我有一些财务数据，并希望只获得特定时间段（小时，天，月......）的最后一笔交易。
示例：
>>df
time price_BRL qt time_dt
1312001297 23.49 1.00 2011-07-30 04:48:17
1312049148 23.40 1.00 2011-07-30 18:05:48
1312121523 23.49 2.00 2011-07-31 14:12:03
1312121523 23.50 6.50 2011-07-31 14:12:03
1312177622 23.40 2.00 2011-08-01 05:47:02
1312206416 23.25 1.00 2011-08-01 13:46:56
1312637929 18.95 1.50 2011-08-06 13:38:49
1312637929 18.95 4.00 2011-08-06 13:38:49
1312817114 0.80 0.01 2011-08-08 15:25:14
1312818289 0.10 0.01 2011-08-08 15:44:49
1312819795 6.00 0.09 2011-08-08 16:09:55
1312847064 16.00 0.86 2011-08-08 23:44:24
1312849282 16.00 6.14 2011-08-09 00:21:22
1312898146 19.90 1.00 2011-08-09 13:55:46
1312915666 6.00 0.01 2011-08-09 18:47:46
1312934897 19.90 1.00 2011-08-10 00:08:17
>>filter_by_last_day(df)
time price_BRL qt time_dt
1312049148 23.40 1.00 2011-07-30 18:05:48
1312121523 23.50 6.50 2011-07-31 14:12:03
1312206416 23.25 1.00 2011-08-01 13:46:56
1312637929 18.95 4.00 2011-08-06 13:38:49
1312847064 16.00 0.86 2011-08-08 23:44:24
1312915666 6.00 0.01 2011-08-09 18:47:46
1312934897 19.90 1.00 2011-08-10 00:08:17
我正在考虑使用groupby()并获取当天的mean()（此解决方案也可用于我的问题，但不完全正确）但不知道如何选择df.groupby('time.day').last() {1}}
答案 0 :(得分：3)
#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
感谢您MaxU寻求其他解决方案 - 为返回as_index=False添加参数DataFrame：
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，但必须按dropna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
您还可以使用dt.month：
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
使用hours有点复杂，如果groupby和date需要hours，则解决方案将替换为minutes和seconds 0之前的astype：
hours = df.time_dt.values.astype('<M8[h]')
print (hours)
['2011-07-30T04' '2011-07-30T18' '2011-07-31T14' '2011-07-31T14'
'2011-08-01T05' '2011-08-01T13' '2011-08-06T13' '2011-08-06T13'
'2011-08-08T15' '2011-08-08T15' '2011-08-08T16' '2011-08-08T23'
'2011-08-09T00' '2011-08-09T13' '2011-08-09T18' '2011-08-10T00']
df = df.groupby(hours).last().reset_index(drop=True)
print (df)
time price_BRL qt time_dt
0 1312001297 23.49 1.00 2011-07-30 04:48:17
1 1312049148 23.40 1.00 2011-07-30 18:05:48
2 1312121523 23.50 6.50 2011-07-31 14:12:03
3 1312177622 23.40 2.00 2011-08-01 05:47:02
4 1312206416 23.25 1.00 2011-08-01 13:46:56
5 1312637929 18.95 4.00 2011-08-06 13:38:49
6 1312818289 0.10 0.01 2011-08-08 15:44:49
7 1312819795 6.00 0.09 2011-08-08 16:09:55
8 1312847064 16.00 0.86 2011-08-08 23:44:24
9 1312849282 16.00 6.14 2011-08-09 00:21:22
10 1312898146 19.90 1.00 2011-08-09 13:55:46
11 1312915666 6.00 0.01 2011-08-09 18:47:46
12 1312934897 19.90 1.00 2011-08-10 00:08:17