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mahaveer hanuman
mahaveer hanuman 2011-4-24
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

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采纳的回答

Walter Roberson
Walter Roberson 2011-4-24
You might find this easier to understand:
w = []; for i = 1:30 w = [w, A(i), B(i), C(i)]; end
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Oleg Komarov
Oleg Komarov 2011-4-24
Seriously :)?
Walter Roberson
Walter Roberson 2011-4-24
Yup. Get it working first, and *then* optimize it.

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更多回答（1 个）

Walter Roberson
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));
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Walter Roberson
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
mahaveer hanuman 2011-4-24
thanks its working

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