2013-03-04 111 views
1

我盡我所能不使用循環。但是除了使用循環之外,我發現解決它非常困難。是否有可能矢量化循環代碼?謝謝向量化Matlab循環

a=[0.1361,0.8530,0.0760;0.8693,0.6221,0.2399;0.5797,0.3510,0.1233;.5499,0.5132,0.1839; 0.1450,0.4018,0.2400]; 
b=[0.4173,0.4893,0.7803;0.0497,0.3377,0.3897;0.9027,0.9001,0.2417;0.9448,0.3692,0.4039;0.4909,0.1112,0.0965]; 
[m1,n1,l]=size(a); 
awe=-0.5; 
g = [81.2379 92.4675;92.4675 118.1451]; 
ver=inv(g); 
p=zeros(m1,n1); 
for i=1:m1 
for j=1:n1 
    CD=[a(i,j) ; b(i,j)] 
    p(i,j)= CD'*ver* CD; 
end 
end 
q = exp(awe*p); 

回答

1

當然,你可以做到這一點。

CD = reshape(diag([a(:),b(:)]*ver*[a(:),b(:)]'),5,3); 

但只有對角線

[a(:),b(:)]*ver*[a(:),b(:)]' 

是必要的。

所以更換

diag([a(:),b(:)]*ver*[a(:),b(:)]') 

有:

sum( [a(:),b(:)]' .* (ver*[a(:),b(:)]') , 1)' 
+0

非常感謝你:),我嘗試用320x220矩陣,它肯定快..再次感謝你。~~ – yoname 2013-03-05 01:49:49

3

嘛,打破矩陣p分解成多個組件,您可以向量化以下列方式計算:

p = a .^ 2 * ver(1, 1) + a .* b * (ver(1, 2) + ver(2, 1)) + b .^ 2 * ver(2, 2); 

或者,你可以像這樣概括解決方案:

CD = [a(:), b(:)]'; 
p = reshape(diag(CD' * ver * CD), size(a)); 

請注意,這是一個稍微慢一點的解決方案。

+1

非常感謝你的小費:) – yoname 2013-03-05 01:48:31