大家早上好,
我写了这段代码:
clc
clear all
close all
%Box size
Nx=4096;
Ny=15;
Nz=15;
%Spatial gird resolution
delta=6;
%WT / turbulence condition
UHub=11.4;
HubHt=90;
z0=0.03;
IECturbC='B';
%%INITIALISATION
% definition of constants
twopi=2*pi;
fourpi=4*pi;
sqrt2=sqrt(2);
%constants and derived parameters from IEC
gamma = 3.9; %IEC, (B.12)
alpha = 0.2; %IEC, sect. 6.3.1.2
%set delta1 according to guidelines (chap.6)
if HubHt<=60,
delta1=0.7*HubHt;
else
delta1=42;
end;
%IEC, Table 1, p.22
if IECturbC == 'A',
Iref=0.16;
elseif IECturbC == 'B',
Iref=0.14;
elseif IECturbC == 'C',
Iref=0.12;
else
error('IECturbC can be equal to A,B or C;adjust the input value')
end;
%IEC, sect. 6.3.1.3
b=6.5;
sigma1=Iref*(0.75*UHub+b);
%derived constants
l=0.8*delta1; %IEC, (B.12)
sigmaiso=0.55*sigma1; %IEC, (B.12)
%%MAIN PROGRAM
Cij=zeros(3,3,Nx,Ny,Nz);
k = zeros(3,1); %current vector k
for ikx=1:(Nx),
m = -1.*Nx/2+ikx;
k(1)=m*l/(Nx*delta)*twopi;
for iky=1:(Ny),
m= -1.*Ny/2+iky;
k(2)=m*l/(Ny*delta)*twopi;
for ikz=1:(Nz),
m= -1.*Nz/2+ikz;
k(3)=m*l/(Nz*delta)*twopi;
if k(1)==0,
Cij(:,:,ikx,iky,ikz)=0;
else
kabs=sqrt(k(1)^2+k(2)^2+k(3)^2);
beta= gamma./(kabs.^(2/3));
k0(3)=k(3)+beta.*k(1);
k0abs=sqrt(k(1)^2+k(2)^2+k0(3)^2);
Ek0=1.453*k0abs^4/(1.+k0abs.^2)^(17/6);
C1=beta.*k(1)^2*( k0abs.^2 - 2*k0(3)^2 + beta.*k(1)*k0(3) )/( kabs.^2*( k(1)^2 + k(2)^2 ));
C2=k(2).*k0abs.^2./ (exp( (3/2).*log( k(1).^2 + k(2).^2 ) )) .* atan2( beta.*k(1).* sqrt( k(1)^2 + k(2)^2 ) ,( k0abs.^2 - k0(3).*k(1).*beta));
xhsi1=C1 - k(2).*C2./k(1);
xhsi2=k(2).*C1./k(1) + C2;
Cij(1,1,ikx,iky,ikz)= sigmaiso*sqrt(twopi*pi*l^3.*Ek0/(Nx*Ny*Nz*delta^3.*k0abs.^4))*( k(2).*xhsi1);
Cij(1,2,ikx,iky,ikz)= sigmaiso*sqrt(twopi*pi*l^3.*Ek0/(Nx*Ny*Nz*delta^3.*k0abs.^4))*( k(3) - k(1).*xhsi1 + beta.*k(1));
Cij(1,3,ikx,iky,ikz)= sigmaiso*sqrt(twopi*pi*l^3.*Ek0/(Nx*Ny*Nz*delta^3.*k0abs.^4))*( -k(2));
Cij(2,1,ikx,iky,ikz)= sigmaiso*sqrt(twopi*pi*l^3.*Ek0/(Nx*Ny*Nz*delta^3.*k0abs.^4))*( k(2).*xhsi2 - k(3) - beta.*k(1));
Cij(2,2,ikx,iky,ikz)= sigmaiso*sqrt(twopi*pi*l^3.*Ek0/(Nx*Ny*Nz*delta^3.*k0abs.^4))*( -k(1).*xhsi2);
Cij(2,3,ikx,iky,ikz)= sigmaiso*sqrt(twopi*pi*l^3.*Ek0/(Nx*Ny*Nz*delta^3.*k0abs.^4))*( k(1));
Cij(3,1,ikx,iky,ikz)= sigmaiso*sqrt(twopi*pi*l^3.*Ek0/(Nx*Ny*Nz*delta^3.*k0abs.^4))*( k0abs.^2.*k(2) ./ (kabs.^2));
Cij(3,2,ikx,iky,ikz)= sigmaiso*sqrt(twopi*pi*l^3.*Ek0/(Nx*Ny*Nz*delta^3.*k0abs.^4))*( -k0abs.^2*k(1) ./ (kabs.^2));
Cij(3,3,ikx,iky,ikz)= 0;
end;
end;
end;
end;
我想问你: 1.有没有更快的方法来获得Cij矩阵?当Nx,Ny,Nz增加时,Cij的计算速度相当慢; 2.有没有办法获得情节(kabs,beta)和情节(kabs,Ek0)?
