Create a matrix from 2 vectors
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I have a vector A =[ 1;3;1;4] and a vector B = [ 8,9,9,6]. The values of A must increase one by one till reaching the value speficied in B, then, how can I obtain the following matrix:
[1 2 3 4 5 6 7 8 0; 3 4 5 6 7 8 9 0 0; 1 2 3 4 5 6 7 8 9; 4 5 6 0 0 0 0 0 0]
without using a for loop?
Thank your very much for your help
采纳的回答
Star Strider
2017-1-8
This creates your matrix with an expressed loop, however there are obviously loops within the functions.
The Code —
A = [ 1;3;1;4];
B = [ 8,9,9,6];
C = ones(size(A,1), max(B));
C = cumsum(C,2);
C = bsxfun(@plus, C, A-1);
I = bsxfun(@le, C, B(:));
C = C.*I % Desired Result
C =
1 2 3 4 5 6 7 8 0
3 4 5 6 7 8 9 0 0
1 2 3 4 5 6 7 8 9
4 5 6 0 0 0 0 0 0
The ‘C’ matrix is (obviously) the output.
6 个评论
Star Strider
2017-1-8
Jose Luis’s ‘Answer’ moved here:
Thank you for your answer, that works. It seems to be a little long, is there any faster way of doing this?
Star Strider
2017-1-8
My pleasure.
A 5-line solution to your problem is long?
That 5-line solution to your problem is the most efficient approach without a loop (that you wrote you wanted to avoid) that exists, and a loop would be less efficient. You could combine the first two lines into one (creating a 4-line solution):
C = cumsum(ones(size(A,1), max(B)),2);
but I doubt that would significantly speed my code. I kept the lines separate in order to make my code easier to read and understand. The rest of the code must remain as it is.
It is fast, because bsxfun is very efficient, and faster than many alternative approaches to do the same things.
If my Answer solves your problem, please Accept it!
Guillaume
2017-1-8
Slight correction to Star's code. It should be:
C = ones(size(A, 1), max(B(:) - A(:) + 1));
C = cumsum(C, 2);
or
C = cumsum(ones(size(A, 1), max(B(:) - A(:) + 1)), 2);
Star Strider
2017-1-8
Thank you, Guillaume.
My code does work, as posted. Your changes make it more robust, an important consideration. (I was just happy to get the logic working correctly yesterday!)
Jose Luis
2017-1-11
编辑：Star Strider
2017-1-11
Star Strider
2017-1-11
As always, my pleasure!
更多回答（1 个）
Guillaume
2017-1-8
There is nothing wrong with using loops when they make the code clearer. This is, arguably, the most efficient loop version:
C = zeros(numel(A), max(B(:) - A(:)) + 1);
for row = 1:numel(A)
C(row, 1:B(row)-A(row)+1) = A(row):B(row);
end
Another option, not using explicit loops, shorter but probably far less efficient than Star's answer:
ncols = max(B(:) - A(:)) + 1;
C = cell2mat(arrayfun(@(s, e) [s:e, zeros(1, ncols-e+s-1)], A, B(:), 'UniformOutput', false))
1 个评论
Star Strider
2017-1-11
Guillaume, thank you for the corrections to my code.
+1
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