Hello everyone, I am facing one problem,
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I have 3 different non linear data. Let's say x,y and z.for different values of Z I have different data set of x and y. If I fix the value of Z I am able to find the relation between x and y, but if I change the value of Z that relation won't work, I need to find relation between x,y and z independently. Can anyone please help! Thank you in advance.
for z=500 I have x = 1145 to 1240 with the gap of 1,y = (17.75,16.2,15.1,14.25,13.45.... respectively
for z=400, x values remains the same 1145-1240 y values differs.
I need to find a relation between x,y and z independently. in which y = fun(x,z). thank you.
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Image Analyst
el 30 de Nov. de 2018
For each z, get or compute the x and y arrays. Then fit a model for y using the x array, and the scalar value of z for those arrays. In pseudocode:
z = [......] % Some list of z values
for k = 1 : length(z)
thisZ = z(k);
xVector = ...% Whatever you need to do to get the x vector for this z
yVector = ....% Whatever you need to do to get the y vector for this z
% Now find some model passing in the xVector and yVector and this z value.
yourModel = FitModel(xVector, yVector, thisZ);
% Now do something with the model, like apply it, save it, plot it, or whatever.....
end
I have no idea what the model for each pair of x and y vectors might be. Would they all be the same, like fitting a line or parabola? Would they be different, like some might be a line and others might need an exponential decay? Would it be empirical so you don't have model parameters, but just a smoothed/fitted y vector as output, like as if you applied smooth() or sgolayfilt() to the raw data? We have no idea since you didn't share that information with us.
3 comentarios
Aakash Nanda
el 1 de Dic. de 2018
Image Analyst
el 1 de Dic. de 2018
Editada: Image Analyst
el 1 de Dic. de 2018
Why do the first two have only 4 points, and the others have 5 or 6 points? Why aren't they all the same size?
Attached is what I have so far.

I didn't go further because I don't think you have enough points in there to do meaninful fits. Attach more data.
Aakash Nanda
el 1 de Dic. de 2018
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