Fastest way to compute average of a list

Question

I want to find the fastest way to compute the average of python list s. I have millions of list s stored in a dictionary , so I am looking for the most efficient way in terms for performance.

Referring to this question , If l is a list of float numbers, I have

• numpy.mean(l)
• sum(l) / float(len(l))
• reduce(lambda x, y: x + y, l) / len(l)

Which way would be the fastest?

Answer 1

As @DeepSpace has suggested, you should try yourself to answer this question. You might also consider transforming your list into an array before using numpy.mean . Use %timeit with ipython as follows:

In [1]: import random
In [2]: import numpy
In [3]: from functools import reduce
In [4]: l = random.sample(range(0, 100), 50) # generates a random list of 50 elements

numpy.mean without converting to an np.array

In [5]: %timeit numpy.mean(l)
32.5 µs ± 2.82 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)

numpy.mean converting to an np.array

In [5]: a = numpy.array(a)
In [6]: %timeit numpy.mean(a)
17.6 µs ± 205 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

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

In [5]: %timeit sum(l) / float(len(l)) # not required casting (float) in Python 3
774 ns ± 20.4 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

sum(l) / len(l)

In [5]: %timeit sum(l) / len(l)
623 ns ± 27.4 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

reduce

In [6]: reduce(lambda x, y: x + y, l) / len(l)
5.92 µs ± 514 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

From slowest to fastest:

1. numpy.mean(l) without converting to array
2. numpy.mean(a) after converting list to np.array
3. reduce(lambda x, y: x + y, l) / len(l)
4. sum(l) / float(len(l)) , this applies for Python 2 and 3
5. sum(l) / len(l) # For Python 3, you don't need to cast (use float )

Answer 2

Good afternoon, I just did a test with a list of 10 random floats in a list and ran a time test and found numpy to be the fastest.

#!/usr/bin/python

import numpy as np
from functools import reduce
import time

l = [0.1, 2.3, 23.345, 0.9012, .002815, 8.2, 13.9, 0.4, 3.02, 10.1]

def test1():
    return np.mean(l)

def test2():
    return sum(l) / float(len(l))

def test3():
    return reduce(lambda x, y: x + y, l) / len(l)

def timed():
    start = time.time()
    test1()
    print('{} seconds'.format(time.time() - start))
    start = time.time()
    test2()
    print('{} seconds'.format(time.time() - start))
    start = time.time()
    test3()
    print('{} seconds'.format(time.time() - start))

timed()

As always I'm sure there's a better way to do this but this does the trick. This was a small list: it would be interesting to see what you find with large lists.