概述
在数据量很大的情况下,用矩阵运算可代替FOR循环,使运算速度大大加快。
我想问的是,如果一个FOR循环中,有很多if、continue等条件语句或循环语句,还能变成矩阵运算吗?
我有个程序需要循环几百万次,速度非常慢,请大家赐教!谢谢。
例如下面的程序,
for i=1:cyc-1
if (Smax(i)>=Smaxol(i));
Smaxol(i+1)=Smax(i);
ol=i;
else
Smaxol(i+1)=Smax(ol);
end
Yre=sqrt((1.0/cos(PI*re/w)));
alpha0=KIc/(sigmau*sqrt(PI*re)*(1.0+Yre/sqrt(2.0)));
if alpha0>sqrt(2.0);
sigmav=sigmau*PI/2.0/acos(1.0/(alpha0^2.0-1.0));
else
error('PLEASE TO CHECK THE VALUE OF ALHPA0!');
break;
end
R(i)=Smin(i)/Smax(i);
Y=sqrt((1.0/cos(PI*a(i)/w)));
Yol=sqrt((1.0/cos(PI*a(ol)/w)));
Kmax(i)=sqrt(PI*re*(1.0/cos(PI*Smax(i)/2.0/sigmav)+1.0))*(1.0+Y*sqrt(a(i)/2.0/re))*Smax(i);
Kmaxol(i)=sqrt(PI*re*(1.0/cos(PI*Smaxol(i+1)/2.0/sigmav)+1.0))*(1.0+Yol*sqrt(a(ol)/2.0/re))*Smaxol(i+1);
deltaK(i)=(1-R(i))*Kmax(i);
alpha_1=1.0/(1.0-2.0*mu)+(1.0-1.0/(1.0-2.0*mu))/(1.0+0.8861*(t/(Kmax(i)/sigmay)^2.0)^3.2251)^0.75952;
AA0=(0.825-0.34*alpha_1+0.05*alpha_1^2.0)*(cos(PI*Smax(i)/2.0/sigmaf))^(1.0/alpha_1);
AA1=(0.415-0.071*alpha_1)*Smax(i)/sigmaf;
AA3=2.0*AA0+AA1-1.0;
AA2=1.0-AA0-AA1-AA3;
if 0<=R(i) && R(i)<1.0;
fop=max(R(i),AA0+AA1*R(i)+AA2*R(i)^2.0+AA3*R(i)^3.0);
else
fop=AA0+AA1*R(i);
end
if R(i)>=-2.0 && R(i)<0.0;
feff(i)=(0.52-0.1*R(i))/(1.0-R(i));
if mat==1.0;
fth(i)=(1.0-R(i))^0.84;
elseif mat==2.0;
fth(i)=(1.0-R(i))^0.5;
end
end
if R(i)>=0.0 && R(i)<0.5;
feff(i)=0.52+0.42*R(i)+0.06*R(i)^2.0;
if mat==1.0;
fth(i)=(1-R(i))^0.7;
elseif mat==2.0 ;
fth(i)=(1-R(i))^0.3;
end
end
if R(i)>=0.5 && R(i)<1.0;
feff(i)=0.52+0.42*R(i)+0.06*R(i)^2.0;
if mat==1.0;
fth(i)=(1.05-1.4*R(i)+0.6*R(i)^2.0)^0.7;
elseif mat==2.0;
fth(i)=(1.05-1.4*R(i)+0.6*R(i)^2.0)^0.3;
end
end
deltaKeffth(i)=feff(i)*fth(i)*deltaKth0;
deltaKu=Y*sqrt(PI*a(i))*(Smin(i)-Smin(i+1));
alpha1=(1.0-1.65*mu)^2.0/5.0-1.0/(20.0*n0)*(1.0-1.65*mu)^(2.0/n0)+(1.0/PI-1.0/(2.2*n0)*(1.0/PI)^(1.0/n0)-((1.0-1.65*mu)^2.0/5.0-1.0/(20.0*n0)*((1.0-1.65*mu)^2.0)^(1.0/n0)))/(1.0+(t/(Kmax(i)/sigmay)^2.0)/(1.0+1.0/n0))^(1.6+1.0/n0);
alpha2=(1.0-1.65*mu)^2.0/5.0-1.0/(20.0*n0)*(1.0-1.65*mu)^(2.0/n0)+(1.0/PI-1.0/(2.2*n0)*(1.0/PI)^(1.0/n0)-((1.0-1.65*mu)^2.0/5.0-1.0/(20.0*n0)*((1.0-1.65*mu)^2.0)^(1.0/n0)))/(1.0+(t/(Kmaxol(i)/sigmay)^2.0)/(1.0+1.0/n0))^(1.6+1.0/n0);
ry(i)=alpha1*(Kmax(i)/sigmay)^2.0;
rol(i)=alpha2*(Kmaxol(i)/sigmay)^2.0;
deltar=alpha1*(deltaKu/sigmay)^2.0;
lamda=(1.0-1.65*mu)^2.0/5.0-1.0/(20.0*n0)*(1.0-1.65*mu)^(2.0/n0)+(1.0/PI-1.0/(2.2*n0)*(1.0/PI)^(1.0/n0)-((1.0-1.65*mu)^2.0/5.0-1.0/(20.0*n0)*((1.0-1.65*mu)^2.0)^(1.0/n0)))/(1.0+(t/(Kmax(i)/sigmay)^2.0)/(1.0+1.0/n0))^(1.6+1.0/n0);
Kc=(((1.0-2.0*mu)^2.0-sqrt(1.0-mu^2.0))*PI*lamda/((1.0-2.0*mu)^2.0-1.0)/(1.0-2.0*mu)^2.0+(sqrt(1.0-mu^2.0)-1.0)/((1.0-2.0*mu)^2.0-1.0))*KIc;
if a(i)+ry(i)
mf=((a(ol)+rol(i)-ry(i)-deltar)/a(i))^c;
deltaKeff(i)=Kmax(i)*(1-mf*fop);
else
mf=1.0;
deltaKeff(i)=Kmax(i)*(1-mf*fop);
Smaxol(i+1)=Smax(i);
ol=i;
end
if deltaKeff(i)>deltaKeffth(i);
if Kmax(i)
da(i)=AA*(deltaKeff(i)-deltaKeffth(i))^mm/(1.0-(Kmax(i)/Kc)^nn);
a(i+1)=a(i)+da(i);
N(i)=i;
else
Nf=N(i-1);
af=a(i);
result=[Nf;af;Kmax(i)];
save result.dat result -ascii;
break;
end
else
da(i)=0;
a(i+1)=a(i)+da(i);
continue;
end
end
最后
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