??? Error using ==> eq Matrix dimensions must agree

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Giannakis Stoukas
Giannakis Stoukas 2015 年 4 月 15 日
コメント済み: John D'Errico 2015 年 4 月 15 日
[row, column]=find(My_matrix== pnode); I try to use this command and i get this error
??? Error using ==> eq Matrix dimensions must agree
What should i do?How i should write the syntaxis of it?
  3 件のコメント
Giannakis Stoukas
Giannakis Stoukas 2015 年 4 月 15 日
pnode is a variable and it has a value and the matrix has dimensions 10X14,this is why it is weird
Giannakis Stoukas
Giannakis Stoukas 2015 年 4 月 15 日
Actually the previous orders are min_cost = min(x(:)); [pnode, node] = find(x == min_cost); [row, column]=find(My_matrix == pnode); [rows, columns]=find(My_matrix==node);
and the min_cost instead of taking one value it takes an array

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回答 (1 件)

John D'Errico
John D'Errico 2015 年 4 月 15 日
編集済み: John D'Errico 2015 年 4 月 15 日
Your matrices are not compatible in size. READ THE ERROR MESSAGE!
Try this:
whos MY_matrix pnode
What does it tell you?
  3 件のコメント
Giannakis Stoukas
Giannakis Stoukas 2015 年 4 月 15 日
編集済み: John D'Errico 2015 年 4 月 15 日
Actually the previous orders are
min_cost = min(x(:));
[pnode, node] = find(x == min_cost);
[row, column]=find(My_matrix == pnode);
[rows, columns]=find(My_matrix==node);
and the min_cost instead of taking one value it takes an array
John D'Errico
John D'Errico 2015 年 4 月 15 日
I see that the last line has
find(My_matrix==node);
Even if pnode is a scalar, node may well not be so, since it was created from a previous call to find.

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