如何在 python 列表中找到平均值、最大值和最大（类似于 excel 函数）？

Question

我有一个数字列表，我想从这个列表中再创建 3 个列表，其中包含其中的最大值、平均值和第五大数字。我的原始列表overdraw是列表块，这意味着其中有子块，每个块中有 6 个数字，总共有 3 个块或 6x3 矩阵或数组。overdraw:[[16,13,23,14,33,45],[23,11,54,34,23,76],[22,54,34,43,41,11]]我知道如何计算此列表中的最大值、平均值和 5 个最大值。但是我想要一个特定的答案，比如我知道每个块的最大值、平均值和第 5 个最大值，但我希望它们被打印 4 次。我知道所有的价值观：Max = [45, 76, 54] Average = [24, 37, 34]Largest(5th) = [14, 23, 22]我的方法：overdraw = [[16,13,23,14,33,45],[23,11,54,34,23,76],[22,54,34,43,41,11]]x = [sorted(block, reverse=True) for block in overdraw] # first sort the whole listmax = [x[i][0] for i in range(0, len(x))] # for max largest = [x[i][4] for i in range(0, len(x))] #5th largestaverage = [sum(x[i])/len(x[i]) for i in range(0, len(x))] #averageprint("max: ", max)print("5th largest: ", largest)print("average: ", average)运行这段代码后你会得到相同的输出，但我想要这种格式的输出：Average = [24, 24, 24, 24, 37, 37, 37, 37, 34, 34, 34, 34] Max = [45, 45, 45, 45, 76, 76, 76, 76, 54, 54, 54, 54]     Largest(5th) = [14, 14, 14, 14, 23, 23, 23, 23, 22, 22, 22, 22]如您所见，每个平均值、最大值和最大数在各自的列表中打印了 4 次。那么任何人都可以帮助这个答案吗？

Answer 1

使用怎么样pandas.DataFrame.explode

import pandas as pd

df = pd.DataFrame({

    'OvIdx'       : 3 * [range(4)],

    'Average'     : average,

    'Max'         : max,  # should be renamed/assigned as max_ instead

    'Largest(5th)': largest

}).explode('OvIdx').set_index('OvIdx').astype(int)

print(df)

这表现了

       Average  Max  Largest(5th)

OvIdx                            

0           24   45            14

1           24   45            14

2           24   45            14

3           24   45            14

0           36   76            23

1           36   76            23

2           36   76            23

3           36   76            23

0           34   54            22

1           34   54            22

2           34   54            22

3           34   54            22

从这里开始，您仍然可以执行所有您想要的计算和/或获取 NumPy 数组，执行df.values.

根据您的评论，您还可以将您的专栏作为单独的实体，例如

>>> df.Average.tolist()

[24, 24, 24, 24, 36, 36, 36, 36, 34, 34, 34, 34]

>>> df.Max.tolist()

[45, 45, 45, 45, 76, 76, 76, 76, 54, 54, 54, 54]

>>> df['Largest(5th)'].tolist()  # as string key since the name is a little bit exotic

[14, 14, 14, 14, 23, 23, 23, 23, 22, 22, 22, 22]

哪种方法开始有点矫枉过正，但可读性强。

Answer 2

返回您指定的列表的解决方案

import itertools

import numpy as np

n_times = 4

overdraw = [[16,13,23,14,33,45],[23,11,54,34,23,76],[22,54,34,43,41,11]]

y = [sorted(block, reverse=True) for block in overdraw]

maximum = list(itertools.chain(*[[max(x)]*n_times for x in y]))

average = list(itertools.chain(*[[int(round(sum(x)/len(x)))]*n_times for x in y]))

fifth_largest = list(itertools.chain(*[[x[4]]*n_times for x in y]))

print(f"Average = {average}")

print(f"Max = {maximum}")

print(f"Largest(5th): {fifth_largest}")

输出：

Average = [24, 24, 24, 24, 37, 37, 37, 37, 34, 34, 34, 34]

Max = [45, 45, 45, 45, 76, 76, 76, 76, 54, 54, 54, 54]

Largest(5th): [14, 14, 14, 14, 23, 23, 23, 23, 22, 22, 22, 22]