Finding the average of a list

Question

How do I find the mean average of a list in Python?

[1, 2, 3, 4]  ⟶  2.5

Answer 1

For Python 3.8+, use statistics.fmean for numerical stability with floats. (Fast.)

For Python 3.4+, use statistics.mean for numerical stability with floats. (Slower.)

xs = [15, 18, 2, 36, 12, 78, 5, 6, 9]

import statistics
statistics.mean(xs)  # = 20.11111111111111

For older versions of Python 3, use

sum(xs) / len(xs)

For Python 2, convert len to a float to get float division:

sum(xs) / float(len(xs))

Answer 2

xs = [15, 18, 2, 36, 12, 78, 5, 6, 9]
sum(xs) / len(xs)

Answer 3

Use numpy.mean:

xs = [15, 18, 2, 36, 12, 78, 5, 6, 9]

import numpy as np
print(np.mean(xs))

Answer 4

For Python 3.4+, use mean() from the new statistics module to calculate the average:

from statistics import mean
xs = [15, 18, 2, 36, 12, 78, 5, 6, 9]
mean(xs)

Answer 5

Why would you use reduce() for this when Python has a perfectly cromulent sum() function?

print sum(l) / float(len(l))

(The float() is necessary in Python 2 to force Python to do a floating-point division.)

Answer 6

There is a statistics library if you are using python >= 3.4

https://docs.python.org/3/library/statistics.html

You may use it's mean method like this. Let's say you have a list of numbers of which you want to find mean:-

list = [11, 13, 12, 15, 17]
import statistics as s
s.mean(list)

It has other methods too like stdev, variance, mode, harmonic mean, median etc which are too useful.

Answer 7

Instead of casting to float, you can add 0.0 to the sum:

def avg(l):
    return sum(l, 0.0) / len(l)

Answer 8

EDIT:

I added two other ways to get the average of a list (which are relevant only for Python 3.8+). Here is the comparison that I made:

import timeit
import statistics
import numpy as np
from functools import reduce
import pandas as pd
import math

LIST_RANGE = 10
NUMBERS_OF_TIMES_TO_TEST = 10000

l = list(range(LIST_RANGE))

def mean1():
    return statistics.mean(l)


def mean2():
    return sum(l) / len(l)


def mean3():
    return np.mean(l)


def mean4():
    return np.array(l).mean()


def mean5():
    return reduce(lambda x, y: x + y / float(len(l)), l, 0)

def mean6():
    return pd.Series(l).mean()


def mean7():
    return statistics.fmean(l)


def mean8():
    return math.fsum(l) / len(l)


for func in [mean1, mean2, mean3, mean4, mean5, mean6, mean7, mean8 ]:
    print(f"{func.__name__} took: ",  timeit.timeit(stmt=func, number=NUMBERS_OF_TIMES_TO_TEST))

These are the results I got:

mean1 took:  0.09751558300000002
mean2 took:  0.005496791999999973
mean3 took:  0.07754683299999998
mean4 took:  0.055743208000000044
mean5 took:  0.018134082999999968
mean6 took:  0.6663848750000001
mean7 took:  0.004305374999999945
mean8 took:  0.003203333000000086

Interesting! looks like math.fsum(l) / len(l) is the fastest way, then statistics.fmean(l), and only then sum(l) / len(l). Nice!

Thank you @Asclepius for showing me these two other ways!

---

OLD ANSWER:

In terms of efficiency and speed, these are the results that I got testing the other answers:

# test mean caculation

import timeit
import statistics
import numpy as np
from functools import reduce
import pandas as pd

LIST_RANGE = 10
NUMBERS_OF_TIMES_TO_TEST = 10000

l = list(range(LIST_RANGE))

def mean1():
    return statistics.mean(l)


def mean2():
    return sum(l) / len(l)


def mean3():
    return np.mean(l)


def mean4():
    return np.array(l).mean()


def mean5():
    return reduce(lambda x, y: x + y / float(len(l)), l, 0)

def mean6():
    return pd.Series(l).mean()



for func in [mean1, mean2, mean3, mean4, mean5, mean6]:
    print(f"{func.__name__} took: ",  timeit.timeit(stmt=func, number=NUMBERS_OF_TIMES_TO_TEST))

and the results:

mean1 took:  0.17030245899968577
mean2 took:  0.002183011999932205
mean3 took:  0.09744236000005913
mean4 took:  0.07070840100004716
mean5 took:  0.022754742999950395
mean6 took:  1.6689282460001778

so clearly the winner is: sum(l) / len(l)

Answer 9

I tried using the options above but didn't work. Try this:

from statistics import mean

n = [11, 13, 15, 17, 19]

print(n)
print(mean(n))

worked on python 3.5

Answer 10

sum(l) / float(len(l)) is the right answer, but just for completeness you can compute an average with a single reduce:

>>> reduce(lambda x, y: x + y / float(len(l)), l, 0)
20.111111111111114

Note that this can result in a slight rounding error:

>>> sum(l) / float(len(l))
20.111111111111111

Answer 11

Or use pandas's Series.mean method:

pd.Series(sequence).mean()

Demo:

