"Assignment has more non-singleton rhs dimensions than non-singleton subscripts"

Hey all, having the problem "Assignment has more non-singleton rhs dimensions than non-singleton subscripts" as stated above.
Firstly, I know what this problem means and where the error is occuring, I just cannot figure out a solution for it.
My code is as followed:
a0 = [0 0 0];
b0 = a0/3 .* [1 1 1];
a1 = a0/2 .* [0 1 1];
a2 = a0/2 .* [1 0 1];
a3 = a0/2 .* [1 1 0];
In = n .*a1 + m .*a2 + l .*a3;
P = n .*a1 + n .*a2 + l .*a3 + b0;
for n = 1:5
for m = 1:5
for l = 1:5
In(n+1,m+1,l+1,:) = a0;
P(n+1,m+1,l+1,:) = a0 + b0;
end
end
end
The error occurs in:
In(n+1,m+1,l+1,:) = a0;

댓글 수: 3

I've solved it!
Working code is:
a0 = [0 0 0];
b0 = a0/3 .* [1 1 1];
a1 = a0/2 .* [0 1 1];
a2 = a0/2 .* [1 0 1];
a3 = a0/2 .* [1 1 0];
In = n .*a1 + m .*a2 + l .*a3;
P = n .*a1 + n .*a2 + l .*a3 + b0;
for n = 1:5
for m = 1:5
for l = 1:5
In(n+1,m+1,l+1,1:3) = a0';
P(n+1,m+1,l+1,1:3) = a0' + b0';
end
end
end
Note that the lines
In = n .*a1 + m .*a2 + l .*a3;
P = n .*a1 + n .*a2 + l .*a3 + b0;
are superfluos and that you can get the same result for In and P if you just initialize them to zero:
In = zeros(6,6,6,3);
P = zeros(6,6,6,3);
Best wishes
Torsten.
Thanks Torsten, I'm always looking for ways to improve my coding!

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도움말 센터File Exchange에서 Creating and Concatenating Matrices에 대해 자세히 알아보기

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2016년 6월 20일

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2016년 6월 20일

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