table data by hand and leave blank cells in the down right corner of the matrix and the cellc automatically have accepted value of 0.

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