Unable to perform assignment because the left and right sides have a different number of elements.

1 次查看（过去 30 天）

显示更早的评论

sci hub 2021-7-19

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/881648-unable-to-perform-assignment-because-the-left-and-right-sides-have-a-different-number-of-elements

评论： sci hub 2021-7-20

在 MATLAB Online 中打开

Hi all,

I am trying to divide a two line (two times) based on different length as the following code:

but I can not

clear all 
clc
t                                       =       [2 0.40 1.2];        
t_ID                                    =       [3  1];                 %  Each line index    
Z                                       =       zeros(2+1,1);
for j = 1:2
  if (j == 1)
        t1 = 0; t2 = t(t_ID(j));
        Z(j) = linspace(t1,t2,2);
   else
        t1 = Z(2*j-1);
        t2 = t1+t_av(t_ID(j));
        Z(2*j-1:2*j) = linspace(t1,t2,2);
  end
end
% Z should be [0 0.2 0.3250]

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

请先登录，再回答此问题。

回答（1 个）

Walter Roberson 2021-7-19

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/881648-unable-to-perform-assignment-because-the-left-and-right-sides-have-a-different-number-of-elements#answer_749718

在 MATLAB Online 中打开

First get rid of the "clear all". Your code relies on variables being in memory, but the clear destroys them.

Z(j) = linspace(t1,t2,2);

linspace requesting two output elements is never going to fit a scalar destination. Also linspace requesting two output elements would just be the vector [t1, t2] so just code that instead of confusing things.

When j becomes 2 then 2*j-1 would be 3 so you would be writing to locations 3 and 4 in the vector. Your code will never create a vector of odd length.