Thread Subject: Array of Repeated Values

Hi,

I am currently working on a project that has a doubly-
nested for loop. I wish to vectorize the loop as its takes
too long, but am having problems doing so.

Here is a snippet of the code

for theta1 = -pi/3: pi/40: pi/3
    for theta2 = -2*pi/3: pi/40: 2*pi/3
        for theta3 = -pi/2: pi/40: pi/2

         x = 0.4*cos(theta1) + 0.4*cos(theta1 + theta2)...
             + 0.2*cos(theta1 + theta2 + theta3);
         y = 0.4*sin(theta1) + 0.4*sin(theta1 + theta2)...
             + 0.2*sin(theta1 + theta2 + theta3);
         xvec = [xvec,x];
         yvec = [yvec,y];

        end
    end
end

My idea is to create theta1, theta2, theta3 as arrays with
all necessary values so I can just calculate the vectors in
one line. Now, as far as I know, vector operations are
element-wise, so to create the three arrays they would need
an element for each potential value (in this case array
length of 56889) and the values of the array would be
repeated values in match the size.

If the previous paragraph is confusing consider the
following example:

i = [1, 1, 1, 1, 1, 1, 1, 1, 1,...
     5, 5, 5, 5, 5, 5, 5, 5, 5,...
     9, 9, 9, 9, 9, 9, 9, 9, 9]

j = [1, 1, 1, 5, 5 ,5, 9, 9, 9,...
     1, 1, 1, 5, 5 ,5, 9, 9, 9,...
     1, 1, 1, 5, 5 ,5, 9, 9, 9]

k = [1, 5, 9, 1, 5, 9, 1, 5, 9,...
     1, 5, 9, 1, 5, 9, 1, 5, 9,...
     1, 5, 9, 1, 5, 9, 1, 5, 9]

where i, j, k are all potential values (that the for loop
could have)

Any advice on how to create the three arrays would be
greatly appreciated

Thanks Braddon

"Brandon Navra" <Braddon@sogetthis.com> wrote in message <fsb57i$4sg
$1@fred.mathworks.com>...
> Hi,
>
> I am currently working on a project that has a doubly-
> nested for loop. I wish to vectorize the loop as its takes
> too long, but am having problems doing so.
>
> Here is a snippet of the code
>
> for theta1 = -pi/3: pi/40: pi/3
> for theta2 = -2*pi/3: pi/40: 2*pi/3
> for theta3 = -pi/2: pi/40: pi/2
>
> x = 0.4*cos(theta1) + 0.4*cos(theta1 + theta2)...
> + 0.2*cos(theta1 + theta2 + theta3);
> y = 0.4*sin(theta1) + 0.4*sin(theta1 + theta2)...
> + 0.2*sin(theta1 + theta2 + theta3);
> xvec = [xvec,x];
> yvec = [yvec,y];
>
> end
> end
> end
>
> My idea is to create theta1, theta2, theta3 as arrays with
> all necessary values so I can just calculate the vectors in
> one line. Now, as far as I know, vector operations are
> element-wise, so to create the three arrays they would need
> an element for each potential value (in this case array
> length of 56889) and the values of the array would be
> repeated values in match the size.
>
> If the previous paragraph is confusing consider the
> following example:
>
> i = [1, 1, 1, 1, 1, 1, 1, 1, 1,...
> 5, 5, 5, 5, 5, 5, 5, 5, 5,...
> 9, 9, 9, 9, 9, 9, 9, 9, 9]
>
> j = [1, 1, 1, 5, 5 ,5, 9, 9, 9,...
> 1, 1, 1, 5, 5 ,5, 9, 9, 9,...
> 1, 1, 1, 5, 5 ,5, 9, 9, 9]
>
> k = [1, 5, 9, 1, 5, 9, 1, 5, 9,...
> 1, 5, 9, 1, 5, 9, 1, 5, 9,...
> 1, 5, 9, 1, 5, 9, 1, 5, 9]
>
> where i, j, k are all potential values (that the for loop
> could have)
>
> Any advice on how to create the three arrays would be
> greatly appreciated
>
> Thanks Braddon
----------
  Use ndgrid:

 [t3,t2,t1] = ...
  ndgrid(-pi/2:pi/40:pi/2,-2*pi/3:pi/40:2*pi/3,-pi/3:pi/40:pi/3);
 t1 = t1(:).'; t2 = t2(:).'; t3 = t3(:).'; % Make row vectors
 xvec = 0.4*cos(t1)+0.4*cos(t1+t2)+0.2*cos(t1+t2+t3);
 yvec = 0.4*sin(t1)+0.4*sin(t1+t2)+0.2*sin(t1+t2+t3);

  You should note that, whereas the highest theta3 (t3) here will be pi/2, the
highest values of theta1 and theta2 will not be pi/3 and 2*pi/3, resp.,
because 40 is not divisible by 3.

Roger Stafford

> Use ndgrid:
>
> [t3,t2,t1] = ...
> ndgrid(-pi/2:pi/40:pi/2,-2*pi/3:pi/40:2*pi/3,-pi/3:pi/
40:pi/3);
> t1 = t1(:).'; t2 = t2(:).'; t3 = t3(:).'; % Make row
vectors
> xvec = 0.4*cos(t1)+0.4*cos(t1+t2)+0.2*cos(t1+t2+t3);
> yvec = 0.4*sin(t1)+0.4*sin(t1+t2)+0.2*sin(t1+t2+t3);
>
> You should note that, whereas the highest theta3 (t3)
here will be pi/2, the
> highest values of theta1 and theta2 will not be pi/3 and
2*pi/3, resp.,
> because 40 is not divisible by 3.
>
> Roger Stafford
>

Roger,

Thank you for your help. This is exactly what I wanted.
Since the speed up is so substantial I can adjust the
intervals to pi/120,pi/90 and pi/180 (to get better
accuracy).

Brandon