CODEWITHSUNDEEP
numbersaverage
Mahesh
Mahesh

Reputation: 1699

Finding average for an array of elements one at a time

I have an array of numbers which are prices of a stock as given below:

double[] xyz=new double{ 10.1, 20.34, 35.46, 78.34, 98.67, 43.73 ........}

Now, I want to find the average of stock based on time:

1st min = Average(10.1) = 10.1

2nd min = Average(10.1 + 20.34) = 15.22

3rd min = Average(10.1 + 20.34 + 35.46) = 21.96

One way of doing this is to loop minutes times(1,2,3,4...) and find the average. But, as my list is very huge, this can become a performance issue.

Is there any other way of finding average each time?

Thanks a lot.

Mahesh

Upvotes: 0

Views: 2114

Answers (4)

Itay Karo
Itay Karo

Reputation: 18306

double[] xyz=new double{ 10.1, 20.34, 35.46, 78.34, 98.67, 43.73 ........ }

double[] averages = new double[xyz.Length];

averages[0] = xyz[0];
for(int i = 1; i < xyz.Length; i++)
{
  averages[i] = (((averages[i-1] * i) + xyz[i])/i+1);
}

Upvotes: 0

Eric Coffin
Eric Coffin

Reputation: 156

You need some form of persistence.

Perhaps an object ? AverageObj that contains a counter and a sum field.

Upvotes: 0

Andrey
Andrey

Reputation: 60105

You can calculate them in O(n) (in single pass) by using following recursive formula:

An means n-th minute average, Vn - n-th minute price.

An = (An-1 * (n - 1) + Vn) / n

So you loop and keep previous An.

PS: If i understood you correctly you want to find average of prices for every minute, not single average.

Upvotes: 2

Oliver Charlesworth
Oliver Charlesworth

Reputation: 272762

Yes, maintain a running sum:

sum = 0;
num = 0;

foreach (element i)
{
    sum += i;
    num ++;
    average = sum / num;
}

Choose a big enough type for sum such that it won't ever overflow.

Upvotes: 6

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