How to creating 3 nested for loop to obtain 3d matrix?
조회 수: 3 (최근 30일)
이전 댓글 표시
Hi Everyone,
I am trying to run the following code and obtain a 3d matrix but i got following error.
"Subscript indices must either be real positive integers or logicals.
Error in coilsens (line 13)
B(z,alpha,beta) = z-y*tand(beta)+x*tand(alpha); "
How can i fix this problem?
Thanks.
clc
clear all
u0 = 4*pi*(10e-7);
y = 10e-3;
x = 10e-3;
B = zeros([11,31,31]);
for z = 2e-3:0.1e-3:3e-3
for alpha = 1:0.1:3
for beta = 1:0.1:3
B(z,alpha,beta) = z-y*tand(beta)+x*tand(alpha);
end
end
end
댓글 수: 0
채택된 답변
Andrei Bobrov
2018년 12월 21일
편집: Andrei Bobrov
2018년 12월 21일
y = 10e-3;
x = 10e-3;
z = 2e-3:0.1e-3:3e-3;
alpha = 1:0.1:3;
beta = 1:0.1:3;
[zz,alp,bet] = ndgrid(z,alpha,beta);
B = zz-y*tand(bet)+x*tand(alp);
or
y = 10e-3;
x = 10e-3;
z = 2e-3:0.1e-3:3e-3;
alpha = 1:0.1:3;
beta = 1:0.1:3;
B = zz(:)-y*tand(reshape(bet,1,1,[]))+x*tand(alp(:)');
댓글 수: 3
Luna
2018년 12월 21일
@Andrei someone please teach me how to use this reshape function right on point everytime like he did!! geniusly avoid for loops amazing!
추가 답변 (1개)
madhan ravi
2018년 12월 21일
clc
clear all
u0 = 4*pi*(10e-7);
y = 10e-3;
x = 10e-3;
z = 2e-3:0.1e-3:3e-3;
alpha = 1:0.1:3;
beta = 1:0.1:3;
B=zeros(numel(z),numel(alpha),numel(beta)); % preallocate
for z1=1:numel(z)
for alpha1 = 1:numel(alpha)
for beta1 = 1:numel(beta)
B(z1,alpha1,beta1) = z(z1)-y*tand(beta(beta1))+x*tand(alpha(alpha1));
end
end
end
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