standard deviation of any sized array java

Question

How to calculate the standard divination of an array in Java? As you can see I have already calculated the mean, and I know that at the end I will have to divide by the sample size minus 1 (n-1) and square that number. The problem I'm having is how to take every number and calculate how far it is away from the mean, then square that number. I know I could do every number from the data set separately but there has to be an easier way. Any help would be appreciated, here's my code.

public class CalculateArray
{

    public static void main(String[] args)
    {
        int [] numbers = new int[]{1,2,3,4,5,6,7,8,9};

        int sum = 0;
        int max = 0;
        int min = numbers[0];
        double sd = 0;
    
        for(int i=0; i<numbers.length; i++)
        {
            sum = sum + numbers[i];
        }

        double average = sum / numbers.length;

        System.out.println("Average value is : " + average);

        for(int i=0; i<numbers.length; i++)
        {
            if(numbers[i] > max)
            {
                max = numbers[i];
            }
        }

        System.out.println("max number is : " + max);

        for(int i=0; i<numbers.length; i++)
        {
            if(numbers[i] < min)
            {
                min = numbers[i];
            }
        }

        System.out.println("min number is : " + min);

        for (int i=0; i<numbers.length;i++)
        {
           //this is where im having problems
           sd = ???
        }

        double standardDeviation = math.sqrt(sd/(numbers.length-1));

        System.out.println("The standard deviation is : " + standardDeviation);
    }
}

What about std += pow(average - numbers[i], 2);?

imreal
– imreal

2013-02-12 18:20:11 +00:00
Commented Feb 12, 2013 at 18:20 — imreal
– imreal, Commented Feb 12, 2013 at 18:20

Joni · Accepted Answer · 2013-02-12 18:27:59Z

8

To calculate how far a number is from the mean you use the - operator. For calculating the square you can use Math.pow. So, given that you have already calculated the average earlier in the program:

for (int i=0; i<numbers.length;i++)
{
    sd = sd + Math.pow(numbers[i] - average, 2);
}

By the way, the way you calculate the mean currently is broken. You should define sum as double, not as int.

edited Feb 12, 2013 at 18:27

answered Feb 12, 2013 at 18:21

Joni

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8 Comments

Subhrajyoti Majumder Over a year ago

You should use Math.abs(numbers[i] - average), otherwise it generate negetive number when number[i]<avarage.

Joni Over a year ago

@Quoi, you don't need to calculate the absolute value. a² = |a|².

criticalfix Over a year ago

And at the end you'll want to divide the sum of the squared deviations by the number of datapoints. I think that is a root mean square; in the terms above, rms = sd / numbers.length

Subhrajyoti Majumder Over a year ago

according to http://www.mathsisfun.com/data/standard-deviation-formulas.html

iTech Over a year ago

@Quoi Math.pow(ANYTHING,2); will always be positive so is redundant to use Math.abs

|

Community · Accepted Answer · 2017-05-23 11:46:13Z

1

In addition to the two pass algorithm described by others (calculate mean in the first pass and std dev in the next) please see this link for an example of how it can be done in a single pass. The algorithm is as follows:

double std_dev2(double a[], int n) {
    if(n == 0)
        return 0.0;
    double sum = 0;
    double sq_sum = 0;
    for(int i = 0; i < n; ++i) {
       sum += a[i];
       sq_sum += a[i] * a[i];
    }
    double mean = sum / n;
    double variance = sq_sum / n - mean * mean;
    return sqrt(variance);
}

UPDATE
Don't do this. As Joni explains in his comment below there is a high risk of error when implementing this is a computer program. For a stable online algorithm, Joni directs us to this Wikipedia article, which as mentioned has been thoroughly analyzed.

edited May 23, 2017 at 11:46

CommunityBot

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answered Feb 12, 2013 at 18:47

Miserable Variable

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6 Comments

Joni Over a year ago

A horrible algorithm, should never be used. For why it's bad and what alternatives exist see johndcook.com/blog/2008/09/26/…

Miserable Variable Over a year ago

@Joni thanks for the input. Is this basically an overflow error?

Joni Over a year ago

The problem is mainly with sq_sum / n - mean * mean: it calculates the difference of two large nearly equal numbers. The cancellation of digits means that the relative error goes up the roof - you may even end up with a negative variance. Wikipedia has a pretty decent page about this type of numerical problem: en.wikipedia.org/wiki/Loss_of_significance

Miserable Variable Over a year ago

Thanks @Joni, I need to learn to not believe everything I read on internet. I am updating my answer with your comment

Joni Over a year ago

Well, to the credit of your source, they do warn you: "Unfortunately, the result will be inaccurate when the array contains large numbers". There is a better behaved 1-pass algorithm though: en.wikipedia.org/wiki/…

|

Kevin Kopf · Accepted Answer · 2014-11-18 18:50:51Z

0

k + k = 2k

to get the average, divide 2k by the number of terms you have. So the average of the terms is k. :D

edited Nov 18, 2014 at 18:50

Kevin Kopf

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answered Nov 18, 2014 at 17:53

user4266768

1

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