Unable to perform assignment because the indices on the left side are not compatible with the size of the right side

Question

Nima 2020-7-9

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/562157-unable-to-perform-assignment-because-the-indices-on-the-left-side-are-not-compatible-with-the-size-o

编辑： Subhadeep Koley 2020-7-9

采纳的回答： Subhadeep Koley

在 MATLAB Online 中打开

Can anyone solve the problem in my code that returns the error by

?

close all
clear all
clc
%%
mu0 = 4*pi*1e-7; % Vs/Am
M0 = 1e3; % A/m
maxnum = 31;
rho1_min = 0;
rho1_max = 0.25;
xlim = [-1, 1];
ylim = xlim;
zlim = xlim;
x = linspace(min(xlim), max(xlim), maxnum);
y = linspace(min(ylim), max(ylim), maxnum);
z = linspace(min(zlim), max(zlim), maxnum);
[Xg, Yg, Zg] = ndgrid(x, y, z);
rho = sqrt(Xg.^2 + Yg.^2 + Zg.^2);
phi = angle(Xg + 1i*Yg);
theta = angle(Zg + 1i*sqrt(Xg.^2 + Yg.^2));
%%
RHO = sqrt(x.^2 + y.^2 + z.^2);
THETA = linspace(0, pi, 31);                          % Trapz
PHI = linspace(0, 2*pi, 31);                          % Trapz
%% 
for i=1:numel(RHO)
    for j=1:numel(THETA)
        for k=1:numel(PHI)
        
        
F_x{i,j,k} = (RHO(i)>= rho1_max) .* 2/3*M0*mu0 .* sin(theta) .* (RHO(i) .* (sin(THETA(j)) .* cos(theta) .* cos(PHI(k)-phi) - cos(THETA(j)) .* sin(theta)) ./ ...
                                   (RHO(i).^2 + rho1_max.^2 - 2.*RHO(i) .* rho1_max .* (sin(THETA(j)) .* sin(theta) .* cos(PHI(k)-phi) + cos(THETA(j)).* cos(theta))).^3/2) .* rho1_max.^2 .* sin(theta);
                                 
B1x(i,j,k) = -trapz(PHI,trapz(THETA,F_x{i,j,k},2)) ; 
        end
     end 
end

1 个评论
显示 -1更早的评论隐藏 -1更早的评论

madhan ravi 2020-7-9

Couple of suggestions:

1) Never name a variable which is the same as MATLAB’s in - built function (xlim.., etc)

2) i and j are imaginary units use ii and jj instead.

3) preallocation is really significant

4) Use cell arrays for preallocation which avoids ambiguities

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Subhadeep Koley 2020-7-9

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/562157-unable-to-perform-assignment-because-the-indices-on-the-left-side-are-not-compatible-with-the-size-o#answer_463358

编辑：Subhadeep Koley 2020-7-9

在 MATLAB Online 中打开

Pre allocate B1x as cell array instead of numeric array to solve the problem.

close
clear
clc
%%
mu0 = 4*pi*1e-7; % Vs/Am
M0 = 1e3; % A/m
maxnum = 31;
rho1_min = 0;
rho1_max = 0.25;
xlimit = [-1, 1];
ylimit = xlimit;
zlimit = xlimit;
x = linspace(min(xlimit), max(xlimit), maxnum);
y = linspace(min(ylimit), max(ylimit), maxnum);
z = linspace(min(zlimit), max(zlimit), maxnum);
[Xg, Yg, Zg] = ndgrid(x, y, z);
rho = sqrt(Xg.^2 + Yg.^2 + Zg.^2);
phi = angle(Xg + 1i*Yg);
theta = angle(Zg + 1i*sqrt(Xg.^2 + Yg.^2));
%%
RHO = sqrt(x.^2 + y.^2 + z.^2);
THETA = linspace(0, pi, 31);                          % Trapz
PHI = linspace(0, 2*pi, 31);                          % Trapz
%%
% Pre-allocate F_x and B1x as cell array
F_x = cell(numel(RHO), numel(THETA), numel(PHI));
B1x = cell(numel(RHO), numel(THETA), numel(PHI));
for ii = 1:numel(RHO)
    for jj = 1:numel(THETA)
        for kk = 1:numel(PHI)
            F_x{ii, jj, kk} = (RHO(ii)>= rho1_max) .* 2/3*M0*mu0 .* sin(theta) .* (RHO(ii) .* (sin(THETA(jj)) .* cos(theta) .* cos(PHI(kk)-phi) - cos(THETA(jj)) .* sin(theta)) ./ ...
                (RHO(ii).^2 + rho1_max.^2 - 2.*RHO(ii) .* rho1_max .* (sin(THETA(jj)) .* sin(theta) .* cos(PHI(kk)-phi) + cos(THETA(jj)).* cos(theta))).^3/2) .* rho1_max.^2 .* sin(theta);
            B1x{ii, jj, kk} = squeeze(-trapz(PHI,trapz(THETA,F_x{ii,jj,kk},2)));
        end
    end
end