How to swap matrix ?? Anyone can help me to improve my code.?? I want 12No of 30's & 8 Nos of 0's in every iteration.
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M=[30 30 30 30 30 30 0 0 0 0 0 0 0 0 30 30 30 30 30 30];
p = M(:,:,1);
Np = length(M);
for p = Np/2: -1:2
for CP = p-1:-1:1
if M(1,CP) ~= M(1,CP+1)
M(1,CP) = M(1,CP+1)
end
end
end
%I want to swap looks like
%[30 30 30 30 30 30 30 30 30 30 30 30 0 0 0 0 0 0 0 0]
%"linalg::swapCol" command only works in MuPad
1 Kommentar
Guillaume
am 20 Jan. 2016
Please, use the body (rather than the title) to actually ask your question.
It's not clear at all what you're asking. You've posted some code, half of which is commented out and would do something useful it wasn't, and the other half of which does something useless (just copy the middle element to all elements before). And in the end, you show a desired output which just appears to be the input sorted in descending order.
Antworten (1)
Guillaume
am 20 Jan. 2016
If you're asking for a better implementation of your commented out code above:
M = [30 30 30 30 30 30 0 0 0 0 0 0 0 0 30 30 30 30 30 30];
M2 = dec2bin(0 : pow2(numel(M)) - 1) - '0';
M2 = M2(sum(M2, 2) == nnz(M), :);
values = unique(M);
out = values(M2 + 1)
%check that it generated the right number of permutations:
assert(size(out, 1) == factorial(numel(M)) / (factorial(nnz(M)) * factorial(numel(M) - nnz(M))))
3 Kommentare
Guillaume
am 20 Jan. 2016
That code makes no sense to me, nor do the explanation
p = M(:, :, 1);
That doesn't make much sense when M is a row vector. It is equivalent to:
p = M;
Similarly,
Np = length(M)
only makes sense if M is a row vector.
The rest of the code (the dual loop) is equivalent to:
M(1:floor(Np/2) = M(floor(Np/2));
and most likely not what you want to do.
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