how to vectorise this for loop?

2018 8 14
1 个回答
更新时间：2018 8 14
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0 个投票

for i = 2:numHarm
h1HatN = h1HatN + alpha(i-1)*sin(i*2*pi*Jint*n/M) + beta(i-1)*cos(i*2*pi*Jint*n/M);
h1Hat2N = h1Hat2N + alpha(i-1)*sin(i*2*pi*Jnew2*n/M) + beta(i-1)*cos(i*2*pi*Jnew2*n/M);
end

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Stephen23
Stephen23 2018-8-14
编辑：Stephen23 2018-8-14
What are n, M, Jint, alpha and beta? What are the initial values of h1HatN and h1Hat2N? What value does numHarm have?
n = 262144 x 1 array M = 262144 x 1 array Jint = 263 alpha = 1 x 400 array beta = 1 x 400 array
Initial values of h1HatN and h1Hat2N will be an array of size 262144 x 1
Jan
Jan 2018-8-14
@AKHILESH KESAVANUNNITHAN: Please post a piece of code, which is running by copy&paste.
alpha = rand(1, numHarm ); beta = rand(1, numHarm ); % h1HatN = rand(z, 1 ); % h1Hat2N = rand(z, 1 );
alpha = zeros(1,numHarm);
beta = zeros(1,numHarm);
for i = 2:numHarm
k = round(i*Jnew2);
cons1 = sin(pi*(i*Jnew2-k))/sin(pi*(i*Jnew2-k)/M);
cons2 = sin(pi*(i*Jnew2+k))/sin(pi*(i*Jnew2+k)/M);
theta1 = pi*(M-1)*(i*Jnew2-k)/M;
theta2 = pi*(M-1)*(i*Jnew2+k)/M;
cons3 = ((sin(pi*(Jnew2-k))/sin(pi*(Jnew2-k)/M))*(exp((1i)*(a*(Jnew2-k)+phi))) + (sin(pi*(Jnew2+k))/sin(pi*(Jnew2+k)/M))*(exp(-(1i)*(a*(Jnew2+k)+phi))))*(A/(2*M));
Y = [cons1*sin(theta1)+cons2*sin(theta2) cons1*cos(theta1)+cons2*cos(theta2); cons2*cos(theta2)-cons1*cos(theta1) cons1*sin(theta1)-cons2*sin(theta2)];
Z = [2*M*real((yfft(mod(round(i*Jnew2),M)+1)-cons3)) 2*M*imag((yfft(mod(round(i*Jnew2),M)+1)-cons3))]';
% Y inversion start
detY = (Y(1,1)*Y(2,2)-Y(1,2)*Y(2,1));
invY = [Y(2,2) -Y(1,2); -Y(2,1) Y(1,1)]/detY;
% Y inversion end
SecParams = (invY)*Z;
alpha(i-1) = SecParams(1);
beta(i-1) = SecParams(2);
end
% ----- Construct the non-coherent and coherent fundamental -----
n=ones(size(dataRec));
n=cumsum(n)-1;
h1HatN = Amp*cos((2*pi()*Jint*n/M)+phi); % Coherent fundamental
h1Hat2N = Amp*cos((2*pi()*Jnew2*n/M)+phi); % Non-coherent fundamental
dataProc2Ne = dataRec - h1Hat2N; % Error containing the information of harmonics and noise.
% ----- Add the non-coherent harmonics to the non-coherent fundamental
% and coherent harmonics to the coherent fundamental -------------
for i = 2:numHarm
h1HatN = h1HatN + alpha(i-1)*sin(i*2*pi*Jint*n/M) + beta(i-1)*cos(i*2*pi*Jint*n/M);
h1Hat2N = h1Hat2N + alpha(i-1)*sin(i*2*pi*Jnew2*n/M) + beta(i-1)*cos(i*2*pi*Jnew2*n/M);
end

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回答（1 个）

OCDER
OCDER 2018-8-14

0 个投票

Jint = 263;
Jnew2 = 360;
z = 262144;
numHarm = 400;
n = rand(z, 1);
M = rand(z, 1);
alpha = rand(1, numHarm);
beta = rand(1, numHarm);
h1HatN = rand(z, 1);
h1Hat2N = rand(z, 1);
i = [2:numHarm];
h1HatN = h1HatN + sum(alpha(i-1).*sin(i*2*pi*Jint.*n./M) + beta(i-1).*cos(i*2*pi*Jint.*n./M), 2);
h1Hat2N = h1Hat2N + sum(alpha(i-1).*sin(i*2*pi*Jnew2.*n./M) + beta(i-1).*cos(i*2*pi*Jnew2.*n./M), 2);
%check this n/M vs n./M . n/M for 262144x1 / 262144x1 matrix will generate a 512 GB matrix!

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