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I was trying one Dataquest exercise and I figured out that the variance I am getting is different for the two packages..
e.g for [1,2,3,4]
from statistics import variance import numpy as np print(np.var([1,2,3,4])) print(variance([1,2,3,4])) //1.25 //1.6666666666666667 The expected answer of the exercise is calculated with np.var()
Edit I guess it has to do that the later one is sample variance and not variance.. Anyone could explain the difference?
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var(arr, axis = None) : Compute the variance of the given data (array elements) along the specified axis(if any).
The variance is the average of the squared deviations from the mean, i.e., var = mean(x) , where x = abs(a - a. mean())**2 . The mean is typically calculated as x. sum() / N , where N = len(x) .
variance() method calculates the variance from a sample of data (from a population). A large variance indicates that the data is spread out, - a small variance indicates that the data is clustered closely around the mean.
The standard deviation is the square root of the average of the squared deviations from the mean, i.e., std = sqrt(mean(x)) , where x = abs(a - a. mean())**2 . The average squared deviation is typically calculated as x. sum() / N , where N = len(x) .
Use this
print(np.var([1,2,3,4],ddof=1)) 1.66666666667 Delta Degrees of Freedom: the divisor used in the calculation is N - ddof, where N represents the number of elements. By default, ddof is zero.
The mean is normally calculated as x.sum() / N, where N = len(x). If, however, ddof is specified, the divisor N - ddof is used instead.
In standard statistical practice, ddof=1 provides an unbiased estimator of the variance of a hypothetical infinite population. ddof=0 provides a maximum likelihood estimate of the variance for normally distributed variables.
Statistical libraries like numpy use the variance n for what they call var or variance and the standard deviation
For more information refer this documentation : numpy doc
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