Matrix Average of neighbour values

6 次查看（过去 30 天）
Kiwala Martin
Kiwala Martin 2020-2-19
编辑： Kiwala Martin 2020-2-19
After creating a matrix of order MxM it will then output a new matrix in which each element is calculated from the original matrix by finding the average of each element adjacent (horizontally, vertically, excluding diagonal elements) to the original element including the element itself.
For example
M=3
then matrix is
[1,2,3;6,5,4;7,8,9]
element
�(1,1) = (1 + 2 + 6 )/ 3 = 3
�
�(1,2) = (1 + 2 + 3 + 5 ) /4 = 3
  1 个评论
Kiwala Martin
Kiwala Martin 2020-2-19
I have tried using :
Result = conv2(M, ones(3), 'same')./conv2(ones(3),ones(3), 'same')
But this approach includes the diagonal neighbours, how can i exclude them.
Thanks in advance

请先登录，再进行评论。

请先登录，再回答此问题。

采纳的回答

Marta G
Marta G 2020-2-19
Hello!
I think using conv2 was a great idea. However, you should use it like this
DIM=3;
aa=[0 1 0;1 1 1;0 1 0];
matrix=[1,2,3;6,5,4;7,8,9];
conv2(matrix, aa, 'same')./conv2(ones(DIM),aa, 'same')
This way of convolution does not account for diagonal neighbours. Afterwards, if you do not want decimals, use ceil or round depending on your desire.
Hope this helps :)
  3 个评论
显示 1更早的评论
Marta G
Marta G 2020-2-19
编辑：Marta G 2020-2-19
You are welcome! The good thing is that you dont need to change de aa matrix.
M=4
N=3
matrix=[1,2,3;6,5,4;7,8,9;7,8,9]
aa=[0 1 0;1 1 1;0 1 0];
conv2(matrix, aa, 'same')./conv2(ones(M,N),aa, 'same')
Since you are using 'same' : returns the central part of the convolution that is the same size as matrix.
You still have the same dimensions as before while using aa still. With matrix being:
1 2 3
6 5 4
7 8 9
7 8 9
you get
ans =
3.0000 2.7500 3.0000
4.7500 5.0000 5.2500
7.0000 7.4000 7.5000
7.3333 8.0000 8.6667
Kiwala Martin
Kiwala Martin 2020-2-19
编辑：Kiwala Martin 2020-2-19
Thank you very much Marta.
This has been very helpful.

请先登录，再进行评论。

更多回答（0 个）

类别

在 Help Center 和 File Exchange 中查找有关 Matrix Indexing 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by