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calling seperate members of a vector function

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João Teixeira
João Teixeira 2021 年 10 月 25 日
コメント済み: Matt J 2021 年 10 月 28 日
I want to call seperate members of my defined vector function and I don't know how to do it.
For example I defined the vector function as shown:
F=@(x,y,z) [x- y, y^2 , z+2] which shows:
@(x, y, z) [x - y, y ^ 2, z + 2]
how do i call only one member, like the first one "x-y"? Because if I do F(1) I'm defining x as equal to 1.
  1 件のコメント
Ive J
Ive J 2021 年 10 月 25 日
Why not this?
F = @(x) [x(1) - x(2), x(2)^2 , x(3) + 2];
res = F(1:3);
ans = -1



Matt J
Matt J 2021 年 10 月 26 日
編集済み: Matt J 2021 年 10 月 26 日
There's no way to do that. You could do it if the components are held in struct form, e .g.,
F=@(x,y) struct('a',x-y,'b', y^2);
ans = 2
  2 件のコメント
Matt J
Matt J 2021 年 10 月 28 日
You're welcome, but if you consider your question answered, please Accept-click the answer.


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