您可以在dt.date之前使用groupby并在last之前聚合：

#if necessery convert to datetime

df.time_dt = pd.to_datetime(df.time_dt)

df = df.groupby(df.time_dt.dt.date).last().reset_index(drop=True)

print (df)

time price_BRL qt time_dt

0 1312049148 23.40 1.00 2011-07-30 18:05:48

1 1312121523 23.50 6.50 2011-07-31 14:12:03

2 1312206416 23.25 1.00 2011-08-01 13:46:56

3 1312637929 18.95 4.00 2011-08-06 13:38:49

4 1312847064 16.00 0.86 2011-08-08 23:44:24

5 1312915666 6.00 0.01 2011-08-09 18:47:46

6 1312934897 19.90 1.00 2011-08-10 00:08:17

谢谢MaxU的另一个解决方案 – 为返回DataFrame添加参数as_index = False：

df = df.groupby(df.time_dt.dt.date,as_index=False).last()

print (df)

time price_BRL qt time_dt

0 1312049148 23.40 1.00 2011-07-30 18:05:48

1 1312121523 23.50 6.50 2011-07-31 14:12:03

2 1312206416 23.25 1.00 2011-08-01 13:46:56

3 1312637929 18.95 4.00 2011-08-06 13:38:49

4 1312847064 16.00 0.86 2011-08-08 23:44:24

5 1312915666 6.00 0.01 2011-08-09 18:47:46

6 1312934897 19.90 1.00 2011-08-10 00:08:17

使用resample的解决方案,但必须在dropna之前删除NaN行：

df = df.resample('d',on='time_dt').last().dropna(how='all').reset_index(drop=True)

#cast column time to int

df.time = df.time.astype(int)

print (df)

time price_BRL qt time_dt

0 1312049148 23.40 1.00 2011-07-30 18:05:48

1 1312121523 23.50 6.50 2011-07-31 14:12:03

2 1312206416 23.25 1.00 2011-08-01 13:46:56

3 1312637929 18.95 4.00 2011-08-06 13:38:49

4 1312847064 16.00 0.86 2011-08-08 23:44:24

5 1312915666 6.00 0.01 2011-08-09 18:47:46

6 1312934897 19.90 1.00 2011-08-10 00:08:17

—

df = df.groupby(df.time_dt.dt.month).last().reset_index(drop=True)

print (df)

time price_BRL qt time_dt

0 1312121523 23.5 6.5 2011-07-31 14:12:03

1 1312934897 19.9 1.0 2011-08-10 00:08:17

有几个小时它有点复杂,如果需要groupby按日期和小时一起,解决方案是用astype将分钟和秒数替换为0：

hours = df.time_dt.values.astype('