I am very new to MATLAB but I have to deliver an assignment. How do I solve matrix dimension issue in line 10 and 11.
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Frederick Kwame
2024 年 6 月 11 日
コメント済み: Frederick Kwame
2024 年 6 月 12 日
Given the following matrix dimensions
L=39360*39360
y=39360*1
q=1*39360
E= 2.8709
X=39360*39360
T=39360*39360
code
1. dq=L*y;
2. dtr=dq.*q'/E;
3. ql = q * L;
4. ldf = L * y;
5. dt = zeros(39360, 39360);
6. dtr = zeros(39360, 39360);
7. for a = 1:39360
8. for b = 1:39360
9. if X(b, :) ~= 0
10. dt(a,b) = ql(:,a)*ldf(b,:) /X(b,:);
11. dtr(a,b) = (ql(:,a)*ldf(b,:)* T))/(X(b,:)*E);
else
dt(a, b) = 0;
dtr(a, b) = 0;
end
end
end
1 件のコメント
Walter Roberson
2024 年 6 月 11 日
N = 39360;
L = rand(N, N);
y = rand(N, 1);
q = rand(1, N);
E = 2.8709;
X = rand(N, N);
T = rand(N, N);
dq=L*y;
dtr=dq.*q'/E;
ql = q * L;
ldf = L * y;
dt = zeros(39360, 39360);
dtr = zeros(39360, 39360);
for a = 1:39360
for b = 1:39360
if X(b, :) ~= 0
dt(a,b) = ql(:,a)*ldf(b,:) /X(b,:);
dtr(a,b) = (ql(:,a)*ldf(b,:)* T))/(X(b,:)*E);
1 0 1 2 1 2 1 0? 0 1 0 ?
else
dt(a, b) = 0;
dtr(a, b) = 0;
end
end
end
You have too many )
採用された回答
Walter Roberson
2024 年 6 月 11 日
移動済み: Walter Roberson
2024 年 6 月 11 日
Your ldf is 39360 x 1. Your ql is 1 x 3936. Select rows and columns out of those and you get 1 x 1 and 1 x 1. Use * to multiply two 1 x 1 and you get 1 x 1. You cannot / between 1 x 1 and a 1 x something (not unless the "something" is 1)
%N = 39360; %too large for actual test
N = 3936;
L = rand(N, N);
y = rand(N, 1);
q = rand(1, N);
E = 2.8709;
X = rand(N, N);
T = rand(N, N);
dq=L*y;
dtr=dq.*q'/E;
ql = q * L;
ldf = L * y;
whos ql ldf
dt = zeros(N, N);
dtr = zeros(N, N);
for a = 1:N
for b = 1:N
if X(b, :) ~= 0
temp1 = ql(:,a);
temp2 = ldf(b,:);
temp3 = temp1 * temp2;
temp4 = X(b,:);
whos temp1 temp2 temp3 temp4
dt(a,b) = temp3 / temp4;
temp4 = (ql(:,a)*ldf(b,:)* T);
temp5 = (X(b,:)*E);
whos temp4 temp5
dtr(a,b) = (ql(:,a)*ldf(b,:)* T)/(X(b,:)*E);
else
dt(a, b) = 0;
dtr(a, b) = 0;
end
end
end
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