Coud anyone help me to solve the issue.

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jaah navi
jaah navi am 2 Aug. 2019
Kommentiert: jaah navi am 5 Aug. 2019
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 Kommentare
jaah navi
jaah navi am 2 Aug. 2019
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 am 2 Aug. 2019
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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Andrei Bobrov
Andrei Bobrov am 2 Aug. 2019
Bearbeitet: Andrei Bobrov am 2 Aug. 2019
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
  8 Kommentare
jaah navi
jaah navi am 5 Aug. 2019
Thanks a lot.
jaah navi
jaah navi am 5 Aug. 2019
thanks a lot.

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