How to speed up a for loop ?

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Sleh Eddine Brika
Sleh Eddine Brika 2016-10-6
编辑： elias GR 2016-10-6
I have a matrix A n*3 of normal, I want to calculate the angles as shown in the code
angle=zeros(length(A),1);
for i = 1 : length(A)
n=A(i,:);
angle(i)=asin((n(3))/(sqrt(n(1).^2+n(2).^2+n(3).^2)));
end
It works but since I am dealing with really big matrices I need to speed this up. I tried this way, but it doesn't works.
angle=asin(A(:,3))/(sqrt(A(:,3).^2+A(:,2).^2+A(:,1).^2));

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采纳的回答

George
George 2016-10-6
Are you sure this is correct?
angle=asin(A(:,3))/(sqrt(A(:,3).^2+A(:,2).^2+A(:,1).^2));
That's doing matrix division. In your example, because of your loop, you are doing elementwise division. Fso
angle=asin(A(:,3)) ./ (sqrt(A(:,3).^2+A(:,2).^2+A(:,1).^2));

更多回答（3 个）

Massimo Zanetti
Massimo Zanetti 2016-10-6
Operate on rows, not columns:
angle=asin(A(3,:))/(sqrt(A(3,:).^2+A(2,:).^2+A(1,:).^2));
This will work.
  1 个评论
Guillaume
Guillaume 2016-10-6
编辑：Guillaume 2016-10-6
No it won't. The / should be ./
There is also no issue operating on columns or rows (whatever that mean).

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Guillaume
Guillaume 2016-10-6
编辑：Guillaume 2016-10-6
It looks like A is a 2D matrix with a variable number of rows and 3 columns. If so, does not use length for getting the number of rows as it will return the number of columns if you have less than 3 rows. Use size(A, 1) to get the number of rows.
No loop is needed to get your result:
angle = asin(A(:, 3) ./ sqrt(sum(A.^2, 2)))
Your issue is that you want to do elementwise division so you need ./ instead of /.
I've also simplified your square root expression.
elias GR
elias GR 2016-10-6
编辑：elias GR 2016-10-6
If A have 3 rows and n columns, try that:
angle=asin(A(3,:))./(sqrt(A(3,:).^2+A(2,:).^2+A(1,:).^2));
  2 个评论
Sleh Eddine Brika
Sleh Eddine Brika 2016-10-6
Sorry just a typo, A is n*3 matrix
elias GR
elias GR 2016-10-6
编辑：elias GR 2016-10-6
Furthermore, I think that the equation that you use is not correct for 3D vectors.

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