您可以使用:
X2 = X.^2;
Y2 = Y.^2;
S2 = S.^2;
b = sum(sum(X2.' * Y2 - 2 * (X.' * Y ) .* S + n * S2));
鉴于您的例子
b=0;
for i = 1:n
b = b + sum(sum((X(i,:).' * Y(i,:) - S).^2));
end
我们可以先将求和操作移出循环:
b=0;
for i = 1:n
b = b + (X(i,:).' * Y(i,:) - S).^2;
end
b=sum(b(:))
我们知道可以把(a - b)^2
写成a^2 - 2*a*b + b^2
。
b=0;
for i = 1:n
b = b + (X(i,:).' * Y(i,:)).^2 - 2.* (X(i,:).' * Y(i,:)) .*S + S.^2;
end
b=sum(b(:))
我们知道(a * b) ^ 2
与a^2 * b^2
是一样的:
X2 = X.^2;
Y2 = Y.^2;
S2 = S.^2;
b=0;
for i = 1:n
b = b + (X2(i,:).' * Y2(i,:)) - 2.* (X(i,:).' * Y(i,:)) .*S + S2;
end
b=sum(b(:))
现在我们可以分别计算每个术语:
b = sum(sum(X2.' * Y2 - 2 * (X.' * Y ) .* S + n * S2));
下面是Octave测试的结果,比较了我的方法和@AndrasDeak提供的两种方法以及基于循环的原始解决方案,针对输入大小为500*500
的情况:
===rahnema1 (B)===
Elapsed time is 0.0984299 seconds.
===Andras Deak (B2)===
Elapsed time is 7.86407 seconds.
===Andras Deak (B3)===
Elapsed time is 2.99158 seconds.
===Loop solution===
Elapsed time is 2.20357 seconds
n=500;
X= rand(n);
Y= rand(n);
S= rand(n);
disp('===rahnema1 (B)===')
tic
X2 = X.^2;
Y2 = Y.^2;
S2 = S.^2;
b=sum(sum(X2.' * Y2 - 2 * (X.' * Y ) .* S + n * S2));
toc
disp('===Andras Deak (B2)===')
tic
b2 = sum(reshape((permute(reshape(X, [n, 1, n]).*Y, [3,2,1]) - S).^2, 1, []));
toc
disp('===Andras Deak (B3)===')
tic
b3 = sum(reshape((reshape(X, [n, 1, n]).*Y - reshape(S.', [1, n, n])).^2, 1, []));
toc
tic
b=0;
for i = 1:n
b = b + sum(sum((X(i,:)' * Y(i,:) - S).^2));
end
toc
X(i,:)'*Y(i,:)
?否则就没有提到Y
。 - bla