根据 Groupby 函数的输出命名 Pandas 数据框

我有一个数据集,其中包含在多个赛季的大量足球比赛中拍摄的所有镜头。我编写了以下脚本来为每个比赛和相应的赛季制作子集。


import pandas as pd

import csv

shots = pd.read_csv("C:/Users/HJA/Desktop/Betting/understatV0.01/shots.csv", encoding='iso-8859-1')


shots_premier_league = shots.groupby(['Competition']).get_group('Premier_League')

shots_bundesliga = shots.groupby(['Competition']).get_group('Bundesliga')

shots_la_liga = shots.groupby(['Competition']).get_group('La_Liga')

shots_ligue_1 = shots.groupby(['Competition']).get_group('Ligue_1')

shots_serie_a = shots.groupby(['Competition']).get_group('Serie_A')

一切顺利,直到这一点。但是,现在我想将每个比赛细分为每个赛季的样本。我使用以下脚本(在这种情况下,我以英超联赛为例:


shots_premier_league_2014 = shots_premier_league.groupby(['Season']).get_group('2014')

shots_premier_league_2015 = shots_premier_league.groupby(['Season']).get_group('2015')

shots_premier_league_2016 = shots_premier_league.groupby(['Season']).get_group('2016')

shots_premier_league_2017 = shots_premier_league.groupby(['Season']).get_group('2017')

shots_premier_league_2018 = shots_premier_league.groupby(['Season']).get_group('2018')

这导致以下错误:

http://img3.mukewang.com/613c6ab80001fd2418260656.jpg

我 100% 确定 2014 年是一个实际值。另外,如何编写一个以pandas数据框的名义自动包含比赛季节的函数?


一只萌萌小番薯
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1回答

繁星淼淼

我认为问题是2014整数,所以需要删除'':.get_group(2014)但这里更好的是 create dictionary of DataFrameslike,因为不推荐使用全局变量:dfs = dict(tuple(shots_premier_league.groupby(['Season'])))然后通过键选择每个数据帧,例如:print (dfs[2014])print (dfs[2015])如何编写一个以熊猫数据框的名义自动包含比赛和季节的函数?dfs = dict(tuple(shots_premier_league.groupby(['Competition','Season'])))print (dfs[('Bundesliga', 2014)])如果要按字符串选择:d = dict(tuple(df.groupby(['Competition','Season'])))#python 3.6+ solution with f-stringsdfs = {f'{k1}_{k2}' :v for (k1, k2), v in d.items()}#python bellow#dfs = {'{}_{}'.format(k1, k2) :v for (k1, k2), v in d.items()}print (dfs['Bundesliga_2014'])如果想查看数据的所有键:print (dfs.keys())
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