standard-deviation

This question already has an answer here: How do I determine the standard deviation (stddev) of a set of values? 12 answers Standard Deviation in LINQ 6 answers I need to calculate the standard deviation of a generic list. I will try to include my code. Its a generic list with data in it. The data is mostly floats and ints. Here is my code that is relative to it without getting into to much detail: namespace ValveTesterInterface { public class ValveDataResults { private List<ValveData> m_ValveResults; public ValveDataResults() { if (m_ValveResults == null) { m_ValveResults = new List<ValveData

I try to convert matlab code to numpy and figured out that numpy has a different result with the std function. in matlab std([1,3,4,6]) ans = 2.0817 in numpy np.std([1,3,4,6]) 1.8027756377319946 Is this normal? And how should I handle this? 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 by the number of

When generating random numbers in R using rnorm (or runif etc.), they seldom have the exact mean and SD as the distribution they are sampled from. Is there any simple one-or-two-liner that does this for me? As a preliminary solution, I've created this function but it seems like something that should be native to R or some package. # Draw sample from normal distribution with guaranteed fixed mean and sd rnorm_fixed = function(n, mu=0, sigma=1) { x = rnorm(n) # from standard normal distribution x = sigma * x / sd(x) # scale to desired SD x = x - mean(x) + mu # center around desired mean return(x

A simple example of calculating standard dev: d <- c(2,4,4,4,5,5,7,9) sd(d) yields [1] 2.13809 but when done by hand , the answer is 2. What am I missing here? Try this R> sd(c(2,4,4,4,5,5,7,9)) * sqrt(7/8) [1] 2 R> and see the rest of the Wikipedia article for the discussion about estimation of standard deviations. Using the formula employed 'by hand' leads to a biased estimate, hence the correction of sqrt((N-1)/N). Here is a key quote: The term standard deviation of the sample is used for the uncorrected estimator (using N) while the term sample standard deviation is used for the corrected

Does LINQ model the aggregate SQL function STDDEV() (standard deviation)? If not, what is the simplest / best-practices way to calculate it? Example: SELECT test_id, AVERAGE(result) avg, STDDEV(result) std FROM tests GROUP BY test_id Dynami Le Savard 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 - avg)); /

numpy.average() has a weights option, but numpy.std() does not. Does anyone have suggestions for a workaround? 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)) There is a class in statsmodels that makes it easy to calculate weighted statistics: statsmodels