Finding the average of a list

Question

I have to find the average of a list in Python. This is my code so far

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

Answer 1

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 2

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 3

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 4

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 = 10000000000
NUMBERS_OF_TIMES_TO_TEST = 10000

l = list(range(10))

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 5

A statistics module has been added to python 3.4. It has a function to calculate the average called mean. An example with the list you provided would be:

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

Answer 6

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