如何按照id和日期（YYYY-MM-DD）分组生成pandas数据框的时间差列？

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英文:

How to generate pandas dataframe timedelta column grouped by id and date (YYYY-MM-DD)?

问题

  1. id	datetime	            datetime_baseline	    timedelta
  2. a1	2016-01-01 00:01:00.156	2016-01-01 00:01:00.156	0
  3. a1	2016-01-01 12:00:00.425	2016-01-01 00:01:00.156	719
  4. a1	2016-01-02 00:59:00.123	2016-01-02 00:59:00.123	0
  5. a1	2016-01-02 14:16:00.548	2016-01-02 00:59:00.123	797
  6. a2	2016-01-01 12:00:00.147	2016-01-01 12:00:00.147	0
  7. a2	2016-01-01 13:59:00.123	2016-01-01 12:00:00.147	119
  8. a2	2016-01-02 08:01:00.147	2016-01-02 08:01:00.147	0
  9. a2	2016-01-02 18:49:00.123	2016-01-02 08:01:00.147	648
  10. a3	2016-02-01 12:00:00.147	2016-02-01 12:00:00.147	0
  11. a3	2016-02-01 13:59:00.123	2016-02-01 12:00:00.147	119
  12. a3	2016-02-02 08:01:00.147	2016-02-02 08:01:00.147	0
  13. a3	2016-02-02 18:49:00.123	2016-02-02 08:01:00.147	648
英文:

Suppose I have a dataframe with id and datetime columns:

  1. df = pd.DataFrame({"id": ["a1", "a1", "a1", "a1", "a2", "a2", "a2", "a2", "a3", "a3", "a3", "a3"],
  2.                    "datetime": ["2016-01-01 00:01:00.156",
  3.                                 "2016-01-01 12:00:00.425",
  4.                                 "2016-01-02 00:59:00.123",
  5.                                 "2016-01-02 14:16:00.548",
  6.                                 "2016-01-01 12:00:00.147",
  7.                                 "2016-01-01 13:59:00.123",
  8.                                 "2016-01-02 08:01:00.147",
  9.                                 "2016-01-02 18:49:00.123",
  10.                                 "2016-02-01 12:00:00.147",
  11.                                 "2016-02-01 13:59:00.123",
  12.                                 "2016-02-02 08:01:00.147",
  13.                                 "2016-02-02 18:49:00.123"]})
  14. df["datetime"] = pd.to_datetime(df["datetime"])
  15. df

Here is the dataframe:

  1.     id	datetime
  2. 0	a1	2016-01-01 00:01:00.156
  3. 1	a1	2016-01-01 12:00:00.425
  4. 2	a1	2016-01-02 00:59:00.123
  5. 3	a1	2016-01-02 14:16:00.548
  6. 4	a2	2016-01-01 12:00:00.147
  7. 5	a2	2016-01-01 13:59:00.123
  8. 6	a2	2016-01-02 08:01:00.147
  9. 7	a2	2016-01-02 18:49:00.123
  10. 8	a3	2016-02-01 12:00:00.147
  11. 9	a3	2016-02-01 13:59:00.123
  12. 10	a3	2016-02-02 08:01:00.147
  13. 11	a3	2016-02-02 18:49:00.123

I want to generate column timedelta that has a timedelta value. This is the output I expect to get:

  1.     id	datetime	            datetime_baseline	    timedelta
  2. 0	a1	2016-01-01 00:01:00.156	2016-01-01 00:01:00.156	0
  3. 1	a1	2016-01-01 12:00:00.425	2016-01-01 00:01:00.156	719
  4. 2	a1	2016-01-02 00:59:00.123	2016-01-02 00:59:00.123	0
  5. 3	a1	2016-01-02 14:16:00.548	2016-01-02 00:59:00.123	797
  6. 4	a2	2016-01-01 12:00:00.147	2016-01-01 12:00:00.147	0
  7. 5	a2	2016-01-01 13:59:00.123	2016-01-01 12:00:00.147	119
  8. 6	a2	2016-01-02 08:01:00.147	2016-01-02 08:01:00.147	0
  9. 7	a2	2016-01-02 18:49:00.123	2016-01-02 08:01:00.147	648
  10. 8	a3	2016-02-01 12:00:00.147	2016-02-01 12:00:00.147	0
  11. 9	a3	2016-02-01 13:59:00.123	2016-02-01 12:00:00.147	119
  12. 10	a3	2016-02-02 08:01:00.147	2016-02-02 08:01:00.147	0
  13. 11	a3	2016-02-02 18:49:00.123	2016-02-02 08:01:00.147	648

