# How to speed up a for loop ?

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Sleh Eddine Brika 2016 年 10 月 6 日

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));

### 採用された回答

George 2016 年 10 月 6 日
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));

#### 1 件のコメント

Sleh Eddine Brika 2016 年 10 月 6 日
Thanks, I didn't thought about that

### その他の回答 (3 件)

Massimo Zanetti 2016 年 10 月 6 日
Operate on rows, not columns:
angle=asin(A(3,:))/(sqrt(A(3,:).^2+A(2,:).^2+A(1,:).^2));
This will work.

#### 1 件のコメント

Guillaume 2016 年 10 月 6 日
No it won't. The / should be ./
There is also no issue operating on columns or rows (whatever that mean).

Guillaume 2016 年 10 月 6 日

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.

#### 0 件のコメント

elias GR 2016 年 10 月 6 日

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 件のコメント

Sleh Eddine Brika 2016 年 10 月 6 日
Sorry just a typo, A is n*3 matrix
elias GR 2016 年 10 月 6 日
Furthermore, I think that the equation that you use is not correct for 3D vectors.