Question

我拉出了我的小组的短信，它看起来类似于下表（不包括第三列）。

如何根据时间将第一个对话与第二个对话分开，并将每条消息分配给一个session_id？如果有帮助，我很高兴地假设，如果未响应任何消息一个小时后，下一封邮件会开始新的对话。

如果在前两栏中给出第一列，那么我理想地能够在会话长度不同的近45,000条消息中找出第三列。

一旦我的短信分成对话，我想我可以训练一个聊天机器人来参与我们的聊天！我不知道我在这里做什么，所以不胜感激：）

| created_at          | message                              | conversation_id |
| 2018-07-03 02:12:33 | knock knock                          | 1               |
| 2018-07-03 02:12:35 | who's there                          | 1               |
| 2018-07-03 02:12:40 | Europe                               | 1               |
| 2018-07-03 02:12:45 | Europe who?                          | 1               |
| 2018-07-03 02:12:48 | No - you're a poo                    | 1               |
| 2018-07-03 03:15:17 | knock knock                          | 2               |
| 2018-07-03 03:15:20 | who's there                          | 2               |
| 2018-07-03 03:15:23 | the KGB                              | 2               |
| 2018-07-03 03:15:28 | the KGB who?                         | 2               |
| 2018-07-03 03:15:33 | SLAP the KGB will ask the questions! | 2               |

Answer 1

这应该可以解决问题。我认为您在遍历数据帧以分配ID方面不会有太大改进，因为它们基于与列中先前值的链接连接：

d = {"created_at": pd.to_datetime(["2018-07-03 02:12:33", "2018-07-03 02:12:35","2018-07-03 02:12:40","2018-07-03 02:12:45","2018-07-03 02:12:48","2018-07-03 03:15:17","2018-07-03 03:15:20","2018-07-03 03:15:23","2018-07-03 03:15:28","2018-07-03 03:15:33","2018-08-03 09:00:00","2018-09-03 10:15:00"]),
     "message": ["knock knock","who's there","Europe","Europe who?","No - you're a poo","knock knock","who's there","the KGB","the KGB who?","SLAP the KGB will ask q's!","Hello?","Hello, again?"]}

import pandas as pd
import numpy as np

#60mins in secs
thresh = 60*60

df = pd.DataFrame(data=d)

#Creating time delta from previous message
df["delta"] = df["created_at"].diff().fillna(0).dt.total_seconds()

#Normalising delta based on threshold as a flag for new convos
df["id"] = np.where(df["delta"] < thresh, 0, 1)
df = df.drop(["delta"], axis=1)

#Assigning ID's to each convo
for i in range(1, len(df)):
    df.loc[i, 'id'] += df.loc[i-1, 'id']

print(df)

            created_at                     message  id
0  2018-07-03 02:12:33                 knock knock   0
1  2018-07-03 02:12:35                 who's there   0
2  2018-07-03 02:12:40                      Europe   0
3  2018-07-03 02:12:45                 Europe who?   0
4  2018-07-03 02:12:48           No - you're a poo   0
5  2018-07-03 03:15:17                 knock knock   1
6  2018-07-03 03:15:20                 who's there   1
7  2018-07-03 03:15:23                     the KGB   1
8  2018-07-03 03:15:28                the KGB who?   1
9  2018-07-03 03:15:33  SLAP the KGB will ask q's!   1
10 2018-08-03 09:00:00                      Hello?   2
11 2018-09-03 10:15:00               Hello, again?   3

如何根据时间差异分离大熊猫数据框？

1 个答案: