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我试图编写一个代码，它是一个更大程序的一部分，它将z在s. 但是，当我运行代码时，我只得到z=0，或者如果 lastelse被忽略，则代码返回零向量。

有人知道我在哪里犯错了吗？我已使用此来源的方法 1 。任何帮助将不胜感激，我几个月来一直在努力完成这项工作。

% clc;close all; %// not generally appreciated
%initial values
b=1.25;
h=0.313;

%define the s coordinate
s= 0:0.001:2*(b+h); 

%create zero matrix for speed
z=zeros(size(s));

%calculate z at every point of s coordinate
for i =length(s)
   if 0 <= s(i) && s(i) <=b   %0<=s<=b
       z=0.5*h;

   elseif b <= s(i) && s(i) <=(b+h)   %b<=s<=(b+h)
       z=0.5*h+((-0.5*h)/(b+h-b))*(s-b);

   elseif b <= s(i) && s(i) <=(b+h)    %(h+b)<=s<=(b+h)
       z=-0.5*h;

   elseif b <= s(i) && s(i) <=(b+h)    %(h+2b)<=s<=(2b+2h)
       z=-0.5*h+((-0.5*h)/(b+h-b))*(s-b);
   else z=0;

   end
end

为了进一步参考，这解决了我的问题。谢谢@丹！

 %// initial values
   b=1.25;
   h=0.313;
   %// define the s coordinate
   s= 0:0.001:2*(b+h);
   %// Create z
   z = zeros(size(s));
   idx1 = 0 <= s & s <=b;
   idx2 = b <= s & s <=(b+h);
   idx3 = (b+h) <= s & s <= (2*b+h);
   idx4 = (2*b+h) <= s & s <=(2*b+2*h);
   z(idx1) = 0.5*h;
   z(idx2) = 0.5*h+((-0.5*h-0.5*h)/(b+h-b))*(s(idx2)-b);
   z(idx3) = -0.5*h;
   z(idx4) =-0.5*h+((0.5*h+0.5*h)/((2*b+2*h-b)-(h+b+b)))*(s(idx4)-b)
4

2 回答 2

1

你的代码有很多问题。您需要分配一个索引，z否则您每次都只是覆盖一个标量（即z(i)=...）。你需要循环一个向量，所以fori=1:length(s)你的最后三个循环条件是相同的！

%// initial values
b=1.25;
h=0.313;

%// define the s coordinate
s= 0:0.001:2*(b+h); 

%// create zero matrix for speed
z=zeros(size(s));

%// calculate z at every point of s coordinate
for i = 1:length(s)
   if 0 <= s(i) && s(i) <=b                   %// 0<=s<=b
       z=0.5*h;

   elseif b <= s(i) && s(i) <=(b+h)           %// b<=s<=(b+h)
       z(i)=0.5*h+((-0.5*h)/(b+h-b))*(s-b);

   elseif (b+h) <= s(i) && s(i) <= (2*b+h)    %// (h+b)<=s<=(2b+h)
       z(i)=-0.5*h;

   elseif (2*b+h) <= s(i) && s(i) <=(2*b+2*h) %// (h+2b)<=s<=(2b+2h)
       z(i)=-0.5*h+((-0.5*h)/(b+h-b))*(s-b);
   else z(i)=0;

   end
end

综上所述，在 MATLAB 中，您甚至根本不需要循环来执行此操作，通常最好不要使用循环：

%// initial values
b=1.25;
h=0.313;
%// define the s coordinate
s= 0:0.001:2*(b+h);
%// Create z
z = zeros(size(s));
idx1 = 0 <= s && s <=b;
idx2 = b <= s && s <=(b+h);
idx3 = (b+h) <= s && s <= (2*b+h);
idx4 = (2*b+h) <= s && s <=(2*b+2*h);
z(idx1) = 0.5*h;
z(idx2) = 0.5*h+((-0.5*h)/(b+h-b))*(s(idx2)-b);
z(idx3) = -0.5*h;
z(idx4) = -0.5*h+((-0.5*h)/(b+h-b))*(s(idx4)-b);
于 2016-01-18T08:15:36.373 回答
0 
b=1.25;
h=0.313;

%define the s coordinate
s= 0:0.001:2*(b+h); 

%create zero matrix for speed
z=zeros(size(s));

%calculate z at every point of s coordinate
for ii =1:length(s)
   if 0 <= s(ii) && s(ii) <=b   %0<=s<=b
       z(ii)=0.5*h;

   elseif b <= s(ii) && s(ii) <=(b+h)   %b<=s<=(b+h)
       z(ii)=0.5*h+((-0.5*h)/(b+h-b))*(s(ii)-b);

   elseif b <= s(ii) && s(ii) <=(b+h)    %(h+b)<=s<=(b+h)
       z(ii)=-0.5*h;

   elseif b <= s(ii) && s(ii) <=(b+h)    %(h+2b)<=s<=(2b+2h)
       z(ii)=-0.5*h+((-0.5*h)/(b+h-b))*(s(ii)-b);
   else z(ii)=0;

   end
end

让您的for循环运行超过 1 次迭代，因此for ii = 1:length(s).

使用每个元素的分配，因此z(ii) = some function

不要i用作变量。

于 2016-01-18T08:09:22.707 回答