Python 如何计算Pandas 中具有特定条件的行之间的天差

您可以将正向填充日期减go GroupBy.ffill，如果是Visit，则减go Series.sub，并将时间增量转换为Series.dt.days的天数，最后，如果需要，将0天减go Series.mask:

#convert column to datetimes
df['date'] = pd.to_datetime(df['date'])

m = df['Acton'].eq('Visit')
df['CV_days_after_visit'] = (df['date'].sub(df['date'].where(m).groupby(df['ID']).ffill())
                                       .dt.days
                                       .mask(m))
print (df)
    ID  Acton       date  CV_days_after_visit
0  A2A  Visit 2023-04-01                  NaN
1  A2A     CV 2023-04-05                  4.0
2  A2A     CV 2023-04-07                  6.0
3  BB3  Visit 2023-05-05                  NaN
4  BB3  Visit 2023-05-29                  NaN
5  BB3     CV 2023-05-30                  1.0
6  5EE  Visit 2023-06-01                  NaN
7  5EE     CV 2023-06-10                  9.0

它的工作原理:

df['date'] = pd.to_datetime(df['date'])

m = df['Acton'].eq('Visit')

print (df.assign(m1 = df['date'].where(m),
                 ffill = df['date'].where(m).groupby(df['ID']).ffill(),
                 sub = df['date'].sub(df['date'].where(m).groupby(df['ID']).ffill()),
                 days = df['date'].sub(df['date'].where(m).groupby(df['ID']).ffill())
                                  .dt.days))

    ID  Acton       date         m1      ffill    sub  days
0  A2A  Visit 2023-04-01 2023-04-01 2023-04-01 0 days     0
1  A2A     CV 2023-04-05        NaT 2023-04-01 4 days     4
2  A2A     CV 2023-04-07        NaT 2023-04-01 6 days     6
3  BB3  Visit 2023-05-05 2023-05-05 2023-05-05 0 days     0
4  BB3  Visit 2023-05-29 2023-05-29 2023-05-29 0 days     0
5  BB3     CV 2023-05-30        NaT 2023-05-29 1 days     1
6  5EE  Visit 2023-06-01 2023-06-01 2023-06-01 0 days     0
7  5EE     CV 2023-06-10        NaT 2023-06-01 9 days     9