请耐心等待我,我仍然是matlab世界的新人。
先致谢谢,并致以诚挚的问候, 弗朗西斯
答案 0 :(得分:3)
如果你想在matlab中获得良好的表现,你应该尽量矢量化你的代码。
例如,而不是:
for x=1:n
A(x)=x^2
end
做
x=1:n;
A=x.^2;
如果您有多个索引,则可以使用 ndgrid 。所以不要这样做:
for x=1:nx
for y=1:ny
for z=1:nz
A(x,y,z)=x^2+y-2*z;
end
end
end
DO
[x y z]=ndgrid(1:nx,1:ny,1:nz)
A=x.^2+y-2*z
由于您看起来很努力,我已经为您更改了代码。执行时间现在是0.33秒。归化版本是:
clc
clear all
close all
tic
%Box size
Nx=1024;
Ny=15;
Nz=15;
%Spatial gird resolution
delta=6;
%WT / turbulence condition
UHub=11.4;
HubHt=90;
z0=0.03;
IECturbC='B';
%%INITIALISATION
% definition of constants
twopi=2*pi;
fourpi=4*pi;
sqrt2=sqrt(2);
%constants and derived parameters from IEC
gamma = 3.9; %IEC, (B.12)
alpha = 0.2; %IEC, sect. 6.3.1.2
%set delta1 according to guidelines (chap.6)
if HubHt<=60,
delta1=0.7*HubHt;
else
delta1=42;
end;
%IEC, Table 1, p.22
if IECturbC == 'A',
Iref=0.16;
elseif IECturbC == 'B',
Iref=0.14;
elseif IECturbC == 'C',
Iref=0.12;
else
error('IECturbC can be equal to A,B or C;adjust the input value')
end;
%IEC, sect. 6.3.1.3
b=6.5;
sigma1=Iref*(0.75*UHub+b);
%derived constants
l=0.8*delta1; %IEC, (B.12)
sigmaiso=0.55*sigma1; %IEC, (B.12)
Cij2=zeros(3,3,Nx,Ny,Nz);
[x y z]=ndgrid(1:Nx,1:Ny,1:Nz);
k1=(x-Nx/2)*l/(Nx*delta)*twopi;
k2=(y-Ny/2)*l/(Ny*delta)*twopi;
k3=(z-Nz/2)*l/(Nz*delta)*twopi;
kabs=sqrt(k1.^2+k2.^2+k3.^2);
beta= gamma./(kabs.^(2/3));
k03=k3+beta.*k1;
k0abs=sqrt(k1.^2+k2.^2+k03.^2);
Ek0=1.453*k0abs.^4./(1+k0abs.^2).^(17/6);
C1=beta.*k1.^2.*( k0abs.^2 - 2*k03.^2 + beta.*k1.*k03 )./( kabs.^2.*( k1.^2 + k2.^2 ));
C2=k2.*k0abs.^2./ (exp( (3/2).*log( k1.^2 + k2.^2 ) )) .* atan2( beta.*k1.* sqrt( k1.^2 + k2.^2 ) ,( k0abs.^2 - k03.*k1.*beta));
xhsi1=C1 - k2.*C2./k1;
xhsi2=k2.*C1./k1 + C2;
CC=sigmaiso*sqrt(twopi*pi*l^3.*Ek0./(Nx*Ny*Nz*delta^3.*k0abs.^4));
Cij2(1,1,:,:,:)= CC.*( k2.*xhsi1);
Cij2(1,2,:,:,:)= CC.*( k3 - k1.*xhsi1 + beta.*k1);
Cij2(1,3,:,:,:)= CC.*( -k2);
Cij2(2,1,:,:,:)= CC.*( k2.*xhsi2 - k3 - beta.*k1);
Cij2(2,2,:,:,:)= CC.*( -k1.*xhsi2);
Cij2(2,3,:,:,:)= CC.*( k1);
Cij2(3,1,:,:,:)= CC.*( k0abs.^2.*k2 ./ (kabs.^2));
Cij2(3,2,:,:,:)= CC.*( -k0abs.^2.*k1 ./ (kabs.^2));
答案 1 :(得分:0)
我会更广泛地回答你的问题,所以其他人也可以受益。
您应该对for
循环进行矢量化,以使您的代码更快。而不是像:
for i=1:n
for j=1:m
M(i,j)=sqrt(i) + sqrt(j);
end
end
根据以下代码进行Vectorize循环:
[xi,xj] = ndgrid(1:n,1:m);
M = sqrt(xi)+sqrt(xj);