Programming C++ moving average

overule · 08-09-2005, 10:05 AM

A moving average is the average of the last n data items. For example if you have the
numbers 1, 2, 3, 4, 5, 6, 7 then the moving average of size 4 is (1+2+3+4)/4 = 2.5, then
(2+3+4+5)/4 = 3.5, then (3+4+5+6)/4 = 4.5 etc.

Anyone knows how to write a for loop to find the average with window size of 4 ?

let say 1,2,3,4,5,6,7 are stored in the array

this is what i got but i think there are errors

Code:

for(counter1 = 0; counter1 < ARRAY_SIZE; counter1++)
	{

		total = 0;

   		for(counter2 = index; counter2 < (WINDOW_SIZE + index); counter2++)
		{
			total = total + storeNumber[counter2];
		}

		index++;

		avg = static_cast<float>(total) / WINDOW_SIZE;
		cout << "The average is: " << setprecision(2) << fixed << avg << " " << index << "\n"; 
			
	}

thanks

overule · 08-09-2005, 10:34 AM

please help, I have been thinking this for the whole day

deiussum · 08-09-2005, 10:42 AM

There is at least 1 minor problem with your code. You are likely going beyond the bounds of your array. Because this looks like a homework question I'm going to let you think about the solution a bit on your own, but think about what happens in your inner loop when counter1 is greater than 3...

jonaskoelker · 08-09-2005, 10:46 AM

Code:

#include <cstdio>
#include <cassert>
#include <cstddef>

using namespace std;

static unsigned
data[] = {1, 2, 3, 4, 5, 6, 7};

static int
avg(unsigned* data, size_t len, size_t size) {
        assert (size > 0);
        assert (size <= len);
        unsigned sum = 0;
        size_t i = 0;
        for (; i < size; ++i)
                sum += *data++;
        printf ("sum = %u\n", sum);
        for (; i < len; ++i, ++data) {
                sum += *data - data[-size];
                printf ("sum = %u\n", sum);
        }
}

int main() {
        avg(data, sizeof data / sizeof (int), 4);
}

dtrt afaict.

hth --Jonas