How to vectorize these codes

Dear all,

these days i am dosing some modification on my code with vectorization. For the following codes:

function y=gam(a,m,n)
% cos(m*pi*x/a)*sin(n*pi*x/a)
if m==0
    if n==m
        y=0;
    else
        y=-a*(-1+(-1)^n)/(n*pi);
    end
else
    if n==m
        y=0;
    else
        y=a*n*(-1+(-1)^(m+n))/(pi*(m^2-n^2));
    end
end

I vectorlize it, but when m equal to n, the results is NAN. So please help me on vectorlizing it.

Thank you.

[Code formated, Jan - Please format your code in the future, thanks]

Answer 1

[Jan]

There is no loop and therefore it is not trivial to guess, which operation you want to be vectorized.

[EDITED] Now your problem got clear, thanks.

function y = gam(a,m,n)

y = zeros(size(m));

index    = (m == 0 & n ~= m);
y(index) = -a*(-1+(-1).^n(index)) ./ (n(index)*pi);
index    = (m ~= 0 & n ~= m);
y(index) = a.*n(index).*(-1+(-1).^(m(index)+n(index))) ./ ...
           (pi*(m(index).^2 - n(index) .^ 2));
end

Not tested because I cannot run Matlab currently.

But this is a guess only. I cannot conclude what the vectorized version of "n == m" should be. Is it "ismember(n, m)" or "n == m" or "any(n == m, 2)" etc?

[Xianjie]

m and n are matrix. So i want to vectorize this code.

Answer 2

[Robert Cumming]

help isnan

[Xianjie]

I just find this command. But is there some other method to vectorize these codes without NaN in the results?

function y=gam(a,m,n) % cos(m*pi*x/a)*sin(n*pi*x/a) if m==0 if n==m y=0; else y=-a*(-1+(-1)^n)/(n*pi); end else if n==m y=0; else y=a*n*(-1+(-1)^(m+n))/(pi*(m^2-n^2)); end end

I use this code to vectorize.

y = a*(~(m==n)).*((n.*(-1+(-1).^(m+n)))./(pi*(m.^2-n.^2)));

m and n also the matrix

[Robert Cumming]

its not vectorising thats the issue here - you have 0/0 when m=n => you get a NaN. use a logical check on the answer for isnan to covert it to a zero (as you indicated above).

[Xianjie]

Ok. Thank. Is there some other ways to solve this problem. Because if the dimension of matrix is large, the for loop will cost a lot of time. Thank you

[Robert Cumming]

m = [1 2 0 4 5];
n = [3 3 0 3 3];
y = m./n
y(isnan(y)) = 0

No for loop required - as I said use logical indexing and isnan.