python - 如何使用 Pandas 获取每个组中的上一个和中间日期

Question

我有以下数据框:

df = pd.DataFrame({
'subject_id':[1,1,1,1,1,1,1,2,2,2,2,2],
'time_1' :['2173/04/01 12:35:00','2173/04/01 12:50:00','2173/04/02 12:59:00','2173/04/02 13:14:00','2173/04/04 13:37:00','2173/04/06 13:39:00','2173/04/06 11:30:00','2173/04/08 16:00:00','2173/04/09 22:00:00','2173/04/11 04:00:00','2173/04/13 04:30:00','2173/04/14 08:00:00'],
 'val' :[0,0,0,0,1,0,0,0,0,0,0,0]
})
df['time_1'] = pd.to_datetime(df['time_1'])
df['day'] = df['time_1'].dt.day

我想做的是，根据下面给出的条件，为每个主题分配两天时间

val = 1- 获取前两天(从 val = 1)

val = 0- 获取他持续时间的中间两天

这是我尝试过的方法，但它既不优雅也不准确。

con = lambda x: (x.eq(1))
con_1 = lambda x: (x.eq(0))
df.loc[df.groupby('subject_id')['val'].transform(con)]['time_1'] - timedelta(days = 2)

以下是我的预期输出:

subject = 1 有 val = 1，所以我们从 val = 1 得到他和 subject 的前两天= 2，因为没有 val = 1，我们得到 他持续时间的中间 2 天(第 8 - 14 日)

Answer 1

我相信你需要:

def func(x):
    floor = x['time_1'].dt.floor('d')

    if (x['val'].any()):
        date = x.loc[x['val'] == 1, 'time_1'].iat[0].floor('d')
        new_date = date - pd.Timedelta(3, unit='d')
        return (x[floor.between(new_date, date, inclusive=False)])
    else:
        date = floor.mean().floor('d')
        new_date = date + pd.Timedelta(1, unit='d')
        return (x[floor.between(date, new_date)])

df = df.groupby('subject_id', group_keys=False).apply(func)
print (df)
   subject_id              time_1  val  day
2           1 2173-04-02 12:59:00    0    2
3           1 2173-04-02 13:14:00    0    2
9           2 2173-04-11 04:00:00    0   11