如何在Matlab的lsqcurvefit的成本函数中考虑对数距离？

Question

当我使用lsqcurvefit校准任意函数fun的参数时

fun = @(p,x)(p(1)./x .* 1./(p(3)*sqrt(2*pi)).*exp(-(log(x)-p(2)).^2./(2*p(3)^2)));

对于两种情况p_in ， p_out做

opts = optimoptions('lsqcurvefit','TolX',1e-4,'TolFun',1e-8);
p0 = [1,1,1];
p_in = lsqcurvefit(fun,p0,xR_in,yR_in,[],[],opts);
p_out = lsqcurvefit(fun,p0,xR_out,yR_out,[],[],opts);

乍一看，两条曲线都非常适合

但是，如果我在对数刻度中查看相同的结果，则拟合看起来不再那么好：

这让我想知道是否有可能在lsqcurvefit的成本函数中考虑对数误差，这基本上意味着查看函数每个x值的相对误差。

Matlab 中是否有可能强制执行此操作？

示例数据：

xR_in =[0.0649, 0.0749, 0.0865, 0.1000, 0.1155, 0.1334, 0.1540, 0.1779, 0.2054, 0.2371, 0.2739, 0.3162, 0.3651, ...
        0.4217, 0.4870, 0.5623, 0.6494, 0.7498, 0.8660, 1.0000, 1.1548, 1.3335, 1.5399, 1.7782, 2.0535, 2.3714, ...
        2.7384, 3.1623, 3.6517, 4.2170, 4.8696, 5.6234, 6.4938, 11.5477   13.3351];

xR_out = [0.487, 0.562, 0.649, 0.750, 0.866, 1.000, 1.155, 1.334, 1.540, 1.778, 2.054, 2.371, 2.738, 3.162, 3.652, ...
          4.217, 4.870, 5.623, 6.494, 7.499, 8.660, 10.000, 11.548, 13.335, 15.399, 17.783, 20.535, 23.714, 27.384, ...
          31.623, 36.517, 42.170, 48.697, 56.234, 64.938, 74.989, 86.596, 100.00];

yR_in = [0.00455681508591336, 0.00873409885607543, 0.0181688255259165, 0.0538821352554514, 0.117259031001522, ...
         0.193077037062548, 0.266434471781847, 0.325746362482833, 0.365146728802310, 0.386958383047475, ...
         0.403635741471857, 0.432215334550296, 0.485674792559743, 0.567693518475570, 0.668583592821924, ...
         0.768926132246355, 0.859501365700179, 0.933719545040531, 0.980404014586848, 1, 0.995168235818486, ...
         0.968831382298403, 0.918925456426510, 0.840263581309774, 0.730217907888984, 0.593965562217998, ...
         0.445252129988812, 0.302478487418430, 0.182455631766952, 0.0946623004432102, 0.0395136730456252, ...
         0.0110944993303772, 0.00304918005164052, 0.000176066405779416, 0.000271107188644644];

yR_out = [0.141149758427413, 0.327528725181928, 0.531429476303118, 0.734825800245580, 0.905142105752982, ...
          1, 0.994585150172040, 0.897227892811480, 0.737878031509986, 0.553911757285727, 0.378467001536220, ...
          0.235489445927972, 0.136533896783523, 0.0811622468252899, 0.0597202443876324, 0.0577324907072722, ...
          0.0649847546602922, 0.0739253772095743, 0.0770973088912625, 0.0732042102857309, 0.0642224578204625, ...
          0.0538581912412821, 0.0446084800059044, 0.0369556660640144, 0.0301643235397618, 0.0235320482975639, ...
          0.0170281463058672, 0.0111839380638397, 0.00660341791973352, 0.00352225498991166, 0.00174404920373791, ...
          0.000833106531219361, 0.000466707400714911, 0.000321368144365539, 0.000238498897310069, 0.000187001782051892, ...
          0.000152087422572435, 0.000120039487163363];

Answer 1

我可以为两个数据集找到的最佳峰值方程是具有指数截止方程的幂律，“y = C * pow(x, -1.0 * T) * exp(-1.0 * x / K)”，参见图下面的两个数据集，并注意 x 轴刻度的差异。

对于“in”数据，参数为：

C =  3.7599874401256059E+00
T = -1.4597138492114836E+00
K =  7.5655555953645282E-01

产生 R 平方 = 0.0.989 和 RMSE = 0.037

“输出”数据既有更尖的峰值，也有尖峰底部的“凸点”，没有单独的峰值函数可以建模——实际上，这有一个大峰值加上一个小得多且不太尖锐的第二个峰值。

对于“输出”数据，参数为：

C =  7.5124684986001625E+01
T = -4.9620437310832193E+00
K =  2.2970202935383399E-01

产生 R 平方 = 0.984 和 RMSE = 0.045