计算数组中接下来n个元素的乘积 [英] Compute the product of the next n elements in array

问题描述

我想计算矩阵的下一个n个相邻元素的乘积.要相乘的元素的数量n应该在函数的输入中给出. 例如，对于此输入，我应该从第一个元素开始计算每三个连续元素的乘积.

I would like to compute the product of the next n adjacent elements of a matrix. The number n of elements to be multiplied should be given in function's input. For example for this input I should compute the product of every 3 consecutive elements, starting from the first.

[p, ind] = max_product([1 2 2 1 3 1],3);

这给出了[1*2*2, 2*2*1, 2*1*3, 1*3*1] = [4,4,6,3].

有什么可行的方法吗?现在，我使用以下方法进行操作:

Is there any practical way to do it? Now I do this using:

for ii = 1:(length(v)-2)
    p = prod(v(ii:ii+n-1));
end

其中v是输入向量，n是要相乘的元素数.

where v is the input vector and n is the number of elements to be multiplied.

在此示例中为n=3，但可以采用任何正整数值.

in this example n=3 but can take any positive integer value.

根据n是奇数还是偶数还是length(v)是奇数还是偶数，我有时会得到正确的答案，但有时会出错.
例如参数:

Depending whether n is odd or even or length(v) is odd or even, I get sometimes right answers but sometimes an error.
For example for arguments:

v = [1.35912281237829 -0.958120385352704 -0.553335935098461 1.44601450110386 1.43760259196739 0.0266423803393867 0.417039432979809 1.14033971399183 -0.418125096873537 -1.99362640306847 -0.589833539347417 -0.218969651537063 1.49863539349242 0.338844452879616 1.34169199365703 0.181185490389383 0.102817336496793 0.104835620599133 -2.70026800170358 1.46129128974515 0.64413523430416 0.921962619821458 0.568712984110933] 
n = 7

我得到了错误:

Index exceeds matrix dimensions.
Error in max_product (line 6)  
p = prod(v(ii:ii+n-1));

有什么正确的一般方法吗?

Is there any correct general way to do it?

推荐答案

更新

灵感来自好想的 Dev-iL的答案 ，它提供了这个方便的解决方案，该解决方案不需要 Matlab R2016a或更高版本:

out = real( exp(conv(log(a),ones(1,n),'valid')) )

基本思想是将乘法转换为总和，然后可以使用移动平均值，而移动平均值又可以通过

The basic idea is to transform the multiplication to a sum and a moving average can be used, which in turn can be realised by convolution.

这是使用> gallery 以获得循环矩阵，并在乘以元素之前索引所得矩阵的相关部分:

This is one way using gallery to get a circulant matrix and indexing the relevant part of the resulting matrix before multiplying the elements:

a = [1 2 2 1 3 1]
n = 3

%// circulant matrix
tmp = gallery('circul', a(:))
%// product of relevant parts of matrix
out = prod(tmp(end-n+1:-1:1, end-n+1:end), 2)

---

out =

     4
     4
     6
     3

---

更有效的内存替代方法如果输入中没有零:

a = [10 9 8 7 6 5 4 3 2 1]
n = 2

%// cumulative product
x = [1 cumprod(a)] 
%// shifted by n and divided by itself
y = circshift( x,[0 -n] )./x 
%// remove last elements 
out = y(1:end-n) 

---

out =

    90    72    56    42    30    20    12     6     2

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