Coud anyone help me to solve the issue.

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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.
  3 个评论
jaah navi
jaah navi 2019-8-2
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 2019-8-2
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??

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采纳的回答

Andrei Bobrov
Andrei Bobrov 2019-8-2
编辑:Andrei Bobrov 2019-8-2
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
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