Question

给出以下熊猫数据框：

            timestamp
0     2018-10-05 23:07:02
1     2018-10-05 23:07:13
2     2018-10-05 23:07:23
3     2018-10-05 23:07:36
4     2018-10-05 23:08:02
5     2018-10-05 23:09:16
6     2018-10-05 23:09:21
7     2018-10-05 23:09:39
8     2018-10-05 23:09:47
9     2018-10-05 23:10:01
10    2018-10-05 23:10:11
11    2018-10-05 23:10:23
12    2018-10-05 23:10:59
13    2018-10-05 23:11:03
14    2018-10-08 03:35:32
15    2018-10-08 03:35:58
16    2018-10-08 03:37:16
17    2018-10-08 03:38:04
18    2018-10-08 03:38:30
19    2018-10-08 03:38:36
20    2018-10-08 03:38:42
21    2018-10-08 03:38:52
22    2018-10-08 03:38:57
23    2018-10-08 03:39:10
24    2018-10-08 03:39:27
25    2018-10-08 03:40:47
26    2018-10-08 03:40:54
27    2018-10-08 03:41:02
28    2018-10-08 03:41:12
29    2018-10-08 03:41:32

如何在每行十分钟的时间内标记？例如：

            timestamp       10min_period
0     2018-10-05 23:07:02   period_1
2     2018-10-05 23:07:23   period_1
1     2018-10-05 23:07:13   period_1
2     2018-10-05 23:07:23   period_1
3     2018-10-05 23:07:36   period_1
4     2018-10-05 23:08:02   period_1
5     2018-10-05 23:09:16   period_1
6     2018-10-05 23:09:21   period_1
7     2018-10-05 23:09:39   period_1
8     2018-10-05 23:09:47   period_1
9     2018-10-05 23:10:01   period_1
10    2018-10-05 23:10:11   period_1
11    2018-10-05 23:10:23   period_1
12    2018-10-05 23:10:59   period_1
13    2018-10-05 23:11:03   period_1
14    2018-10-08 03:35:32   period_2
15    2018-10-08 03:35:58   period_2
16    2018-10-08 03:37:16   period_2
17    2018-10-08 03:38:04   period_2
18    2018-10-08 03:38:30   period_2
19    2018-10-08 03:38:36   period_2
20    2018-10-08 03:38:42   period_2
21    2018-10-08 03:38:52   period_2
22    2018-10-08 03:38:57   period_2
23    2018-10-08 03:39:10   period_2
24    2018-10-08 03:39:27   period_2
25    2018-10-08 03:40:47   period_2
26    2018-10-08 04:40:54   period_3
27    2018-10-08 04:41:02   period_3
28    2018-10-08 04:41:12   period_3
29    2018-10-08 04:41:32   period_3

从上面的预期输出中可以看到，每个period_n标签都是通过计算10分钟的时间段来创建的，当datetime系列超过10分钟的阈值时，将创建一个新标签。我尝试使用dt.floor(10Min)对象，但是它无法正常工作，因为它无法跟踪10分钟的开始时间和结束时间。我也尝试过：

a = df['timestamp'].offsets.DateOffset(minutes=10)

但是，它不起作用。是否知道如何在10分钟内对DF进行分段？这个问题与其他问题有所不同，因为我没有指定任何特定的开始计数的时间。也就是说，我从第一个datetime行实例开始计数，并从那开始计算十分钟的时间间隔。

更新：

转换为日期时间对象后，我还尝试了

df['timestamp'].groupby(pd.TimeGrouper(freq='10Min'))

但是，我得到了：

TypeError: Only valid with DatetimeIndex, TimedeltaIndex or PeriodIndex, but got an instance of 'RangeIndex'

Answer 1

使用一些向量化算法，这应该是可能的（并且表现出色）：

# Convert to datetime if not already.
# df['timestamp'] = pd.to_datetime(df['timestamp'], errors='coerce')    
u = (df.assign(timestamp=df['timestamp'].dt.floor('20min'))
       .groupby(pd.Grouper(key='timestamp',freq='10min'))
       .ngroup())

df['10min_period'] = np.char.add('period_', (pd.factorize(u)[0] + 1).astype(str))

不幸的是，这里的缺点是，虽然这将为您的样本数据产生预期的输出，但没有简单的方法来处理10分钟的连续间隔（pd.Grouper从您的列，因此dt.floor('20min')是第一步的必要步骤-这会无意中使“ period_ {i}”下的“ period_ {i + 1}”行中的某些行或大部分行）。

Answer 2

为重现您的问题，我这样做：

index = pd.date_range(datetime.datetime.now().date() - datetime.timedelta(10), periods=100, freq='min')

这样我就有了这个DataFrame：

a = pd.DataFrame(index)
a
                     0
0  2018-10-28 00:00:00
1  2018-10-28 00:01:00
2  2018-10-28 00:02:00
3  2018-10-28 00:03:00
4  2018-10-28 00:04:00
5  2018-10-28 00:05:00
6  2018-10-28 00:06:00
7  2018-10-28 00:07:00
8  2018-10-28 00:08:00
9  2018-10-28 00:09:00
10 2018-10-28 00:10:00
                   ...
[100 rows x 1 columns]

然后，我这样做：

a['period'] = a.apply(lambda x: "perdio_%d"%(int(x[0].minute/10) + 1), axis=1)

我有这个解决方案：

                     0    period
0  2018-10-28 00:00:00  perdio_1
1  2018-10-28 00:01:00  perdio_1
2  2018-10-28 00:02:00  perdio_1
3  2018-10-28 00:03:00  perdio_1
4  2018-10-28 00:04:00  perdio_1
5  2018-10-28 00:05:00  perdio_1
6  2018-10-28 00:06:00  perdio_1
7  2018-10-28 00:07:00  perdio_1
8  2018-10-28 00:08:00  perdio_1
9  2018-10-28 00:09:00  perdio_1
10 2018-10-28 00:10:00  perdio_2
11 2018-10-28 00:11:00  perdio_2
12 2018-10-28 00:12:00  perdio_2
13 2018-10-28 00:13:00  perdio_2
14 2018-10-28 00:14:00  perdio_2
15 2018-10-28 00:15:00  perdio_2
                             ...

我希望它会有所帮助

Answer 3

我已将您的数据框保存在记事本中，并将其命名为timestamp.txt。在记事本中看起来像这样：

然后我写了这个简单的代码：

import pandas as pd

timestamp = pd.read_csv("C:\\...path_of_your_file...\\timestamp.txt")  # read file
timestamp['10_Minute_Period'] = 0  # add column and initilize it to zero
numb_groups = int((timestamp.shape[0])/10)  # calculate number of groups
groups = 1  # initialize number of groups to one

while groups <= numb_groups+1:
   for idx, _ in timestamp.iterrows():  # iterate over row indexes
       # check if current row is below the group and the value is equal to 0
       if idx < groups*10 and timestamp.at[idx,'10_Minute_Period'] == 0:
           # in this case, write corresponding Period
           timestamp.loc[idx,'10_Minute_Period'] = ('Period' + str(groups))
   groups += 1  # increment groups and check while condition

print(timestamp)  # print the final modified timestamp

希望有帮助！

Answer 4

df['timestamp'] = pd.to_datetime(df['timestamp'])
diffs = df['timestamp'] - df['timestamp'].shift()
laps = diffs > pd.Timedelta('10 min')
periods = laps.cumsum().apply(lambda x: 'period_{}'.format(x+1))
df['10min_period'] = periods

按10分钟间隔将熊猫DataFrame分组

4 个答案: