lsqcurvefit question - ydata has two dimension (Function value and YDATA sizes are not equal)

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Yuqing_noob
Yuqing_noob 2023 年 2 月 12 日
コメント済み: Walter Roberson 2023 年 2 月 12 日
Hi,
I am new to matlab and I want to do a curve fitting using lsqcurvefit (following a paper's action). But I got a error called: Function value and YDATA sizes are not equal. Where did i do mistake...? Thanks for help.
xdata is the timepoint while ydata is 2-D position changing with time. The two dimensions of y are not independent of each other, so I don't think they should be fitted separately. This model have three params.
xdata = 1:60;
ydata = [590.25,511.34;590.23,510.52;590.22,509.51;590.18,508.61;
590.13,507.77;590.16,506.05;590.20,504.68;590.15,503.54;
589.98,501.91;589.66,500.71;589.47,499.64;589.27,498.69;
589.00,497.27;588.65,495.75;588.24,494.53;588.00,493.50;
587.57,492.52;587.45,491.43;587.21,490.27;586.96,489.24;
586.63,488.08;586.44,487.03;586.23,485.81;585.94,484.29;
585.57,482.81;585.21,481.34;584.93,480.24;584.59,479.26;
584.30,478.44;583.86,476.75;583.33,475.30;582.87,474.14;
582.52,473.06;582.21,472.03;582.04,471.08;581.62,469.92;
581.29,468.39;580.85,466.46;580.66,464.52;580.43,463.14;
580.11,462.08;579.58,460.73;579.51,458.95;579.41,457.45;
579.49,455.14;579.26,452.99;578.85,451.19;578.71,450.34;
578.53,449.45;578.54,448.45;578.48,447.62;578.30,446.64;
578.20,445.49;577.43,442.85;577.27,440.24;576.85,438.83;
576.67,437.85;576.54,437.19;576.52,435.73;576.35,434.59];
Filter = [0,0.0191620208109086,0.0683763509212163,0.100379351034585,0.0974133620604203,0.0679132261017191,0.0279145555757322,...
-0.00924174097003220,-0.0361988204543976,-0.0510709858726009,-0.0554801938514580,-0.0525520022466199,-0.0454971586440819,...
-0.0368765372303619,-0.0283884375848139,-0.0209563283968831,-0.0149359405672504,-0.0103302318649573,-0.00696094418434395,...
-0.00458437240976897,-0.00295845614268432,-0.00187479263439797,-0.00116876316001411,-0.000717879195806221,...
-0.000435011977115633,-0.000260359690910572,-0.000154065078565133,-9.02143808828862e-05,-5.23151853625161e-05,...
-3.00650758394752e-05,-1.71336401217675e-05,-9.68796610836146e-06,-5.43788818548629e-06,-3.03138411065368e-06,...
-1.67897757316864e-06,-9.24285783937950e-07,-5.05912271159936e-07,-2.75416083714947e-07,-1.49167755170539e-07,...
-8.03986272699545e-08,-4.31338127221703e-08];
% Define cumulative Gaussian function
F = @(mtnd, stnd, t) normpdf(t, mtnd, stnd);
% Define curve-fitting function
fun = @(params, xdata) conv(Filter, F(params(1), params(2), xdata));
% Estimate parameters using least-square fit
params = lsqcurvefit(fun, [0, 1], xdata, ydata);

回答 (1 件)

Torsten
Torsten 2023 年 2 月 12 日
Your function "fun" must return a matrix of size (60x2) like the ydata you provide. This is not the case - thus the error message.
  4 件のコメント
Walter Roberson
Walter Roberson 2023 年 2 月 12 日
They only return a matrix of one dimension (60x1)
No. You conv() a vector of length 41 (Filters) and xdata (length 60). The result is 41+60-1 = 100 .
If you wanted something the size of xdata to be returned you would need code similar to
fun = @(params, xdata) conv(F(params(1), params(2), xdata), Filter, 'same');
But of course this would still be a vector of length 60, not something that you could use with 60 x 2 Ydata.
for col = 1 : size(ydata,2)
params(:,col) = lsqcurvefit(fun, [0, 1], xdata, ydata(:,col).');
end

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