Question

我正在编写一个代码来实现一个基本的优化方法，以保持观察到的和模拟的配置文件之间的最小误差。我这样做是因为简单的均方根误差。我在这里给出的整个代码都是为了让人知道。

    %% Vegetation profile along a hillslope
    % Parameters
    Tv = 1;             % values are {1,2,3,5,10}
    %Kvg = 0.3;          % according to the Tv values Kvg=1/Tv
    Kvg = 1/Tv;
    Kvd = 0.5;          % valuse are {0.1,0.5,1,5,10}
    Tcv = 10;           % values are {20,50,80,130,180}
    q = (1/3);
    Nv = (1:1:10);               %(Kvd*Tcv)/Kvg; range is given for the parameter
    m = 2/3;
    n = 2/3;

    %% Horizontal coordinate and length scale
    x = 0:1:100;
    L = 100;
    x_non = x/L;
    Ne =10;               %L/Tcv;


    %% Vegetation profile on the slope
    v = zeros(length(x_non),length(Nv));
    for i = 1:length(x_non)
    for j = 1:length(Nv)
    v(i,j) = (1/(1+Nv(j)*(Ne^q)*(x_non(i)^q)));
    end
    end

   %% importing the excel file containg real dataset
   obs_data = xlsread('model_data.xlsx'); % 1-distance, 2-elevation, 3-  vegetation, 4-slope


   %% root mean square error between real and synthetic dataset of vegetation

  rmse_v = zeros(length(x_non),length(Nv));
  err_v = zeros(1,length(Nv));
  for j=1:length(Nv)
  rmse_v(:,j)=(obs_data(:,3)-v(:,j)).^2;
  err_v(1,j)= sqrt(mean(rmse_v(:,j)));

  end

现在我必须提取rmse_v最小的Nv值。上述陈述为真的Nv值必须在下一个会话中使用。

    Nt = (1:1:100);
    % Slope profile on the x direction
    s = zeros(length(x_non),length(Nt));
    for i = 1:length(x_non)
    for j = 1:length(Nt)
    s(i,j)=((Ne^q)/Nt(j))*(x_non(i)^(q-m))+(1/(Nt(j)*(x_non(i)^m)*(1+Nv*(Ne^q)...
    *(x_non(i)^q))))^(1/n);
   end
   end

我将非常感激能够找到完成特定事情的方式。

谢谢。此致

如何在循环中找到确切的元素值？

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