python-熊猫-从每个用户检索先前的结果/行

我是熊猫新手.

我有一个看起来像这样的数据框(只有更大)：

    Horses        RaceDate Position
1   RedHorse      1/2/00   2
2   BlueHorse     1/2/00   6
3   YellowHorse   1/2/00   7
4   RedHorse      15/1/00  3

我想为以前的结果添加列.这样我的数据框可能最终看起来像：

    Horses        RaceDate Position   PrevPosition
1   RedHorse      1/2/00   2          3
2   BlueHorse     1/2/00   6          -
3   YellowHorse   1/2/00   7          -
4   RedHorse      15/1/00  3          -

我尝试了以下方法：

def prevRuns(horseName, raceDate):
    horseDf = df.loc[df['Horse'] == horseName]
    currentRace = horseDf.index[horseDf['RaceDate'] == raceDate]

    if len(horseDf.index) >= currentRace:
        return horseDf.at[currentRace+1,'Position']
    else:
        return 0


df['prevRun'] = df['Horse'].apply(prevRuns, raceDate = df['RaceDate'])

但这是行不通的.

ValueError: Can only compare identically-labeled Series objects

为什么不起作用？

有没有更优雅的方式来实现我要完成的任务？

解决方法:

您可以使用groupby shift：

# convert dates to datetime and sort descending
df['RaceDate'] = pd.to_datetime(df['RaceDate'], dayfirst=True)
df = df.sort_values('RaceDate', ascending=False)

# groupby and shift for previous position
df['PrevPosition'] = df.groupby('Horses')['Position'].shift(-1)

print(df)

        Horses   RaceDate  Position  PrevPosition
1     RedHorse 2000-02-01         2           3.0
2    BlueHorse 2000-02-01         6           NaN
3  YellowHorse 2000-02-01         7           NaN
4     RedHorse 2000-01-15         3           NaN

标签：pandas-groupby,pandas,python
来源： https://codeday.me/bug/20191211/2106727.html