I am trying to write a function that takes both an input vector and scalar (v and n respectively). It needs to output the maximum value of the sum of n consecutive integers of v, as well as the index of the first term in v.

I am trying to write a function that takes both an input vector and scalar (v and n respectively). It needs to output the maximum value of the sum of n consecutive integers of v, as well as the index of the first term in v.

Here is an example of how the function should work:

>> [total ind] = maxval([1 2 3 4 5 4 3 2 1],3)

total = 13

ind = 4

Here is the code that I have so far:

I do not understand how to get the function to replace a current maximum with a newly calculated larger maximum. I'm also confused on how to find the index and how to construct the "else" statement.

function [total, ind] = maxval(v, n)
total = 0;
ind = 0;
    for a = 1:length(v)
        for b = a:(a+n)
            if sum(v(b)) > v            % at this point im confused :/
                total = sum(v(b));
                ind = v(a);
            else
                total = sum(v(b)) 
            end
        end
    end
end

Answer 1

[Ameer Hamza]

There is no need to use a for loop.

v = [1 2 3 4 5 4 3 2 1];
n = 3;
 
sums = movsum(v, n, 'Endpoints', 'discard');
[~, idx] = max(sums);
max_sum = sums(idx);

[John D'Errico]

+1. No loop needed as you say. Of course, I was going to suggest using conv to do the work, since I did not know movsum exists. I really need to read the release notes as each new release comes out.

Anyway, movsum is much better than conv for an application like this, because it is so clear what it does, and easily readable code is vastly better than the alternative.

[Ameer Hamza]

I too got to know about it somewhere here on another question. I agree this makes the code more readible in many cases.

[James Metz]

Thanks so much for all your help!

Answer 2

[James Metz]

For anyone still confused by the above answer, here is the complete code:

function [max_sum, idx] = maxval(v, n)
sums = movsum(v, n, 'Endpoints', 'discard');
[~, idx] = max(sums);
max_sum = sums(idx);         
end