how can I do this mathmatical operation?

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Young Lee 2018-10-22

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https://ww2.mathworks.cn/matlabcentral/answers/425325-how-can-i-do-this-mathmatical-operation

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 2018-10-22

编辑：Kevin Chng 2018-10-22

在 MATLAB Online 中打开

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

Thanks that worked out perfectly with little change @@

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采纳的回答

Kevin Chng 2018-10-22

0
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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/425325-how-can-i-do-this-mathmatical-operation#answer_342654

在 MATLAB Online 中打开

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.

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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)).

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