在Cell数组中绘制数字

您需要停止使用单元格数组进行数字数据，并在数组更简单时使用索引变量名称。

我在下面编辑了你的代码来绘制

I1

阵列。

为了使它工作，我几乎将所有单元格数组更改为数字数组并简化了一堆索引。注意初始化现在是

zeros

代替

cell

因此用括号索引

()

不是花括号

{}

。

我没有太多改变结构，因为你的评论表明你正在关注这个布局的一些文献

对于绘图，你试图在循环中绘制单点 - 为此你没有线（点是不同的），所以需要指定一个像

plot(x,y,’o’)

。然而，我所做的只是在循环之后绘制 - 因为你存储了结果

I1

无论如何阵列。

    
      % Initalize arrays for storing data
C = cell(1,100); % Store output vector from floww()
D = zeros(1,6); % User inputted initial point
I1 = zeros(1,100);
I2 = zeros(1,100);
I3 = zeros(1,100);
%Declare alpha and beta variables detailed in Theorem 1 of paper
a1 = 0; a2 = 2; a3 = 4; a4 = 6;
b1 = 2; b2 = 3; b3 = 7; b4 = 10;
% Declare the \lambda_i, i=1,…, 6, variables
L = zeros(1,6);
L(1) = abs((b2 - b3)/(a2 - a3));
L(2) = abs((b1 - b3)/(a1 - a3));
L(3) = abs((b1 - b2)/(a1 - a2));
L(4) = abs((b1 - b4)/(a1 - a4));
L(5) = abs((b2 - b4)/(a2 - a4));
L(6) = abs((b3 - b4)/(a3 - a4)); 
for j = 1:6
   D(j) = input(‘Input in1 through in6: ‘);
end
% Iterate through floww()
for i = 1:100
   C{i} = floww(D(1), D(2), D(3), D(4), D(5), D(6), L); % Output from floww() is a 6-by-1 vector
   for j = 1:6
       D(j) = C{i}(j,1); % Reassign input values to put back into floww()
   end
   % First integrals as described in the paper
   I1(i) = 2(C{i}(1,1)).^2 + 2(C{i}(2,1)).^2 + 2(C{i}(3,1)).^2 + 2(C{i}(4,1)).^2 + 2(C{i}(5,1)).^2 + 2(C{i}(6,1)).^2;
   I2(i) = (-C{i}(3,1))(-C{i}(6,1)) - (C{i}(2,1))(-C{i}(5,1)) + (-C{i}(1,1))(-C{i}(4,1));
   I3(i) = 2L(1)(C{i}(1,1)).^2 + 2L(2)(C{i}(2,1)).^2 + 2L(3)(C{i}(3,1)).^2 + 2L(4)(C{i}(4,1)).^2 + 2L(5)(C{i}(5,1)).^2 + 2L(6)*(C{i}(6,1)).^2;
end
plot(1:100, I1);
% This function will solve the linear system
% Bx^(n+1) = x detailed in the research notes
function [out1] = floww(in1, in2, in3, in4, in5, in6, L)
    % A_ij = (lambda_i - lambda_j)
    % Declare relevant A_ij values
    A32 = L(3) - L(2);
    A65 = L(6) - L(5);
    A13 = L(1) - L(3);
    A46 = L(4) - L(6);
    A21 = L(2) - L(1);
    A54 = L(5) - L(4);
    A35 = L(3) - L(5);
    A62 = L(6) - L(2);
    A43 = L(4) - L(3);
    A16 = L(1) - L(6);
    A24 = L(2) - L(4);
    A51 = L(5) - L(1);
% Declare del(T)
delT = 1;
% Declare the 6-by-6 coefficient matrix B
B = [1, -A32*(delT/2)*in3, -A32*(delT/2)*in2, 0, -A65*(delT/2)*in6, -A65*(delT/2)*in5;
    -A13*(delT/2)*in3, 1, -A13*(delT/2)*in1, -A46*(delT/2)*in6, 0, A46*(delT/2)*in4;
    -A21*(delT/2)*in2, -A21*(delT/2)*in1, 1, -A54*(delT/2)*in5, -A54*(delT/2)*in4, 0;
    0, -A62*(delT/2)*in6, -A35*(delT/2)*in5, 1, -A35*(delT/2)*in3, -A62*(delT/2)*in2;
    -A16*(delT/2)*in6, 0, -A43*(delT/2)*in4, -A43*(delT/2)*in3, 1, -A16*(delT/2)*in1;
    -A51*(delT/2)*in5, -A24*(delT/2)*in4, 0, -A24*(delT/2)*in2, -A51*(delT/2)*in1, 1];
% Declare input vector
N = [in1; in2; in3; in4; in5; in6];
% Solve the system Bx = N for x where x
% denotes the X_i^(n+1) vector in research notes
x = B\N;
% Assign output variables
out1 = x;
end
</code>
  

    
      % Initalize arrays for storing data
C = cell(1,100);  % Store output vector from floww()
D = zeros(1,6);   % User inputted initial point
I1 = zeros(1,100);
I2 = zeros(1,100);
I3 = zeros(1,100);
%Declare alpha and beta variables detailed in Theorem 1 of paper
a = [0, 2, 4, 6];
b = [2, 3, 7, 10];
% Declare the \lambda_i, i=1,…, 6, variables
L = abs( ( b([2 1 1 1 2 3]) - b([3 3 2 4 4 4]) ) ./ …
         ( a([2 1 1 1 2 3]) - a([3 3 2 4 4 4]) ) );  
for j = 1:6
   D(j) = input(‘Input in1 through in6: ‘);
end
% Iterate through floww()
k = 1:100;
for i = k
   C{i} = floww(D(1), D(2), D(3), D(4), D(5), D(6), L); % Output from floww() is a 6-by-1 vector
   D = C{i}; % Reassign input values to put back into floww()
   % First integrals as described in the paper
   I1(i) = 2sum(D.^2);
   I2(i) = sum( D(1:3).D(4:6) );
   I3(i) = 2sum((L.’).D.^2).^2;
end
plot( k, I1 );
</code>