Challenging Question - Finding mean of specific values of matrix and re-entering

Have tried countless times but need help

Writing script which has matrix M and returns a new Matrix where each element of N is the corresponding element of M averaged with its next elements above, below and left and right.

The script must work for any sized square matrix!

Answer 1

[Arnaud]

Fonctionne pour n'importe quel noyau K de taille 3 (facile à passer à une taille quelconque) et matrice M de taille quelconque :

M = [1 2 3; 4 5 6; 7 7 9];
K = [0 1 0;1 1 1;0 1 0];

Y = conv2(padarray(M,[1 1]),K,'same');
OK = conv2(padarray(ones(size(M)),[1 1]),K,'same');
Y = Y(2:end-1,2:end-1)./OK(2:end-1,2:end-1)

[Andrei Bobrov]

+1

Answer 2

[Salaheddin Hosseinzadeh]

Hey Karan,

Sound very easy and typical!

It's nothing but programming and defining some conditions. you need 2 for loops for 2 dimension matrix, one to check for vertical neighbors and one for horizontal neighbors.

Once you're finding the neighbors in for loops you should check not to exceed matrix x or y dimension, and also not getting below 1, there is no index 0 in matlab (C C++ has index 0)

and neighbor definition is the current index - and + 1 as you know.

Find them correctly, add them and put them in a new matrix using the current index.

Oh, BTW, to get the matrix dimension you can use size

size(M,1) or size(M,2) whichever you need!

see MATLAB documentation for size

doc size

Good Luck!

Answer 3

[Andrei Bobrov]

s = size(M);
l = true(s);
l(2:end-1,2:end-1) = false;
l1([1,s(1),numel(M)-[s(1)-1,0]]) = true;
l1 = reshape(l1,s);
l2 = l & ~l1;

N = conv2(M,[0 1 0;1 1 1;0 1 0]/5,'same'); % N = imfilter(M,[0 1 0;1 1 1;0 1 0]/5);
N(l1) = N(l1)*5/3;
N(l2) = N(l2)*5/4;

Answer 4

[stalin]

clear all
A=[1 2 3 4;5 6 7 8; 9 10 11 12; 13 14 15 16]
[l m]=size(A)

for i=1:l
    for j=1:m
        if i==1&&j==1
            B(i,j)=(A(i,j)+A(i+1,j)+A(i,j+1))/3
        elseif i==1&&j==m
            B(i,j)=(A(i,j)+A(i+1,j)+A(i,j-1))/3
        elseif i==l&&j==1
            B(i,j)=(A(i,j)+A(i-1,j)+A(i,j+1))/3
        elseif i==l&&j==m
            B(i,j)=(A(i,j)+A(i-1,j)+A(i,j-1))/3
        elseif i==1
            B(i,j)=(A(i,j)+A(i+1,j)+A(i,j-1)+A(i,j+1))/4
        elseif i==l
            B(i,j)=(A(i,j)+A(i,j+1)+A(i,j-1)+A(i-1,j))/4
        elseif j==1
            B(i,j)=(A(i,j)+A(i-1,j)+A(i+1,j)+A(i,j+1))/4
        elseif j==m
            B(i,j)=(A(i,j)+A(i+1,j)+A(i-1,j)+A(i,j-1))/4
        else
%             B(i,j)=0
           B(i,j)= (A(i,j)+A(i-1,j)+A(i,j+1)+A(i+1,j)+A(i,j-1))/5
        end
    end
end

[Image Analyst]

Oh my gosh. Please take a look at what conv2() does.