Matrix dimensions must agree.

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y0unes
y0unes 2017 年 8 月 2 日
回答済み: Jan 2017 年 8 月 2 日
This function make a coordinates from sensor in another coordintes and scale from mm to m. but it gives me the following error message (Matrix dimensions must agree.) at line: Bix = Bix .* sign(MP).* 1000;
% code
MP = [P0(:,3), P0(:,4), P0(:,5)] ./ 1000;
Bix = [(interp3 (xs, ys, zs, Bxs, abs(MP(:,1)), abs(MP(:,2)), abs(MP(:,3))));...
(interp3 (xs, ys, zs, Bys, abs(MP(:,1)), abs(MP(:,2)), abs(MP(:,3))));...
(interp3 (xs, ys, zs, Bzs, abs(MP(:,1)), abs(MP(:,2)), abs(MP(:,3))))];
Bix = Bix .* sign(MP).* 1000;
Bix = Bix .* [1, -1, -1];
  5 件のコメント
KSSV
KSSV 2017 年 8 月 2 日
@Jose-Luis..Run this
K = rand(1,10).*rand(1,5)
José-Luis
José-Luis 2017 年 8 月 2 日
編集済み: José-Luis 2017 年 8 月 2 日
You're not getting my point, I think.
Arrays no longer need to have the same number of elements for such arithmetic operations to produce results.
Unsolicited disclaimer: I am having a hard time liking that. I'll eventually get on with the times.

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回答 (1 件)

Jan
Jan 2017 年 8 月 2 日
Use the debugger to identify the problem. Set a breakpoint in the failing line:
Bix = Bix .* sign(MP).* 1000;
Then run the code again until it stops. Not check the dimensions:
size(Bix)
size(MP)
For a proper elementwise multiplication, the dimensions must match. Do they?
Note that Matlab >= 2016b can auto-expand dimensions. With older versions, bsxfun helps. Perhaps you want:
Bix = bsxfun(@times, Bix, sign(MP)) * 1000;
Maybe a reshape helps here to adjust the dimensions as wanted.

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