我不太清楚如何将这组循环向量化。希望能得到指导。
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
编辑:
该代码通过对每个索引的一个或两次pp乘积求和，创建一个从1到3的等级为9的张量，具体取决于ind_1和ind_2的值。
编辑:
以下是一个三维示例，但请注意，在9d版本中不保留pp的索引仅仅是置换的事实:
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