关于非线性拟合的参数问题我用1stOpt进行非线性拟合,现在我对于拟合的结果有点疑问,Root of Mean Square Error (RMSE):26.44652781262Sum of Square Error (SSE):1748547.08335921Correlation Coef.(R):0.259670235027033R-Square:0
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![关于非线性拟合的参数问题我用1stOpt进行非线性拟合,现在我对于拟合的结果有点疑问,Root of Mean Square Error (RMSE):26.44652781262Sum of Square Error (SSE):1748547.08335921Correlation Coef.(R):0.259670235027033R-Square:0](/uploads/image/z/10066184-8-4.jpg?t=%E5%85%B3%E4%BA%8E%E9%9D%9E%E7%BA%BF%E6%80%A7%E6%8B%9F%E5%90%88%E7%9A%84%E5%8F%82%E6%95%B0%E9%97%AE%E9%A2%98%E6%88%91%E7%94%A81stOpt%E8%BF%9B%E8%A1%8C%E9%9D%9E%E7%BA%BF%E6%80%A7%E6%8B%9F%E5%90%88%2C%E7%8E%B0%E5%9C%A8%E6%88%91%E5%AF%B9%E4%BA%8E%E6%8B%9F%E5%90%88%E7%9A%84%E7%BB%93%E6%9E%9C%E6%9C%89%E7%82%B9%E7%96%91%E9%97%AE%2CRoot+of+Mean+Square+Error+%28RMSE%29%3A26.44652781262Sum+of+Square+Error+%28SSE%29%3A1748547.08335921Correlation+Coef.%28R%29%3A0.259670235027033R-Square%3A0)
关于非线性拟合的参数问题我用1stOpt进行非线性拟合,现在我对于拟合的结果有点疑问,Root of Mean Square Error (RMSE):26.44652781262Sum of Square Error (SSE):1748547.08335921Correlation Coef.(R):0.259670235027033R-Square:0
关于非线性拟合的参数问题
我用1stOpt进行非线性拟合,现在我对于拟合的结果有点疑问,Root of Mean Square Error (RMSE):26.44652781262
Sum of Square Error (SSE):1748547.08335921
Correlation Coef.(R):0.259670235027033
R-Square:0.0674286309589946
Determination Coef.(DC):-0.000138321931243052
这些参数分别是什么意思,有什么作用,是越大说明拟合度越好呢,还是越小越好,
关于非线性拟合的参数问题我用1stOpt进行非线性拟合,现在我对于拟合的结果有点疑问,Root of Mean Square Error (RMSE):26.44652781262Sum of Square Error (SSE):1748547.08335921Correlation Coef.(R):0.259670235027033R-Square:0
要新建两个文件:
主文件,test.m
------------------------------------------------------------------------------
function test
clear
clc
x = [51.885; 86.4; 134.73; 157.74; 240.6; 321.15; 374.05];
f = [-21.4823; -21.0836; -20.98485; -20.90685; -20.30025; -19.5936; -16.960725];
ft = fittype( 'fit_func( x, t )' );
cfun = fit( x, f, ft, 'StartPoint', -0.01 ); % 开始拟合
cfun % 输出拟合参数等
plot( cfun, x, f ) % 画出拟合后曲线和原始数据点
end
函数文件,用于自定义的函数:fit_func.m
------------------------------------------------------------------------------
function f = fit_func( x, t )
J = 21.18;
Li = 1e-6;
Voc = 0.418;
Ld = t * ( Voc - x );
f = J*Ld ./ ( Li*(1-exp(-Ld/Li)) );
end
------------------------------------------------------------------------------
可以保证运行输出结果,运行后自动绘出曲线和原始数据点,但是结果可能不会令人满意.我不保证编码时完全正确,但基本思想就是这样.你需要仔细检查代码,试着调正fit函数中StartPoint,还有其他选项(参见文档中fitoptions的说明),直到曲线和数据点吻合到满意为止.当然也可能是该函数不是这批数据点的理想模型,换成别的会更好.