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[求助]
二元非线性拟合，请大神帮忙看下程序哪有问题，谢谢！ 已有1人参与
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两个自变量x1、x2，一个因变量y。通过绘图可以看出各个自变量和因变量的关系分布类似于正态分布，因此选用二维正太分布模型拟合。 下面是二维正态分布模型：  http://baike.baidu.com/view/9306579.htm 可详见 http://baike.baidu.com/view/9306579.htm 百度百科 下面是数据： y x1 x2 0 1000 2 0 1200 5 1 1400 15 1 1570 20 2 1720 30 2 1790 40 3 1860 60 4 1880 90 6 1980 172 7 2250 200 7 2530 250 8 2850 350 9 3100 450 10 3300 550 11 3500 650 12 3600 800 13 3800 1100 14 4000 1440 14 4200 1501 14 4500 1563 14 4800 1637 6 6800 2863 下面是函数代码： function yy=fun(beta,x) b1 = beta(1); b2 = beta(2); b3 = beta(3); b4 = beta(4); b5 = beta(5); yy=(2*pi*b3*b4*(1-b5^2)^(1/2))^(-1)*exp(-1/2*(1-b5^2)^(-1)*((x(:,1)-b1)^2*b3^(-2)-2*b5*(x(:,1)-b1)*(x(:,2)-b2)*(b3*b4)^(-1)+(x(:,2)-b2)^2*b4^(-2))); 下面是程序运行代码： clc clear data=load('data.txt'); y=data(:,1); x=data(:,2:3); beta=[1,0.1,3,1,0.5]; [beta_out,r,J,COVB,mse]=nlinfit(x,y,@fun,beta); beta_out; mse; betaci=nlparci(beta_out,r,'Jacobian',J); betaa=[beta_out',betaci]; [yy,delta]=nlpredci(@fun,x,beta_out,r,'Jacobian',J); alpha=0.05; nlintool(x,y,'fun',beta,alpha) r=corrcoef(y,yy); plotregression(y,yy) regstats(y,yy) 运行结果： Warning: Rank deficient, rank = 0, tol = 0.000000e+00. > In nlinfit>LMfit at 351 In nlinfit at 184 In nihe at 7 Warning: Some columns of the Jacobian are effectively zero at the solution, indicating that the model is insensitive to some of its parameters. That may be because those parameters are not present in the model, or otherwise do not affect the predicted values. It may also be due to numerical underflow in the model function, which can sometimes be avoided by choosing better initial parameter values, or by rescaling or recentering. Parameter estimates may be unreliable. > In nlinfit at 253 In nihe at 7 Warning: Matrix is singular to working precision. > In nlinfit at 274 In nihe at 7 Warning: Matrix is singular to working precision. > In nlparci at 104 In nihe at 10 Warning: Matrix is singular to working precision. > In nlpredci at 181 In nihe at 12 Warning: Matrix is singular to working precision. > In nlintool>PredictionsPlusError at 409 In nlintool at 257 In nihe at 14 Warning: Matrix is singular to working precision. > In nlintool>PredictionsPlusError at 409 In nlintool at 257 In nihe at 14 Warning: Matrix is singular to working precision. > In nlintool>PredictionsPlusError at 409 In nlintool at 271 In nihe at 14 Warning: Matrix is singular to working precision. > In regstats at 193 In regstats at 155 In nihe at 17 Warning: Matrix is singular to working precision. > In regstats at 208 In regstats at 155 In nihe at 17 出现很多错误，拟合度很低或者出现NaN。 求大神看是什么问题。 |
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7楼2016-03-21 16:38:05
dingd
铁杆木虫 (职业作家)
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【答案】应助回帖
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感谢参与，应助指数 +1
o452794544: 金币+6, ★★★很有帮助 2016-03-21 16:39:36
感谢参与，应助指数 +1
o452794544: 金币+6, ★★★很有帮助 2016-03-21 16:39:36
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换这个公式要好的多： y=a*exp(-0.5*(((x1-b)/c)^2))+d*exp(-0.5*(((x2-f)/g)^2)); 均方差(RMSE):0.401981621559676 残差平方和(SSE):3.55496292957842 相关系数(R): 0.996661569977283 相关系数之平方(R^2): 0.993334285069583 确定系数(DC): 0.993333686971469 卡方系数(Chi-Square): 0.838652138061828 F统计(F-Statistic): 480.02546329649 参数 最佳估算 ---------- ------------- a -5.90625065613002 b 1204.03546366578 c -526.613682770912 d 14.0704287034134 f 1448.75058061013 g -1080.76690762784 |
2楼2016-03-15 14:50:58
dingd
铁杆木虫 (职业作家)
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3楼2016-03-15 14:51:47
4楼2016-03-15 15:26:48
