How to place NaN at diagonal position in cell array?
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hey all
a={[],-1,-1,0.8,-0.7,[],[]; [],[],0.9,1,[],-0.9,0.6; -1,[],[],0.9,0.2,[],0.8}
how to place diagonal value in each row of 'a'. Diagonal value can be [] or NaN. Like this
out={NaN,[],-1,-1,0.8,-0.7,[],[]; [],NaN,[],0.9,1,[],-0.9,0.6; -1,[],NaN,[],0.9,0.2,[],0.8}
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Andrei Bobrov
2018년 3월 6일
편집: Andrei Bobrov
2018년 3월 6일
a={[],-1,-1,0.8,-0.7,[],[]; [],[],0.9,1,[],-0.9,0.6; -1,[],[],0.9,0.2,[],0.8};
[m,n] = size(a);
z = sort([repmat(1:n,m,1),(1:m)' - .5],2);
t = rem(z.',1) == 0;
out = cell(size(t));
out(t) = a';
out(~t) = {nan};
out = out';
or
a={[],-1,-1,0.8,-0.7,[],[]; [],[],0.9,1,[],-0.9,0.6; -1,[],[],0.9,0.2,[],0.8};
[m,n] = size(a);
out = num2cell(nan(m,n+1));
out(triu(true(m,n+1),1)) = a(triu(true(m,n)));
out(tril(true(m,n+1),-1)) = a(tril(true(m,n),-1));
추가 답변 (3개)
James Tursa
2018년 3월 6일
Assuming there are at least as many columns as rows:
[m,n] = size(a);
out = cell(n+1,m);
x = logical(eye(size(out)));
out(~x) = a';
out(x) = {nan};
out = out';
Jos (10584)
2018년 3월 6일
a={[],-1,-1,0.8,-0.7,[],[]; [],[],0.9,1,[],-0.9,0.6; -1,[],[],0.9,0.2,[],0.8}
sz = size(a)
out = repmat({NaN}, sz + [0 1])
tf = triu(true(sz))
tfc = false(sz(1),1)
out([tfc tf]) = a(tf)
out([~tf tfc]) = a(~tf)
(btw, why use cell arrays?)
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