how can I do this mathmatical operation?
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I wish to do sum and subtract in column 2 of 84x7 matrices between different rows of the element on the same column and produce the answers into an array. example @Column 3, a = [ 1 3 3 3 ; 2 2 2 2 ; 3 4 4 4 ; 4 0 1 0 ; 5 5 5 5 ; 1 1 1 1 ; 7 7 7 7 ] desired outcome: => b = [ 3 7 10 ]
5 个评论
madhan ravi
2018-10-22
编辑：madhan ravi
2018-10-22
b = [ 3 7 10 ] is not clear
Rik
2018-10-22
Could you show more of the calculation steps? The calculation is not clear to me. The sum of the columns is [23 22 23 22], so I don't see how any subtraction would result in your output.
Kevin Chng
2018-10-22
编辑：Kevin Chng
2018-10-22
I guess what you want is
for i=1:2:(length(a(:,3))-2)
b(i)= a(i,3)-a(i+1,3)+(a(i+2,3)-a(i+1,3))
end
b(2:2:end)=[];
Why length(a(:,3)-2)? It is to avoid exceed the dimension.
Young Lee
2018-10-23
采纳的回答
Kevin Chng
2018-10-22
I guess what you want is
for i=1:2:(length(a(:,3))-2)
b(i)= a(i,3)-a(i+1,3)+(a(i+2,3)-a(i+1,3))
end
b(2:2:end)=[];
Why length(a(:,3)-2)? It is to avoid exceed the dimension.
2 个评论
Jan
2018-10-22
Use size(a, 1) instead of length(a(:, 3)), because it is more efficient and nicer.
Rik
2018-10-22
To expand a bit on Jan's comment: using length can get you into trouble, because it is equivalent to max(size(A)). That means that you need to be sure that the dimension that is relevant for you will always be the largest. Using size with a specified dimension will avoid this problem. If you want to iterate through all elements of a vector, it is safest to use numel, which is equivalent to prod(size(A)).
更多回答（1 个）
This works without a loop:
n = size(a, 1);
b = a(1:2:n-2, 3) - 2 * a(2:2:n-1, 3) + a(3:2:n, 3)
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