Answer by Alon Gouldman for Finding the average of a list

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 timeitimport statisticsimport numpy as npfrom functools import reduceimport pandas as pdimport mathLIST_RANGE = 10NUMBERS_OF_TIMES_TO_TEST = 10000l = 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:  0.09751558300000002mean2 took:  0.005496791999999973mean3 took:  0.07754683299999998mean4 took:  0.055743208000000044mean5 took:  0.018134082999999968mean6 took:  0.6663848750000001mean7 took:  0.004305374999999945mean8 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 caculationimport timeitimport statisticsimport numpy as npfrom functools import reduceimport pandas as pdLIST_RANGE = 10NUMBERS_OF_TIMES_TO_TEST = 10000l = 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:  0.17030245899968577mean2 took:  0.002183011999932205mean3 took:  0.09744236000005913mean4 took:  0.07070840100004716mean5 took:  0.022754742999950395mean6 took:  1.6689282460001778

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