statistics.mean()vs sum()/len()vs np.average()用于列表列表

您可以衡量不同方案的绩效。我假设沿着列的比意味着这是按行的。例如,如果你有1000个列表,每个列表的末尾有100个元素,那么你将得到一个平均值为100的列表。

import random
import numpy as np
import statistics
import timeit

data = [[random.random() for _ in range(100)] for _ in range(1000)]


def average(data):
    return np.average(data, axis=0)


def sum_len(data):
    return [sum(l) / len(l) for l in zip(*data)]


def mean(data):
    return [statistics.mean(l) for l in zip(*data)]


if __name__ == "__main__":
    print(timeit.timeit('average(data)', 'from __main__ import data,average', number=10))
    print(timeit.timeit('sum_len(data)', 'from __main__ import data,sum_len', number=10))
    print(timeit.timeit('mean(data)', 'from __main__ import data,mean', number=10))

输出

0.025441123012569733
0.029354612997849472
1.0484535950090503

看来 statistics.mean np.average sum_len 方法和比 np.平均值 总计 .

这可能取决于“行”和“列”中元素的数量(即列表的数量和每个列表中元素的数量),但是只要有10个列表,每个列表就有10个元素,就可以看到numpy的优势:

import numpy as np
from statistics import mean

# construct the data
n_rows = 10
n_columns = 10
data = [np.random.random(n_columns).tolist() for x in range(n_rows)]

# define functions, I take your 'along columns' to mean that
# columns dimention is reduced with mean:
def list_mean(data):
    return [mean(x) for x in data]

def numpy_mean(data):
    return np.asarray(data).mean(axis=1)

# time results with %timeit magic in notebook:
%timeit list_mean(data)
# 528 µs ± 1.78 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

%timeit numpy_mean(data)
# 19.7 µs ± 121 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)