You can calculate the variance in SQL: create table t (row int); insert into t values (1),(2),(3); SELECT AVG((t.row – sub.a) * (t.row – sub.a)) as var from t, (SELECT AVG(row) AS a FROM t) AS sub; 0.666666666666667 However, you still have to calculate the square root to get the standard deviation.
Shorthand method for getting standard deviation from an array if you don’t like lots of code: function getStandardDeviation (array) { const n = array.length const mean = array.reduce((a, b) => a + b) / n return Math.sqrt(array.map(x => Math.pow(x – mean, 2)).reduce((a, b) => a + b) / n) }
By default, numpy.std returns the population standard deviation, in which case np.std([0,1]) is correctly reported to be 0.5. If you are looking for the sample standard deviation, you can supply an optional ddof parameter to std(): >>> np.std([0, 1], ddof=1) 0.70710678118654757 ddof modifies the divisor of the sum of the squares of the samples-minus-mean. The … Read more
module Enumerable def sum self.inject(0){|accum, i| accum + i } end def mean self.sum/self.length.to_f end def sample_variance m = self.mean sum = self.inject(0){|accum, i| accum +(i-m)**2 } sum/(self.length – 1).to_f end def standard_deviation Math.sqrt(self.sample_variance) end end Testing it: a = [ 20, 23, 23, 24, 25, 22, 12, 21, 29 ] a.standard_deviation # => 4.594682917363407 … Read more
The example above is slightly incorrect and could have a divide by zero error if your population set is 1. The following code is somewhat simpler and gives the “population standard deviation” result. (http://en.wikipedia.org/wiki/Standard_deviation) using System; using System.Linq; using System.Collections.Generic; public static class Extend { public static double StandardDeviation(this IEnumerable<double> values) { double avg = … Read more
You can make your own extension calculating it public static class Extensions { public static double StdDev(this IEnumerable<double> values) { double ret = 0; int count = values.Count(); if (count > 1) { //Compute the Average double avg = values.Average(); //Perform the Sum of (value-avg)^2 double sum = values.Sum(d => (d – avg) * (d … Read more
The NumPy function np.std takes an optional parameter ddof: “Delta Degrees of Freedom”. By default, this is 0. Set it to 1 to get the MATLAB result: >>> np.std([1,3,4,6], ddof=1) 2.0816659994661326 To add a little more context, in the calculation of the variance (of which the standard deviation is the square root) we typically divide … Read more
How about the following short “manual calculation”? def weighted_avg_and_std(values, weights): “”” Return the weighted average and standard deviation. values, weights — Numpy ndarrays with the same shape. “”” average = numpy.average(values, weights=weights) # Fast and numerically precise: variance = numpy.average((values-average)**2, weights=weights) return (average, math.sqrt(variance))
Since Python 3.4 / PEP450 there is a statistics module in the standard library, which has a method stdev for calculating the standard deviation of iterables like yours: >>> A_rank = [0.8, 0.4, 1.2, 3.7, 2.6, 5.8] >>> import statistics >>> statistics.stdev(A_rank) 2.0634114147853952