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w= 1 0 1
w= 0 1 1
w= 0 1 0
how came this can be written in straight line
w= 1 0 1 0 1 1 0 1 0
采纳的回答
Walter Roberson
2011-4-24
0 个投票
You might find this easier to understand:
w = []; for i = 1:30 w = [w, A(i), B(i), C(i)]; end
2 个评论
Oleg Komarov
2011-4-24
Seriously :)?
Walter Roberson
2011-4-24
Yup. Get it working first, and *then* optimize it.
更多回答（1 个）
Walter Roberson
2011-4-24
It cannot. The first one assigns the row vector [0 1 0] to w, but the second one assigns a row vector of length 9 to w.
If your original data is a 3 x 3 matrix instead, then
w = reshape(w.',1,numel(w));
7 个评论
mahaveer hanuman
2011-4-24
mahaveer hanuman
2011-4-24
Walter Roberson
2011-4-24
What *exactly* do you have for input? And what *exactly* do you want for output?
w = 1 0 1
is not valid MATLAB input.
mahaveer hanuman
2011-4-24
Oleg Komarov
2011-4-24
Preallocate before the loop:
w = zeros(1,30*3);
for ...
w(i*3-2:i*3) = ([A(i),B(i),Q(i)]);
end
Walter Roberson
2011-4-24
N = 30;
w = reshape([reshape(A(1:N),1,N); reshape(B(1:N),1,N); reshape(C(1:N),1,N)], 1, 3*N);
The above does not assume that A, B, or C are row vectors or column vectors. If the shapes are known and consistent, the code can be simplified -- especially if you are putting together _all_ of the vector instead of just a subset of it.
For example, if they are all row vectors and you are using all of them, then
w = [A;B;C]; w = w(:).';
mahaveer hanuman
2011-4-24
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2011-4-24
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