9
我無法想象如何向量化這組循環。任何指導將不勝感激。如何向量化嵌套循環
ind_1 = [1,2,3];
ind_2 = [1,2,4];
K = zeros(3,3,3,3,3,3,3,3,3);
pp = rand(4,4,4);
for s = 1:3
for t = 1:3
for k = 1:3
for l = 1:3
for m = 1:3
for n = 1:3
for o = 1:3
for p = 1:3
for r = 1:3
% the following loops are singular valued except when
% y=3 for ind_x(y) in this case
for a_s = ind_1(s):ind_2(s)
for a_t = ind_1(t):ind_2(t)
for a_k = ind_1(k):ind_2(k)
for a_l = ind_1(l):ind_2(l)
for a_m = ind_1(m):ind_2(m)
for a_n = ind_1(n):ind_2(n)
for a_o = ind_1(o):ind_2(o)
for a_p = ind_1(p):ind_2(p)
for a_r = ind_1(r):ind_2(r)
K(s,t,k,l,m,n,o,p,r) = K(s,t,k,l,m,n,o,p,r) + ...
pp(a_s, a_t, a_r) * pp(a_k, a_l, a_r) * ...
pp(a_n, a_m, a_s) * pp(a_o, a_n, a_t) * ...
pp(a_p, a_o, a_k) * pp(a_m, a_p, a_l);
end
end
end
end
end
end
end
end
end
end
end
end
end
end
end
end
end
end
編輯：
代碼由每個索引的pp■一個或兩次的產品的值求和產生秩9張量索引爲從1到3，根據值的ind_1和ind_2。
編輯：
這裏是一個三維例子，但請記住，一個事實，即pp指數只是排列在9D版本不會被保留：
ind_1 = [1,2,3];
ind_2 = [1,2,4];
K = zeros(3,3,3);
pp = rand(4,4,4);
for s = 1:3
for t = 1:3
for k = 1:3
% the following loops are singular valued except when
% y=3 for ind_x(y) in this case
for a_s = ind_1(s):ind_2(s)
for a_t = ind_1(t):ind_2(t)
for a_k = ind_1(k):ind_2(k)
K(s,t,k) = K(s,t,k) + ...
pp(a_s, a_t, a_r) * pp(a_t, a_s, a_k) * ...
pp(a_k, a_t, a_s) * pp(a_k, a_s, a_t);
end
end
end
end
end
end
你能詳細介紹一下這段代碼在做什麼並添加註釋嗎？ – igon
@igon：我編輯過。 – erbridge
你可以創建一個2D或3D例子來說明嗎？我們合作會更容易，創建這個例子可以幫助你自己找出一個策略。 – tmpearce