如何在Python和Matlab中简化for循环

Question

我是C和MATLAB的用户。当我开始学习Python时(一周前)，我注意到我没有充分利用MATLAB的潜力，特别是数组运算。我经常使用for循环，可能是因为我学会了用C语言编程。

在上一篇技巧文章中，我学习了使用cumsum和其他高效的数组操作，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);   

有没有可能在Python (Numpy)或MATLAB中简化以下代码？

      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

在有机会在我的机器和你的机器上并排运行后，我修改了下面的代码。还有几个问题(A在最后的循环中会变得更大吗?dz是什么？)你在运行这个之前遇到的问题是，我忘记了idx_matrix必须符合逻辑。

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

要简化循环，可以使用函数spdiags。

http://www.mathworks.fr/help/techdoc/ref/spdiags.html

例如，你的第一个循环可以写成：

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