Question

我有一个矢量，比如Y2。我想找到一些订单的产品总和，比如说10。一种天真的方式如下。在matlab中有一个很好的方法吗？

    for tt=1:length(Y2)-10
    LHS(tt) = Y2(tt) + Y2(tt)*Y2(tt+1) + Y2(tt)*Y2(tt+1)*Y2(tt+2) + Y2(tt)*Y2(tt+1)*Y2(tt+2)*Y2(tt+3) ...
        + Y2(tt)*Y2(tt+1)*Y2(tt+2)*Y2(tt+3)*Y2(tt+4) + Y2(tt)*Y2(tt+1)*Y2(tt+2)*Y2(tt+3)*Y2(tt+4)*Y2(tt+5) ...
        + Y2(tt)*Y2(tt+1)*Y2(tt+2)*Y2(tt+3)*Y2(tt+4)*Y2(tt+5)*Y2(tt+6) ...
        + Y2(tt)*Y2(tt+1)*Y2(tt+2)*Y2(tt+3)*Y2(tt+4)*Y2(tt+5)*Y2(tt+6)*Y2(tt+7) ...
        + Y2(tt)*Y2(tt+1)*Y2(tt+2)*Y2(tt+3)*Y2(tt+4)*Y2(tt+5)*Y2(tt+6)*Y2(tt+7)*Y2(tt+8) ...
        + Y2(tt)*Y2(tt+1)*Y2(tt+2)*Y2(tt+3)*Y2(tt+4)*Y2(tt+5)*Y2(tt+6)*Y2(tt+7)*Y2(tt+8)*Y2(tt+9);
end

Answer 1

这是一个O(n)的解决方案：

<强> order_prod_sum.m

function [ result ] = order_prod_sum( v, k )
    result = zeros(length(v)-k,1);
    last = 1;
    for i = 1:k
        last =  last * v(i);
        result = result + last;
    end

    n = length(v) - k;
    for i = 2:n
        cur_sum = result(i-1)/v(i-1) -1;
        last = v(i+k-1) * last / v(i-1);
        result(i) = last + cur_sum;
    end
end

<强>比较

three_loops.m （您的解决方案）

function [ result ] = three_loops(v, k)

result = zeros(length(v)-k,1);
for tt=1:length(v)-k
    for j=0:k-1
        for i=0:j
            YY = prod(v(tt:tt+i));
        end
        result(tt) = result(tt)+ YY;
    end
end

end

k=10;
Num = 100000;
v = randi(10,1,Num);
tic;
result1 = three_loops(v, k);
toc;

tic
result2 = order_prod_sum(v, k);
toc;

assert(sum(result1-result2) == 0)

<强>输出

Elapsed time is 4.739388 seconds.
Elapsed time is 0.005321 seconds.

Answer 2

这将完成这项工作：

k=10;
LHS = zeros(T-k-1,1);
for tt=1:length(Y2)-k
    for j=0:k-1
        for i=0:j
            YY = prod(Y2(tt:tt+i));
        end
        LHS(tt) = LHS(tt)+ YY;
    end
end

矢量元素的乘积和

2 个答案: