我有一個(gè)數(shù)字列表，我想從這個(gè)列表中再創(chuàng)建 3 個(gè)列表，其中包含其中的最大值、平均值和第五大數(shù)字。我的原始列表overdraw是列表塊，這意味著其中有子塊，每個(gè)塊中有 6 個(gè)數(shù)字，總共有 3 個(gè)塊或 6x3 矩陣或數(shù)組。overdraw:[[16,13,23,14,33,45],[23,11,54,34,23,76],[22,54,34,43,41,11]]我知道如何計(jì)算此列表中的最大值、平均值和 5 個(gè)最大值。但是我想要一個(gè)特定的答案，比如我知道每個(gè)塊的最大值、平均值和第 5 個(gè)最大值，但我希望它們被打印 4 次。我知道所有的價(jià)值觀：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)運(yùn)行這段代碼后你會(huì)得到相同的輸出，但我想要這種格式的輸出：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]如您所見，每個(gè)平均值、最大值和最大數(shù)在各自的列表中打印了 4 次。那么任何人都可以幫助這個(gè)答案嗎？
2 回答
慕哥9229398
TA貢獻(xiàn)1877條經(jīng)驗(yàn) 獲得超6個(gè)贊
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)
這表現(xiàn)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
從這里開始，您仍然可以執(zhí)行所有您想要的計(jì)算和/或獲取 NumPy 數(shù)組，執(zhí)行df.values.
根據(jù)您的評(píng)論，您還可以將您的專欄作為單獨(dú)的實(shí)體，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]
哪種方法開始有點(diǎn)矯枉過(guò)正，但可讀性強(qiáng)。
梵蒂岡之花
TA貢獻(xiàn)1900條經(jīng)驗(yàn) 獲得超5個(gè)贊
返回您指定的列表的解決方案
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}")
輸出：
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]
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