Functions Matlab, how to modify?

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Jenny Andersen
Jenny Andersen 2019 年 11 月 24 日
コメント済み: Jenny Andersen 2019 年 11 月 25 日
Hi,
so I have two vectors, Q = (a,l,m) and Z = (d,b,c). I have created a function to calculate Z, but the problem is that I don't want the answear to be d= something1, b=something2 and c= something3. I want it to be Z = =[something1, something2, something3] when I call the function.
function [a,l,m]=test1(d,b,c)
a=0;
l=b/sqrt(b^2 + c^2);
m=c/sqrt(d^2 + c^2);
a=a*1;
a=a*-1;
l=l*1;
l=l*-1;
m=m*1;
m=m*-1;
Is it possbible to modify the function in some way to achive this?
  2 件のコメント
Turlough Hughes
Turlough Hughes 2019 年 11 月 24 日
You would write
function Z = test1(d,b,c)
% operations to get a,l,m.
Z=[a l m];
end
KALYAN ACHARJYA
KALYAN ACHARJYA 2019 年 11 月 25 日
編集済み: KALYAN ACHARJYA 2019 年 11 月 25 日
function Z = test1(d,b,c)
a=0;
l=b/sqrt(b^2 + c^2);
m=c/sqrt(d^2 + c^2);
a=a*1;
a=a*-1;
l=l*1;
l=l*-1;
m=m*1;
m=m*-1;
Z=[a l m];
fprintf('Z=[%i %i %i]\n',Z);
end

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採用された回答

Turlough Hughes
Turlough Hughes 2019 年 11 月 25 日
編集済み: Turlough Hughes 2019 年 11 月 25 日
Hi Jenny,
You could write your function to have two outputs as follows:
function [Z1,Z2] = test1(d,b,c)
% code to determine a,l,m
Z1=[a,l,m];
Z2=[a,-l,-m];
end
When calling the function you have to specify two outputs if you want them otherwise matlab does not know what to assign the second output to. Basically:
[Z1,Z2]=test1(d,b,c)
Another thing you could do would be to write a single output where you have 6 elements in a single output variable. That would go something like this:
function Z = test1(d,b,c)
% code to get a,l,m
Z=[a,l,m;a,-l,-m];
end
When you call the function now you get a single output with six elements that you are looking for.
I recommend looking at onramp tutorials. Especially part 7 on calling functions as well as vectors and matrices, part 3. Though, if you intend to use matlab much you are as well to go through all of the sections.
  1 件のコメント
Jenny Andersen
Jenny Andersen 2019 年 11 月 25 日
Thank you alot! It seems to be working very well now. One thing that came to mind is that the value differs a bit from be previous expressions, but I gess that is the way functions work.
I will continue with the Onramp course.

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その他の回答 (1 件)

Jenny Andersen
Jenny Andersen 2019 年 11 月 25 日
編集済み: Jenny Andersen 2019 年 11 月 25 日
So I tried Turlough's suggestion again and it seems to be working fine now. But since two vectors will be generated (-d,b,c) and (d,b,-c) I would like to add this information to my code in some way. I have come up with the following, but still I only get one vector.
function [Z] = test1(d,b,c)
a1=0;
l1=b/sqrt(b^2 + c^2);
m1=-c/sqrt(d^2 + c^2);
a=0;
l=-b/sqrt(b^2 + c^2);
m=c/sqrt(d^2 + c^2);
Z=[a1 l1 m1];
Z=[a l m];
end
  1 件のコメント
KALYAN ACHARJYA
KALYAN ACHARJYA 2019 年 11 月 25 日
編集済み: KALYAN ACHARJYA 2019 年 11 月 25 日
Why you have deleted your previous comments, its may confuse to the answer providers? This is answer section, please post your comment in comment section only.

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