基于pandas中的时间戳创建交互会话 [英] Creating interaction sessions based on timestamps in pandas
问题描述
我正在尝试定义在不同的移动使用会话中使用了哪些应用程序.基本上，一个会话是由一个单个用户在一小段时间内(又名会话增量).换句话说，如果在之前的 5 分钟内没有发生任何交互交互，用户的会话被视为关闭.下一个交互被视为一个单独的会话.我想知道数据集中有多少移动会话.另外，我想知道每个会话中启动了哪些应用程序.我的数据框中的所有行都带有时间戳.以下是数据集中的一个示例:
I am trying to define which apps were used in different mobile usage sessions. Basically, a session is a series of actions by a single user made within a small range of time (aka. session delta). In other words, if no interaction happens within 5 minutes of the previous interaction, a user’s session is deemed closed. The next interaction is considered a separate session. I would like to know how many mobile sessions are there in the dataset. Also, I would like to know which apps were launched in each session. All the rows in my data frame are time stamped. Here is an example from the dataset:
timestamp App
6784 2018-04-08 14:31:29.209 Google
6785 2018-04-08 14:58:42.875 Google
6786 2018-04-08 18:18:04.757 Chrome
6787 2018-04-08 21:08:41.368 Google
6788 2018-04-11 10:53:10.744 Google
6789 2018-04-14 19:54:37.441 Google
6790 2018-04-14 19:54:59.833 Google
6791 2018-04-14 19:55:10.844 YouTube
6792 2018-04-14 19:55:34.486 Google
6793 2018-04-14 20:23:00.315 Google
6794 2018-04-15 08:23:44.873 Google
6795 2018-04-15 08:24:07.257 Google
6796 2018-04-16 16:42:35.538 Google
6797 2018-04-16 16:42:48.351 Google
6798 2018-04-17 08:10:54.734 Google
6799 2018-04-17 08:13:28.855 Google
6800 2018-04-17 08:16:49.408 Google
6801 2018-04-17 08:18:55.049 Google
6802 2018-04-17 08:21:04.201 Google
6803 2018-04-17 08:26:14.254 Google
这是所需的输出:
timestamp App SessionID
6784 2018-04-08 14:31:29.209 Google 1
6785 2018-04-08 14:58:42.875 Google 2
6786 2018-04-08 18:18:04.757 Chrome 3
6787 2018-04-08 21:08:41.368 Google 4
6788 2018-04-11 10:53:10.744 Google 5
6789 2018-04-14 19:54:37.441 Google 6
6790 2018-04-14 19:54:59.833 Google 6
6791 2018-04-14 19:55:10.844 YouTube 6
6792 2018-04-14 19:55:34.486 Google 6
6793 2018-04-14 20:23:00.315 Google 7
6794 2018-04-15 08:23:44.873 Google 8
6795 2018-04-15 08:24:07.257 Google 8
6796 2018-04-16 16:42:35.538 Google 9
6797 2018-04-16 16:42:48.351 Google 9
6798 2018-04-17 08:10:54.734 Google 10
6799 2018-04-17 08:13:28.855 Google 10
6800 2018-04-17 08:16:49.408 Google 10
6801 2018-04-17 08:18:55.049 Google 10
6802 2018-04-17 08:21:04.201 Google 10
6803 2018-04-17 08:26:14.254 Google 11
推荐答案
我想你想要 .shift + .cumsum()
+1 是因为您的第一行将始终是 NaT 的差异，其计算结果为 False 进行比较，这将否则总是从 0 开始 SessionID.
The +1 is because your first row will always be NaT for the difference, which evaluates to False for the comparison, which will always start the SessionID from 0 otherwise.
