how can I do this mathmatical operation?

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Young Lee
Young Lee 2018-10-22
评论： Young Lee 2018-10-23
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 ]
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Kevin Chng
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
Young Lee 2018-10-23
Thanks that worked out perfectly with little change @@

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Kevin Chng
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
Jan 2018-10-22
Use size(a, 1) instead of length(a(:, 3)), because it is more efficient and nicer.
Rik
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)).

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更多回答（1 个）

Jan
Jan 2018-10-22
编辑：Jan 2018-10-22
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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