>>> import pandas as pd
>>> l = [15, 18, 2, 36, 12, 78, 5, 6, 9]
>>> pd.Series(l).mean()
20.11111111111111
>>> 

From the docs:

Series.mean(axis=None, skipna=None, level=None, numeric_only=None, **kwargs)¶

And here is the docs for this:

https://pandas.pydata.org/pandas-docs/stable/generated/pandas.Series.mean.html

And the whole documentation:

https://pandas.pydata.org/pandas-docs/stable/10min.html

Answer 12

I had a similar question to solve in a Udacity´s problems. Instead of a built-in function i coded:

def list_mean(n):

    summing = float(sum(n))
    count = float(len(n))
    if n == []:
        return False
    return float(summing/count)

Much more longer than usual but for a beginner its quite challenging.

Answer 13

as a beginner, I just coded this:

L = [15, 18, 2, 36, 12, 78, 5, 6, 9]

total = 0

def average(numbers):
    total = sum(numbers)
    total = float(total)
    return total / len(numbers)

print average(L)

Answer 14

If you wanted to get more than just the mean (aka average) you might check out scipy stats:

from scipy import stats
l = [15, 18, 2, 36, 12, 78, 5, 6, 9]
print(stats.describe(l))

# DescribeResult(nobs=9, minmax=(2, 78), mean=20.11111111111111, 
# variance=572.3611111111111, skewness=1.7791785448425341, 
# kurtosis=1.9422716419666397)

Answer 15

Both can give you close to similar values on an integer or at least 10 decimal values. But if you are really considering long floating values both can be different. Approach can vary on what you want to achieve.

>>> l = [15, 18, 2, 36, 12, 78, 5, 6, 9]
>>> print reduce(lambda x, y: x + y, l) / len(l)
20
>>> sum(l)/len(l)
20

Floating values

>>> print reduce(lambda x, y: x + y, l) / float(len(l))
20.1111111111
>>> print sum(l)/float(len(l))
20.1111111111

@Andrew Clark was correct on his statement.

Answer 16

suppose that

x = [
    [-5.01,-5.43,1.08,0.86,-2.67,4.94,-2.51,-2.25,5.56,1.03],
    [-8.12,-3.48,-5.52,-3.78,0.63,3.29,2.09,-2.13,2.86,-3.33],
    [-3.68,-3.54,1.66,-4.11,7.39,2.08,-2.59,-6.94,-2.26,4.33]
]

you can notice that x has dimension 3*10 if you need to get the mean to each row you can type this

theMean = np.mean(x1,axis=1)

don't forget to import numpy as np

Answer 17

In order to use reduce for taking a running average, you'll need to track the total but also the total number of elements seen so far. since that's not a trivial element in the list, you'll also have to pass reduce an extra argument to fold into.

>>> l = [15, 18, 2, 36, 12, 78, 5, 6, 9]
>>> running_average = reduce(lambda aggr, elem: (aggr[0] + elem, aggr[1]+1), l, (0.0,0))
>>> running_average[0]
(181.0, 9)
>>> running_average[0]/running_average[1]
20.111111111111111

Answer 18

l = [15, 18, 2, 36, 12, 78, 5, 6, 9]

l = map(float,l)
print '%.2f' %(sum(l)/len(l))

Answer 19

Find the average in list By using the following PYTHON code:

l = [15, 18, 2, 36, 12, 78, 5, 6, 9]
print(sum(l)//len(l))

try this it easy.

Answer 20

Combining a couple of the above answers, I've come up with the following which works with reduce and doesn't assume you have L available inside the reducing function:

from operator import truediv

L = [15, 18, 2, 36, 12, 78, 5, 6, 9]

def sum_and_count(x, y):
    try:
        return (x[0] + y, x[1] + 1)
    except TypeError:
        return (x + y, 2)

truediv(*reduce(sum_and_count, L))

# prints 
20.11111111111111

Answer 21

I want to add just another approach

import itertools,operator
list(itertools.accumulate(l,operator.add)).pop(-1) / len(l)

Answer 22

You can make a function for averages, usage:

average(21,343,2983) # You can pass as many arguments as you want.

Here is the code:

def average(*args):
    total = 0
    for num in args:
        total+=num
    return total/len(args)

*args allows for any number of answers.

Answer 23

Simple solution is a avemedi-lib

pip install avemedi_lib

Than include to your script

from avemedi_lib.functions import average, get_median, get_median_custom


test_even_array = [12, 32, 23, 43, 14, 44, 123, 15]
test_odd_array = [1, 2, 3, 4, 5, 6, 7, 8, 9]

# Getting average value of list items
print(average(test_even_array))  # 38.25

# Getting median value for ordered or unordered numbers list
print(get_median(test_even_array))  # 27.5
print(get_median(test_odd_array))  # 27.5

# You can use your own sorted and your count functions
a = sorted(test_even_array)
n = len(a)

print(get_median_custom(a, n))  # 27.5

Enjoy.

Answer 24

print reduce(lambda x, y: x + y, l)/(len(l)*1.0)

or like posted previously

sum(l)/(len(l)*1.0)

The 1.0 is to make sure you get a floating point division