import pandas as pd
df['SessionID'] = (df.timestamp-df.timestamp.shift(1) > pd.Timedelta(5, 'm')).cumsum()+1
timestamp App SessionID
6784 2018-04-08 14:31:29.209 Google 1
6785 2018-04-08 14:58:42.875 Google 2
6786 2018-04-08 18:18:04.757 Chrome 3
6787 2018-04-08 21:08:41.368 Google 4
6788 2018-04-11 10:53:10.744 Google 5
6789 2018-04-14 19:54:37.441 Google 6
6790 2018-04-14 19:54:59.833 Google 6
6791 2018-04-14 19:55:10.844 YouTube 6
6792 2018-04-14 19:55:34.486 Google 6
6793 2018-04-14 20:23:00.315 Google 7
6794 2018-04-15 08:23:44.873 Google 8
6795 2018-04-15 08:24:07.257 Google 8
6796 2018-04-16 16:42:35.538 Google 9
6797 2018-04-16 16:42:48.351 Google 9
6798 2018-04-17 08:10:54.734 Google 10
6799 2018-04-17 08:13:28.855 Google 10
6800 2018-04-17 08:16:49.408 Google 10
6801 2018-04-17 08:18:55.049 Google 10
6802 2018-04-17 08:21:04.201 Google 10
6803 2018-04-17 08:26:14.254 Google 11
如果您还有 UserID，那么您可以实现逻辑，当时间超过 5 分钟或 userID 更改时，您会增加 ID.这是通过以下方式完成的:
If you also have UserID you can then implement logic where you increment the ID when either the time is greater than 5 minutes, OR the userID changes. That is accomplished by:
import pandas as pd
data = '''\
1,2018-04-08T09:48:17.573,YouTube
1,2018-04-08T09:47:57.849,Chrome
1,2018-04-08T09:48:28.538,Instagram
1,2018-04-08T09:48:37.381,Maps
2,2018-04-08T09:48:46.680,Netflix
2,2018-04-08T09:48:56.672,Google Play Store
1,2018-04-08T09:56:58.880,Google
1,2018-04-08T09:57:25.461,DB Navigator
1,2018-04-08T11:28:38.762,Google
1,2018-04-08T12:58:31.455,Google
1,2018-04-08T14:31:18.131,Google
1,2018-04-08T14:31:29.209,Google
1,2018-04-08T14:58:42.875,Google
1,2018-04-08T18:18:04.757,Chrome
1,2018-04-08T21:08:41.368,Google
1,2018-04-11T10:53:10.744,Google
1,2018-04-14T19:54:37.441,Google
1,2018-04-14T19:54:59.833,Google
1,2018-04-14T19:55:10.844,YouTube
1,2018-04-14T19:55:34.486,Google
1,2018-04-14T20:23:00.315,Google
2,2018-04-15T08:23:44.873,Google
2,2018-04-15T08:24:07.257,Google'''
df = pd.read_csv(pd.compat.StringIO(data), names=['userID','timestamp','App'],
parse_dates=[1])
df.sort_values(by=['userID','timestamp'], inplace=True)
cond1 = df.timestamp-df.timestamp.shift(1) > pd.Timedelta(5, 'm')
cond2 = df.userID != df.userID.shift(1)
df['SessionID'] = (cond1|cond2).cumsum()
返回:
userID timestamp App SessionID
1 1 2018-04-08 09:47:57.849 Chrome 1
0 1 2018-04-08 09:48:17.573 YouTube 1
2 1 2018-04-08 09:48:28.538 Instagram 1
3 1 2018-04-08 09:48:37.381 Maps 1
6 1 2018-04-08 09:56:58.880 Google 2
7 1 2018-04-08 09:57:25.461 DB Navigator 2
8 1 2018-04-08 11:28:38.762 Google 3
9 1 2018-04-08 12:58:31.455 Google 4
10 1 2018-04-08 14:31:18.131 Google 5
11 1 2018-04-08 14:31:29.209 Google 5
12 1 2018-04-08 14:58:42.875 Google 6
13 1 2018-04-08 18:18:04.757 Chrome 7
14 1 2018-04-08 21:08:41.368 Google 8
15 1 2018-04-11 10:53:10.744 Google 9
16 1 2018-04-14 19:54:37.441 Google 10
17 1 2018-04-14 19:54:59.833 Google 10
18 1 2018-04-14 19:55:10.844 YouTube 10
19 1 2018-04-14 19:55:34.486 Google 10
20 1 2018-04-14 20:23:00.315 Google 11
4 2 2018-04-08 09:48:46.680 Netflix 12
5 2 2018-04-08 09:48:56.672 Google Play Store 12
21 2 2018-04-15 08:23:44.873 Google 13
22 2 2018-04-15 08:24:07.257 Google 13
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