Matrix dimensions must agree

Question

Carolina Homem de Gouveia 2016-12-15

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/317068-matrix-dimensions-must-agree

编辑： Robert 2016-12-15

Here is my code

t = 0:pi/300:pi;
F = sin(t).*(sin(5*t)./(5*t));
% b) Polar
r = abs(F);
% c) Discrete
t1 = 0:pi/30:pi;
dis_F = sin(t1).*(sin(5*t1)./(5*t1));
figure(1);
subplot(3,1,1), plot(t,F), title('Cartesian');
subplot(3,1,2), polar(t,r), title('Polar');
subplot(3,1,3), stem(t1,dis_F), title('Discrete');
% 2. Tridimensional
% a) Cartesian coordinates
t2 = 0:pi/100:pi;
p2 = 0:2*pi/100:2*pi;
[t2,p2] = meshgrid(t2,p2);
F2 = (sin(10*sin(t2).*sin(p2)-3)./((10*sin(t2)).*sin(p2)-3)).*((sin(10*cos(t2)-5))./(10*cos(t2)-5));
% b) Spherical coordinates
x = r.*sin(t2).*cos(p2);
y = r.*sin(t2).*sin(p2);
z = r.*cos(t2);

The rest of my code doesn't run because of my x. It says the matrix dimensions must agree but I don't see how I can change it according to the error. Please help. Thanks.

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。

Answer 1

Robert 2016-12-15

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/317068-matrix-dimensions-must-agree#answer_247370

编辑：Robert 2016-12-15

Sorry i looked into it further. p2 and t2 are the same size

the problem is r r is 1 by 301

p2 and t2 are 101X101

2 个评论
显示无隐藏无

Walter Roberson 2016-12-15

In R2016b at least, t2 and p2 are the same size. It would still be a good idea to use linspace though.

Carolina Homem de Gouveia 2016-12-15

编辑：Carolina Homem de Gouveia 2016-12-15

t2 = linspace(0,pi,100);

p2 = linspace(0,2*pi,100)

[t2,p2] = meshgrid(t2,p2);

F2 = (sin(10*sin(t2).*sin(p2)-3)./((10*sin(t2)).*sin(p2)-3)).*((sin(10*cos(t2)-5))./(10*cos(t2)-5));

% b) Gráfico em coordenadas esféricas x = r.*sin(t2).*cos(p2);

I used linspace but it stayed the same. Besides that the t2 and p2 variables show on the workspace as 101*101double, both of them. Thank you.

请先登录，再进行评论。

Answer 2

Walter Roberson 2016-12-15

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/317068-matrix-dimensions-must-agree#answer_247374

在 MATLAB Online 中打开

Your r was created in your first step based upon t = 0:pi/300:pi; which is length 301. Your t2 and p2 are created in your second step based upon t2 = 0:pi/100:pi; and p2 = 0:2*pi/100:2*pi; which are each length 101, and then you meshgrid() each of those so your t2 and p2 are 101 x 101 by the time you try to combine x = r.*sin(t2).*cos(p2); which is then (1 by 301) .* (101 by 101) .* (101 by 101), which fails.

If you were to create t as the same length as t2 starts, so both length 101, then you would have

(1 by 101) .* (101 by 101) .* (101 by 101)

In any version earlier than R2016b that would not be permitted. In R2016b the first term would be automatically replicated to 101 by 101, giving you (101 by 101) .* (101 by 101) .* (101 by 101) which would be legal. In versions before R2016b you would need to do the replication yourself, either by using repmat() or by using bsxfun()

1 个评论
显示 -1更早的评论隐藏 -1更早的评论

Robert 2016-12-15

编辑：Robert 2016-12-15

I also saw the problem above with the R. I changed my comment above. Although walter beat me to it with a more comprehensive answer but yes the problem is the size of R Make it a habit to keep track of your variable type and size in the base workspace.

请先登录，再进行评论。