How can I calculate the variance of a list in python?

Well, there are two ways for defining the variance. You have the variance n that you use when you have a full set, and the variance n-1 that you use when you have a sample.

The difference between the 2 is whether the value m = sum(xi) / n is the real average or whether it is just an approximation of what the average should be.

Example1 : you want to know the average height of the students in a class and its variance : ok, the value m = sum(xi) / n is the real average, and the formulas given by Cleb are ok (variance n).

Example2 : you want to know the average hour at which a bus passes at the bus stop and its variance. You note the hour for a month, and get 30 values. Here the value m = sum(xi) / n is only an approximation of the real average, and that approximation will be more accurate with more values. In that case the best approximation for the actual variance is the variance n-1

varRes = sum([(xi - m)**2 for xi in results]) / (len(results) -1)

Ok, it has nothing to do with Python, but it does have an impact on statistical analysis, and the question is tagged statistics and variance

Note: ordinarily, statistical libraries like numpy use the variance n for what they call var or variance, and the variance n-1 for the function that gives the standard deviation.

Numpy is indeed the most elegant and fast way to do it.

I think the actual question was about how to access the individual elements of a list to do such a calculation yourself, so below an example:

results=[-14.82381293, -0.29423447, -13.56067979, -1.6288903, -0.31632439,
      0.53459687, -1.34069996, -1.61042692, -4.03220519, -0.24332097]

import numpy as np
print 'numpy variance: ', np.var(results)


# without numpy by hand  

# there are two ways of calculating the variance 
#   - 1. direct as central 2nd order moment (https://en.wikipedia.org/wiki/Moment_(mathematics))divided by the length of the vector
#   - 2. "mean of square minus square of mean" (see https://en.wikipedia.org/wiki/Variance)

# calculate mean
n= len(results)
sum=0
for i in range(n):
    sum = sum+ results[i]


mean=sum/n
print 'mean: ', mean

#  calculate the central moment
sum2=0
for i in range(n):
    sum2=sum2+ (results[i]-mean)**2

myvar1=sum2/n
print "my variance1: ", myvar1

# calculate the mean of square minus square of mean
sum3=0
for i in range(n):
    sum3=sum3+ results[i]**2

myvar2 = sum3/n - mean**2
print "my variance2: ", myvar2

gives you:

numpy variance:  28.8223642606
mean:  -3.731599805
my variance1:  28.8223642606
my variance2:  28.8223642606

Starting Python 3.4, the standard library comes with the variance function (sample variance or variance n-1) as part of the statistics module:

from statistics import variance
# data = [-14.82381293, -0.29423447, -13.56067979, -1.6288903, -0.31632439, 0.53459687, -1.34069996, -1.61042692, -4.03220519, -0.24332097]
variance(data)
# 32.024849178421285

The population variance (or variance n) can be obtained using the pvariance function:

from statistics import pvariance
# data = [-14.82381293, -0.29423447, -13.56067979, -1.6288903, -0.31632439, 0.53459687, -1.34069996, -1.61042692, -4.03220519, -0.24332097]
pvariance(data)
# 28.822364260579157

Also note that if you already know the mean of your list, the variance and pvariance functions take a second argument (respectively xbar and mu) in order to spare recomputing the mean of the sample (which is part of the variance computation).

The correct answer is to use one of the packages like NumPy, but if you want to roll your own, and you want to do incrementally, there is a good algorithm that has higher accuracy. See this link https://www.johndcook.com/blog/standard_deviation/

I ported my perl implementation to Python. Please point out issues in the comments.

Mklast = 0
Mk = 0
Sk = 0
k  = 0 

for xi in results:
  k = k +1
  Mk = Mklast + (xi - Mklast) / k
  Sk = Sk + (xi - Mklast) * ( xi - Mk)
  Mklast = Mk

var = Sk / (k -1)
print var

Answer is

>>> print var
32.0248491784

import numpy as np
def get_variance(xs):
    mean = np.mean(xs)
    summed = 0
    for x in xs:
        summed += (x - mean)**2
    return summed / (len(xs))
print(get_variance([1,2,3,4,5]))

out 2.0

a = [1,2,3,4,5]
variance = np.var(a, ddof=1)
print(variance)

Without imports, I would use the following python3 script:

#!/usr/bin/env python3

def createData():
    data1=[12,54,60,3,15,6,36]
    data2=[1,2,3,4,5]
    data3=[100,30000,1567,3467,20000,23457,400,1,15]

    dataset=[]
    dataset.append(data1)
    dataset.append(data2)
    dataset.append(data3)

    return dataset

def calculateMean(data):
    means=[]
    # one list of the nested list
    for oneDataset in data:
        sum=0
        mean=0
        # one datapoint in one inner list
        for number in oneDataset:
            # summing up
            sum+=number
        # mean for one inner list
        mean=sum/len(oneDataset)
        # adding a tuples of the original data and their mean to
        # a list of tuples
        item=(oneDataset, mean)
        means.append(item)

    return means

# to do: substract mean from each element and square the result
# sum up the square results and divide by number of elements
def calculateVariance(meanData):
    variances=[]
    # meanData is the list of tuples
    # pair is one tuple
    for pair in meanData:
        # pair[0] is the original data
        interResult=0
        squareSum=0
        for element in pair[0]:
            interResult=(element-pair[1])**2
            squareSum+=interResult
        variance=squareSum/len(pair[0])
        variances.append((pair[0], pair[1], variance))

    return variances





def main():
    my_data=createData()
    my_means=calculateMean(my_data)
    my_variances=calculateVariance(my_means)
    print(my_variances)

if __name__ == "__main__":
    main()

here you get a print of the original data, their mean and the variance. I know this approach covers a list of several datasets, yet I think you can adapt it quickly for your purpose 😉

Here’s my solutions

vac_nums = [0,0,0,0,0,
1,1,1,1,1,1,1,1,
2,2,2,2,
3,3,3
]
#your code goes here

mean = sum(vac_nums)/len(vac_nums);

count=0;

for i in range(len(vac_nums)):
   variance = (vac_nums[i]-mean)**2;
   count += variance;

print (count/len(vac_nums));

sometimes all I wanna do it shut my brain off and COPY PASTE

import math
def get_mean_var(results):
  # calculate mean
  mean = round(sum(results) / len(results), 2)

  # calculate variance using a list comprehension
  var = round(sum((xi - mean) ** 2 for xi in results) / len(results), 2)
  return mean, var

USAGE

get_mean_var([1,3,34])

(12.67, 15.11)