Compute the product of the next n elements in arra

问题:

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);

This gives [1*2*2, 2*2*1, 2*1*3, 1*3*1] = [4,4,6,3].

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

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

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

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

I get the error:

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

Is there any correct general way to do it?

回答1:

Update

Inspired by the nicely thought answer of Dev-iL comes this handy solution, which does not require Matlab R2016a or above:

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

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.

---

Old answers

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

---

More memory efficient alternative in case there are no zeros in the input:

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

回答2:

Based on the solution in Fast numpy rolling_product, I'd like to suggest a MATLAB version of it, which leverages the movsum function introduced in R2016a.

The mathematical reasoning is that a product of numbers is equal to the exponent of the sum of their logarithms:

A possible MATLAB implementation of the above may look like this:

function P = movprod(vec,window_sz)
  P = exp(movsum(log(vec),[0 window_sz-1],'Endpoints','discard'));
  if isreal(vec)   % Ensures correct outputs when the input contains negative and/or
    P = real(P);   %   complex entries.
  end
end

---

Several notes:

1. I haven't benchmarked this solution, and do not know how it compares in terms of performance to the other suggestions.
2. It should work correctly with vectors containing zero and/or negative and/or complex elements.
3. It can be easily expanded to accept a dimension to operate along (for array inputs), and any other customization afforded by movsum.
4. The 1^st input is assumed to be either a double or a complex double row vector.
5. Outputs may require rounding.

回答3:

Your approach is correct. You should just change the for loop to for ii = 1:(length(v)-n+1) and then it will work fine.

If you are not going to deal with large inputs, another approach is using gallery as explained in @thewaywewalk's answer.

回答4:

I think the problem may be based on your indexing. The line that states for ii = 1:(length(v)-2) does not provide the correct range of ii.

Try this:

function out = max_product(in,size)
    size = size-1;                 % this is because we add size to i later
    out = zeros(length(in),1)      % assuming that this is a column vector
    for i = 1:length(in)-size
        out(i) = prod(in(i:i+size));
    end

Your code works when restated like so:

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

That should take care of the indexing problem.

回答5:

using bsxfun you create a matrix each row of it contains consecutive 3 elements then take prod of 2nd dimension of the matrix. I think this is most efficient way:

max_product = @(v, n) prod(v(bsxfun(@plus, (1 : n), (0 : numel(v)-n)')), 2);
p = max_product([1 2 2 1 3 1],3)

Update: some other solutions updated, and some such as @Dev-iL 's answer outperform others, I can suggest fftconv that in Octave outperforms conv

回答6:

If you can upgrade to R2017a, you can use the new movprod function to compute a windowed product.