Converting the 3D Matrix to 2D Matrix
1 回表示 (過去 30 日間)
古いコメントを表示
Marimuthu Ananthavelu
2018 年 4 月 16 日
コメント済み: Marimuthu Ananthavelu
2018 年 4 月 18 日
I have a matrix outputs from my function in the following form;
zeros(3,5,5)
Basically, its a result of pairwise comparison of a matrix
(5 x 5)
for
3
different modalities (types).
I wish to convert this matrix with each pairwise comparison matrix separately with respect to different modalities.i.e I wish to have the result as below;
1 2 3
i1 j1 k1
i2 j2 k2
i3 . .
i4 . .
i5 . .
i6 . .
. . .
. . .
. . .
. . .
i25 j25 k25
I checked the example using Reshape and could not get much from it. Can anyone please help me through? Thanks
2 件のコメント
Jan
2018 年 4 月 16 日
I do not understand, what "i1, j1, k1" means. Where are they found in the original data? The problem can be solved by:
y = reshape(permute(reshape(x, Size1), Order2), Size3)
All you need is to find the matching parameters Size1, Order2 and Size3.
採用された回答
Jan
2018 年 4 月 16 日
編集済み: Jan
2018 年 4 月 16 日
The general reordering of elements can be done by:
y = reshape(permute(reshape(x, Size1), Order2), Size3)
In your case (perhaps - still not sure what "i1, j1, k1 etc" means):
x = zeros(3,5,5); x(:) = 1:3*5*5; % Some test data y = reshape(permute(reshape(x, 3, 25), [2, 1]), [25, 3])
If this is the wanted output, there is an abbreviation:
y = reshape(x, 3, 25).'
permute(Matrix, [2,1]) is the same as mtranspose() and the 2nd reshape is not required here. But it is worth to keep the general transformation in mind.
その他の回答 (0 件)
参考
カテゴリ
Help Center および File Exchange で Logical についてさらに検索
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!