How I equate the arrangement of elements two matrices???

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Immanuel Saragih
Immanuel Saragih 2018 年 6 月 24 日
コメント済み: Walter Roberson 2018 年 6 月 24 日
I have two matrices with the same size 5x5, the element of the two matrices is 0 and 1 but arrange of the element is different, and now i want to equate the element of the two matrices, arrangement of elements in matrices B must same with arrangement of elements in matrices A , without eliminate the elements in matrices B
If you know the way please help me....
  2 件のコメント
Walter Roberson
Walter Roberson 2018 年 6 月 24 日
What would the expected answer be?
At first I thought perhaps you were asking for the permutation matrix between the two, but we can see that B is not a permutation of A, because A has both rows and columns that are all 1's but B has no rows or columns that are all 1's.
Immanuel Saragih
Immanuel Saragih 2018 年 6 月 24 日
Yes, A and B is not related matrix, it is different, i just want the position of 0 and 1 in matrix B same with matrix A eventhough not all element

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回答 (2 件)

Walter Roberson
Walter Roberson 2018 年 6 月 24 日
[R, C] = find(A == B);
  5 件のコメント
Image Analyst
Image Analyst 2018 年 6 月 24 日
"i expected B same with A eventhough not all it's element....." I hope you can see that this makes no sense whatsoever. If it's the same, all elements match. If all elements don't match, it's not the same.
Immanuel Saragih
Immanuel Saragih 2018 年 6 月 24 日
So how if the number of each element 0 and 1 is same in A and B.... could you told me the way??

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Image Analyst
Image Analyst 2018 年 6 月 24 日
Why can't you simply do
B = A;
This will make the "arrangement of elements in matrices B must same with arrangement of elements in matrices A" just like you asked for.
  5 件のコメント
Walter Roberson
Walter Roberson 2018 年 6 月 24 日
編集済み: Walter Roberson 2018 年 6 月 24 日
More likely
B(A==1) = 1;
Walter Roberson
Walter Roberson 2018 年 6 月 24 日
The Crystal Ball Toolbox is saying, "Concentrate and ask again."

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