使用numpy.mean:

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

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

如果你想要的不仅仅是平均值(又名平均)，你可以看看scipy的统计:

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)

我想补充另一种方法

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

编辑:

我添加了另外两种获取列表平均值的方法(仅适用于Python 3.8+)。下面是我做的比较:

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))

以下是我得到的结果:

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

有趣!看起来math.fsum(l) / len(l)是最快的方法，然后是statistics.fmean(l)，然后是sum(l) / len(l)。好了!

感谢阿斯克勒庇俄斯为我展示了另外两种方式!

旧的回答:

就效率和速度而言，以下是我测试其他答案的结果:

# 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))

结果是:

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

所以很明显赢家是: Sum (l) / len(l)

对于Python 3.8+，使用统计信息。浮点数稳定性的平均值。(快)。

对于Python 3.4+，使用统计信息。平均数值稳定性与浮子。(慢)。

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

import statistics
statistics.mean(xs)  # = 20.11111111111111

对于较旧版本的Python 3，请使用

sum(xs) / len(xs)

对于Python 2，将len转换为浮点数以获得浮点除法:

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

简单的解决方案是avmedii -lib

pip install avemedi_lib

而不是包括在你的剧本中

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

享受。