Coud anyone help me to solve the issue.

I am having a matrix

A=[3.5204    3.7294    3.9112    4.0754    4.2294    4.3787;
         0         0         0         0         0         0;
         0         0         0         0         0         0;
         0         0         0         0         0         0;
    0.4337    0.4255    0.4162    0.4065    0.3967    0.3871]

i want to rearrange the matrix in such a way that the sum of (A,1) and sum of (A,2) should not be equal to zero.

Also the number of non zero values present in each row or column can be more than one.

[madhan ravi]

Show how your desired matrix should look like.

[jaah navi]

A=[3.5204      0       3.9112         0               0            0;
0            0       0              4.0754          0            0.3871;
0            3.7294  0              0               0.3967       0;
0.4337       0       0.4162         0               4.2294       4.3787;
0            0.4255  0              0.4065          0            0]

with respect to this output sum(A,1) and sum(A,2) contain non zero values.

[madhan ravi]

madhan ravi:

A(~sum(A,2),:)=[];
A(:,~sum(A,1))=[]

jaah navi:

I want to have the output in the following manner

A=[3.5204      0       3.9112         0               0            0;
0            0       0              4.0754          0            0.3871;
0            3.7294  0              0               0.3967       0;
0.4337       0       0.4162         0               4.2294       4.3787;
0            0.4255  0              0.4065          0            0]

madhan ravi:

Mind explaining in which logic they are rearranged??

Answer 1

[Andrei Bobrov]

One variant:

[m,n] = size(A);
[~,ii] = sort(rand(m-1,n));
B = A(2:end,:);
An = [A(1,:);B(ii + (m-1)*(0:n-1))];% ATTENTION! If MATLAB < R2016b then use: An = [A(1,:);B(bsxfun(@plus,ii,(m-1)*(0:n-1)))];
jj = mod((1:m)' - (1:n),m) + 1; % for MATLAB < R2016b: jj = mod(bsxfun(@minus,(1:m)',1:n),m) + 1;
jj = jj(:,randperm(n));
out = An(sub2ind([m,n],jj,repmat(1:n,m,1)));

general case:

[m,n] = size(A);
[k,f] = max([m,n]);
p = numel(A);
V = A(randperm(p));
ii = find(V ~= 0, k, 'first');
W = [V(ii),V(setdiff(1:p,ii))];
M = reshape(W,k,[]);
if f == 1
    t = n;
else
    t = m;
end
MM = M(k*mod((1:t) - (1:k)',t) + (1:k)');% ATTENTION! If MATLAB < R2016b then use: MM = M(k*mod(bsxfun(@minus,1:t,(1:k)'),t) + (1:k)');
out = MM(randperm(k),:);
if f == 2
    out = out';
end

[jaah navi]

It show error in line An = [A(1,:);B(ii + (m-1)*(0:n-1))]; stating Error using +

Matrix dimensions must agree.

[madhan ravi]

Andrei +1

@jaaj navi: Use bsxfun() instead of +

[Andrei Bobrov]

On my desktop:

>> A
A =
       3.5204       3.7294       3.9112       4.0754       4.2294       4.3787
            0            0            0            0            0            0
            0            0            0            0            0            0
            0            0            0            0            0            0
       0.4337       0.4255       0.4162       0.4065       0.3967       0.3871
>> [m,n] = size(A);
[~,ii] = sort(rand(m-1,n));
B = A(2:end,:);
An = [A(1,:);B(ii + (m-1)*(0:n-1))];
jj = mod((1:m)' - (1:n),m) + 1;
jj = jj(:,randperm(n));
out = An(sub2ind([m,n],jj,repmat(1:n,m,1)))
out =
       0.4337            0       0.4162       4.0754       4.2294       0.3871
       3.5204            0            0            0            0            0
            0            0            0       0.4065       0.3967       4.3787
            0       3.7294            0            0            0            0
            0       0.4255       3.9112            0            0            0

[madhan ravi]

Andrei the reason jaah is receiving is because he is using version prior to 2016b , so to use implicit expansion , he needs to use bsxfun()

[Andrei Bobrov]

Thank you Madhan!

I forgot about it again :)

[madhan ravi]

: ) , anyhow you're solution is brilliant!

[jaah navi]

Thanks a lot.

[jaah navi]

thanks a lot.