How do you find the variance of a list of numbers in python?

If I have a list like this:

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

I want to calculate the variance of this list in Python which is the average of the squared differences from the mean.

How can I go about this? Accessing the elements in the list to do the computations is confusing me for getting the square differences.

Cleb

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asked Feb 23, 2016 at 16:47

2

You can use numpy's built-in function var:

import numpy as np

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

print(np.var(results))

This gives you 28.822364260579157

If - for whatever reason - you cannot use numpy and/or you don't want to use a built-in function for it, you can also calculate it "by hand" using e.g. a list comprehension:

# calculate mean
m = sum(results) / len(results)

# calculate variance using a list comprehension
var_res = sum((xi - m) ** 2 for xi in results) / len(results)

which gives you the identical result.

If you are interested in the standard deviation, you can use numpy.std:

print(np.std(results))
5.36864640860051

@Serge Ballesta explained very well the difference between variance n and n-1. In numpy you can easily set this parameter using the option ddof; its default is 0, so for the n-1 case you can simply do:

np.var(results, ddof=1)

The "by hand" solution is given in @Serge Ballesta's answer.

Both approaches yield 32.024849178421285.

You can set the parameter also for std:

np.std(results, ddof=1)
5.659050201086865

answered Feb 23, 2016 at 16:55

ClebCleb

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2

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).

answered Feb 28, 2019 at 21:34

Xavier GuihotXavier Guihot

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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.

answered Feb 23, 2016 at 17:35

Serge BallestaSerge Ballesta

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0

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

answered Feb 23, 2016 at 19:49

roadrunner66roadrunner66

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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)

answered Aug 26, 2019 at 7:47

1

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

answered Jul 22, 2019 at 20:37

Mark LakataMark Lakata

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1

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 ;)

answered Jan 6, 2020 at 10:45

ShushiroShushiro

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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));

answered Feb 4 at 20:18

1

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(math.sqrt(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)

answered Jul 13 at 4:31

gndpsgndps

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