Question

I m a C and MATLABuser. 当我开始学习时(一周前)我注意到,我没有充分利用MATLAB的全部潜力,特别是阵列行动。我经常使用假体,可能是因为我学习了C的节目。

在上一个图普,我学会使用<代码>cumsum和其他高效阵列操作,例如:

alpha = [1e-4,1e-3,1e-4,1e-1,1e-2,1e-3,1e-6,1e-3];
zeta = alpha / (dz*dz)
nz = 101
l=[0.3,0.1,0.2,0.1,0.1,0.1,0.2];
wz = cumsum(l*(nz-1));
nl = lenght(l);

是否有可能简化以下法规:

      A = zeros(nz,nz);
      i=1;
      for j = 2:wz(i)-1
        A(j,j-1) = zeta(1,1);
        A(j,j) = -2*zeta(1,1);
        A(j,j+1) = zeta(1,1); % layer 1 nodes 
      end

      %cicle to n-layers
      for i=2:nl
          for j=wz(i-1):wz(i-1)
              A(j,j-1) = zeta(1,i-1);
              A(j,j) = -zeta(1,i-1)-zeta(1,i);
              A(j,j+1) = zeta(1,i); 
          end

          for j=wz(i-1)+1:wz(i)
              A(j,j-1) = zeta(1,i);
              A(j,j) = -2*zeta(1,i);
              A(j,j+1) = zeta(1,i);
          end

      end
end

Answer 1

我在有机会与你一道在我机器旁边管理该守则之后,修改了该守则。仍然存在几个问题(在最后的路口里,是否大得多? 在此之前,你面临的问题是,我原谅的id子必须合乎逻辑。

dz=0.1;
alpha = [1e-4,1e-3,1e-4,1e-1,1e-2,1e-3,1e-6,1e-3];
zeta = alpha / (dz*dz);
nz = 101;
l=[0.3,0.1,0.2,0.1,0.1,0.1,0.2];
wz = cumsum(l*(nz-1));
nl = length(l);

A = zeros(nz);
i=1;

%replaces 1st loop
j_start = 2;
j_end = wz(i)-1;

idx_matrix = false(size(A));
idx_matrix(j_start:j_end,j_start:j_end) = eye(j_end-j_start+1);
A(idx_matrix) = -2*zeta(1,1);

idx_matrix(idx_matrix) = false;
idx_matrix(j_start:j_end,j_start-1:j_end-1) = eye(j_end-j_start+1);
A(idx_matrix) = zeta(1,1);

idx_matrix(idx_matrix) = false;
idx_matrix(j_start:j_end,j_start+1:j_end+1) = eye(j_end-j_start+1);
A(idx_matrix) = zeta(1,1);

%cicle to n-layers
for i=2:nl

    %replaces 3rd loop
    j_start = wz(i-1);
    A(j_start,j_start) = -zeta(1,i-1)-zeta(1,i);
    A(j_start,j_start-1) = zeta(1,i-1);
    A(j_start,j_start+1) = zeta(1,i);

    %replaces 4th loop
    j_start = wz(i-1)+1;
    j_end = min(wz(i),size(A,2)-1);
    idx_matrix = false(size(A));
    idx_matrix(j_start:j_end,j_start:j_end) = eye(j_end-j_start+1);
    A(idx_matrix) = -2*zeta(1,i);

    idx_matrix(idx_matrix) = false;
    idx_matrix(j_start:j_end,j_start-1:j_end-1) = eye(j_end-j_start+1);
    A(idx_matrix) = zeta(1,i);

    idx_matrix(idx_matrix) = false;
    idx_matrix(j_start:j_end,j_start+1:j_end+1) = eye(j_end-j_start+1);
    A(idx_matrix) = zeta(1,i);

end

Answer 2

为简化您的行程,您可使用<>条码>。

rel=“nofollow” http://www.mathwork.fr/help/techdoc/ref/spdiags.html

例如,你的第一例假可以写成:

A=full(spdiags(repmat([zeta(1,1),-2*zeta(1,1),zeta(1,1)],wz(i),1),[-1 0 1],wz(i),wz(i)))

友情链接