I have 744 hourly wind data points for the month of january i.e 24 data points for each day (744 for 31 days) For each day I need to fit the 24 hourly data points with the non linear function C(1)+C(2).​*sin(2*pi/​24.*(x-C(3​)))+C(4).*​sin(2*pi/1​2.*(x-

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Gourav Mitra
Gourav Mitra 2020-6-12
回答： David Goodmanson 2020-6-13
I have 744 hourly wind data points for the month of january i.e 24 data points for each day (744 for 31 days) for 7 different heights(basically I have a wind data matrix of 744X 8 dimension). For each day and for a particular height I need to fit the 24 hourly data points with the non linear function C(1)+C(2).*sin(2*pi/24.*(x-C(3)))+C(4).*sin(2*pi/12.*(x-C(5)))+C(6).*sin(2*pi/8.*(x-C(7))). These 7 value of coefficients which I am getting after using fitnlm needs to be stored in a matrix, as I want to apply a for loop to find and store the coefficient value of each day and a particular height in a 3d matrix , so that I get everything in one place for for easy analysis. Please help me out in applying the for loop and storing the coefficients in best way possible. Thanks in advance.

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回答（2 个）

Ameer Hamza
Ameer Hamza 2020-6-13
编辑：Ameer Hamza 2020-6-13
You can do something like this
M = % 744x8 matrix
M_parts = mat2cell(M, 24*ones(31, 1), ones(1, 8)); % divide the matrix M for each day an height in a cell array
Coeffs = zeros(31, 8, 7);
for day=1:31
for height=1:8
data = Coeffs{day, height};
c = % apply fitlm
Coeffs(day, height, :) = c;
end
end
Coeffs is a 3D array with days along the first dimension (rows), heights along 2nd dimension (columns) and coefficients along the 3rd (pages).
David Goodmanson
David Goodmanson 2020-6-13
Hi Gourav,
this is not really a nonlinear problem. That's because dropping some parenthesis)
C2*sin(2*pi/24*(x-C3)) = -C2*[ sin(2*pi/24*C3)*cos(2*pi/24*x) +cos(2*pi/24*C3)*sin(2*pi/24*x) ]
Sine and cosine are linearly independent functions, so if you can do a linear fit
data = D2*cos(2*pi/24*x) + D3*sin(2*pi/24*x)
then
D2 = -C2*sin(2*pi/24*C3)
D3 = C2*cos(2*pi/24*C3)
meaning that
C2 = sqrt(D2^2 + D3^2)
2*pi/24*C3 = atan2(-D2,D3)
so if you know C you know D and vice versa. Same for the other two sine functions as well, using D4,D5 and D6,D7.
If you set up a wind data matrix that is 24x31 (hours down, days across) x 8 for heights then you can fit the entire month in one go, and loop over heights. In the code below, C1 and D1 are the same.
D = zeros(7,31,8);
x = (0:23)'; % 1:24 instead?
% set up matrix of 7 linearly independent columns
M = [ones(size(x)) cos(2*pi/24*x) sin(2*pi/24*x)...
cos(2*pi/12*x) sin(2*pi/12*x)...
cos(2*pi/ 8*x) sin(2*pi/ 8*x)];
Wind = rand(24,31,8);
for k = 1:8
D(:,:,k) = M\Wind(:,:,k);
end
The result is a 7x31x8 matrix of coefficients.

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