Question

df1 =

Date     Team1     Team2
6/1      Boston    New York
6/13     New York  Boston
6/27     Boston    New York

我正在尝试计算自波士顿上次出现在任一列以来的天数，但我只能使用df1 ['波士顿踢球后的天数'] = df1来计算如何在一个列中查找该天数。 groupby（'Team1'）['Date']。diff（）。fillna（0）

我希望输出为：

Date     Team1     Team2      Days since Boston played
6/1      Boston    New York   0
6/13     New York  Boston     12
6/27     Boston    New York   14

编辑-扩展数据框，以了解如何将其应用于所有团队，而不仅仅是一个团队 我希望输出为：

Date     Team1     Team2      Days since **Team1** played
6/1      Boston    New York   0
6/13     New York  Chicago    12
6/27     Boston    New York   14
6/28     Chicago   Boston     15

Answer 1

使用isin检查team1或team2中是否存在Boston并找到timedelta

df['Date'] = pd.to_datetime(df['Date'], format = '%m/%d')

df.loc[df.isin(['Boston']).any(1),'Days since Boston played'] = df.loc[df.isin(['Boston']).any(1), 'Date'].diff().dt.days


    Date        Team1       Team2       Days since Boston played
0   1900-06-01  Boston      New York    NaN
1   1900-06-13  New York    Boston      12.0
2   1900-06-27  Boston      New York    14.0

如果您希望日期列恢复为原始格式，则可以使用strftime

df['Date'] = df['Date'].dt.strftime('%m/%d')

    Date    Team1       Team2       Days since Boston played
0   06/01   Boston      New York    NaN
1   06/13   New York    Boston      12.0
2   06/27   Boston      New York    14.0

Answer 2

您可以在Team1上进行分组，然后在日期上进行区别：

# Note: you should post a creatable example in your post next time 
data = {
    'Date': ['2018-06-01', '2018-06-13', '2018-06-27'],
    'Team1':['Boston', 'New York', 'Boston'], 
    'Team2':['New York', 'Boston', 'New York']
}
df = pd.DataFrame(data)
df['Date'] = pd.to_datetime(df['Date'])
df['Time between games'] = df.groupby('Team1')['Date'].diff().fillna(0)

这实际上将为您提供所有主队比赛之间的区别。

查找多个列中条目之间的日期差异

2 个答案: