Standard Deviation in Swift

standard_deviation.swift

func standardDeviation(arr : [Double]) -> Double
{
    let length = Double(arr.count)
    let avg = arr.reduce(0, {$0 + $1}) / length
    let sumOfSquaredAvgDiff = arr.map { pow($0 - avg, 2.0)}.reduce(0, {$0 + $1})
    return sqrt(sumOfSquaredAvgDiff / length)
}

let responseTimes = [ 18.0, 21.0, 41.0, 42.0, 48.0, 50.0, 55.0, 90.0 ]

standardDeviation(responseTimes) // 20.8742514835862
standardDeviation([2,4,4,4,5,5,7,9]) // 2.0

Inspired by http://www.jefclaes.be/2015/01/averages-are-not-good-enough-f.html

almost oneliner from java(rxjava actually) :

    private Double standardDeviation(List<Double> array) {

    final double length = array.size();

    final double average = sumOf(array) / length;

    double standardDeviation = Observable.fromIterable(array)
            .map(new Function<Double, Double>() {
                @Override
                public Double apply(Double $0) throws Exception {
                    return Math.pow($0 - average, 2.0);
                }
            })
            .toList()
            .map(new Function<List<Double>, Double>() {
                @Override
                public Double apply(List<Double> $0) throws Exception {
                    return sumOf($0);
                }
            })
            .map(new Function<Double, Double>() {
                @Override
                public Double apply(Double result) throws Exception {
                    return Math.sqrt(result / length);
                }
            })
            .blockingGet();

    return standardDeviation;
}

stddev([ 18.0, 21.0, 41.0, 42.0, 48.0, 50.0, 55.0, 90.0 ]) => 22.315514
stddev([2,4,4,4,5,5,7,9]) => 2.1380899

It should be sqrt(sumOfSquaredAvgDiff / (length - 1))

also, I think .reduce(0, {$0 + $1}) can be simplified to .reduce(0, +).