How to fit a general curve on different datasets ?

Question

yj 2019-9-11

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/479937-how-to-fit-a-general-curve-on-different-datasets

编辑： Jon 2019-9-23

采纳的回答： Jeff Miller

在 MATLAB Online 中打开

Hi,

I have different datasets that I would like to fit together in order to create a general equation.

Yet I have fitted each dataset independently as : y_i=a_i + b_i * log10(x)

%% Example of one dataset
x= [1 10 100]';
y= [165 145 124]';
fo = fitoptions('Method','NonlinearLeastSquares',...
               'Lower',[1],...
               'Upper',[200],...
               'StartPoint',[180 180]);
ft = fittype('a + b*log10(x)','dependent',{'y'},'independent',{'x'},'coefficients',{'a','b'},'options',fo);
myfit=fit(x(:),y(:),ft);
plot(x,y, 'ko', 'LineWidth',1.5, 'MarkerSize',7)
hold on
plot(myfit, 'k--')

But now I would like to test this thing : y_i = a_i + b * log10(x) with a general b for all the dataset. So I need to fit this new model into my whole dataset to get an estimation of b that match with every dataset.

Does anyone as an idea on how to do this in matlab ?

Thank you very much for your help.

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Answer 1

Jeff Miller 2019-9-13

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在 MATLAB Online 中打开

You need to set up separate independent variables for each dataset, something like this:

%% Example of two datasets (3 scores each)
i1= [1 1 1 0 0 0]; % 1 indicates member of 1st data set
i2= [0 0 0 1 1 1]; % 1 indicates member of 2nd data set
x= [1 10 100 1 10 100]';  % concatenated x's for both data sets
y= [165 145 124 155 135 114]'; % concatenated y's for both data sets
% Then fit a model like this
'a1*i1 + a2*i2 + b*log10(x)'

But I don't think the 'fit' function will do that--at least not for more than 2 datasets. You could do it with regress, at least if you make a new independent variable:

log10x = log10(x);
% Then fit 
'a1*i1 + a2*i2 + b*log10x'

MATLAB probably has some other nonlinear fitting routines that would also work, using the same basic indicator variable trick.

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Jeff Miller 2019-9-20

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A small simulation shows that John is exactly right:

% Simulation with equal x's in all groups, but random y's
i1= [1 1 1 0 0 0]'; % 1 indicates member of 1st data set
i2= [0 0 0 1 1 1]'; % 1 indicates member of 2nd data set
x = [1 10 100 1 10 100]';  % concatenated x's for both data sets
log10x = log10(x);
NTries = 10;
for i=1:NTries
    y1 = randn(3,1);
    a1 = regress(y1,[i1(1:3) log10x(1:3)]);
    y2 = randn(3,1);
    a2 = regress(y2,[i2(4:6) log10x(4:6)]);
    y = [y1; y2];
    a = regress(y,[i1 i2 log10x]);
    dif = abs(a(3) - (a1(2) + a2(2))/2)
end
% The resulting difs are all miniscule, so in this case the slope of the
% combined data set matches the average of the individual slopes.
% Simulation with different x's in all groups, but random y's
for i=1:NTries
    x = randn(6,1).^2;
    log10x = log10(x);
    y1 = randn(3,1);
    a1 = regress(y1,[i1(1:3) log10x(1:3)]);
    y2 = randn(3,1);
    a2 = regress(y2,[i2(4:6) log10x(4:6)]);
    y = [y1; y2];
    a = regress(y,[i1 i2 log10x]);
    dif2 = abs(a(3) - (a1(2) + a2(2))/2)
end
% The resulting dif2s are much larger, so in this case the slope of the
% combined data set need not match the average of the individual slopes.

yj 2019-9-20

Yes thank you very much for your answers !!

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Answer 2

Jon 2019-9-11

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If I am understanding what you are trying to do correctly, it seems like you should be able to concatenate all of your individual x values into one overall x vector, and similarly concatenate all of your y values into one overall y vector and then perform the curve fit on the combined set.

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Jon 2019-9-13

编辑：Jon 2019-9-13

在 MATLAB Online 中打开

I did not initially understand the problem you are trying to solve. Now I get that you want to fit an equation that possibly has different values of "a" for each individual data set but a common value of b. Conceptually, this suggests to me that there should be an additional, numerical parameter value (e.g. voltage, pressure, temperature, valve position ...), which characterizes each of the individual tests. Let's call this parameter, z, so we have have a vector z, the same length as x and y (the overall concatenated vectors) with the values of the test parameter corresponding to each experiment in the overall concatenated data set (so it repeats for the points that fall within one test). The simplest possibility would be that your test dependent parameter "a" depends linearly on z, so we would have a = c*(z - z0) where z0 is a nominal value for z. Putting it all together you want to fit y = c*(z - z0) + b*log10(x), or expanding y = cz + b*log10(x) + d, where d = -c*z0. You can solve this as a linear regression to solve for c,b, and d by putting it into the form:

[y(1);y(2);..y(n)] = [z(1) log10(x(1) 1;z(2) log10(x(2)) 1; ... z(n) log10(x(n)) 1]*[c;b;d]

which has the form y = A*p, where p are the unknown parameters c,b, and d you can solve this using

p = A\y;

If you needed the values of "a" you could then compute them afterwards using a = c*(z-c0)

Note that in the above x,y,and z are the concatenated values for all of the tests. In particular z will have repeated values, as it will be the same for each individual test.

John D'Errico 2019-9-20

Actually, this is the incorrect answer, since it will also use a common value for a.

Jon 2019-9-23

编辑：Jon 2019-9-23

I don't think that the approach I outlined will result in a common value of a, but I'm glad that you found a solution. (The approach I outlined may be buried under the older comments, you may not have seen it.)

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