Here is how the timedelta values should be calculated: 1) the code needs to identify the FIRST datetime within the same id and date ('YYYY-MM-DD'), and 2) use it as baseline (datetime_baseline) to compute the timedelta (in minutes) w.r.t. other datetimes within same id and same date. For id='a1' and date='2016-01-01', the datetime_baseline='2016-01-01 00:01:00.156'. So, at index=0, timedelta has value=0 because '2016-01-01 00:01:00.156' - datetime_baseline=0. Meanwhile, at index=1, timedelta has value=719 because '2016-01-01 12:00:00.425' - datetime_baseline=719 (minutes). At index=2, id is the same as before but date is now '2016-01-02', so a new baseline will be used: '2016-01-02 00:59:00.123'. timedelta='2016-01-02 00:59:00.123' - datetime_baseline=0. At index=3, timedelta='2016-01-02 14:16:00.548' - datetime_baseline=797.

Although I see how the timedelta values should be calculated (timedelta=datetime-datetime_baseline), I don't know how to have the baseline values identified (i.e. how to generate datetime_baseline column). Please, let me know if you need any further explanation.

ps> the actual dataframe has +500 thousand rows.

答案1

得分: 2

使用GroupBy.transform进行基线计算：

  1. df["datetime_baseline"] = (df.groupby(["id", df["datetime"].dt.date])
  2.                                        ["datetime"].transform("first"))

并使用dt.total_seconds计算时间差：

  1. df["timedelta"] = ((df["datetime"].sub(df["datetime_baseline"]))
  2.                               .dt.total_seconds().div(60).round(0).astype(int))

输出：

  1. print(df)
  2.     
  3.     id                datetime       datetime_baseline  timedelta
  4. 0   a1 2016-01-01 00:01:00.156 2016-01-01 00:01:00.156          0
  5. 1   a1 2016-01-01 12:00:00.425 2016-01-01 00:01:00.156        719
  6. 2   a1 2016-01-02 00:59:00.123 2016-01-02 00:59:00.123          0
  7. 3   a1 2016-01-02 14:16:00.548 2016-01-02 00:59:00.123        797
  8. 4   a2 2016-01-01 12:00:00.147 2016-01-01 12:00:00.147          0
  9. 5   a2 2016-01-01 13:59:00.123 2016-01-01 12:00:00.147        119
  10. 6   a2 2016-01-02 08:01:00.147 2016-01-02 08:01:00.147          0
  11. 7   a2 2016-01-02 18:49:00.123 2016-01-02 08:01:00.147        648
  12. 8   a3 2016-02-01 12:00:00.147 2016-02-01 12:00:00.147          0
  13. 9   a3 2016-02-01 13:59:00.123 2016-02-01 12:00:00.147        119
  14. 10  a3 2016-02-02 08:01:00.147 2016-02-02 08:01:00.147          0
  15. 11  a3 2016-02-02 18:49:00.123 2016-02-02 08:01:00.147        648
英文:

With GroupBy.transform to make the baseline :

  1. df["datetime_baseline"] = (df.groupby(["id", df["datetime"].dt.date])
  2. ["datetime"].transform("first"))

And dt.total_seconds to compute the timedelta :

  1. df["timedelta"] = ((df["datetime"].sub(df["datetime_baseline"]))
  2. .dt.total_seconds().div(60).round(0).astype(int))

​
Output :

  1. print(df)
  2. id                datetime       datetime_baseline  timedelta
  3. 0   a1 2016-01-01 00:01:00.156 2016-01-01 00:01:00.156          0
  4. 1   a1 2016-01-01 12:00:00.425 2016-01-01 00:01:00.156        719
  5. 2   a1 2016-01-02 00:59:00.123 2016-01-02 00:59:00.123          0
  6. 3   a1 2016-01-02 14:16:00.548 2016-01-02 00:59:00.123        797
  7. 4   a2 2016-01-01 12:00:00.147 2016-01-01 12:00:00.147          0
  8. 5   a2 2016-01-01 13:59:00.123 2016-01-01 12:00:00.147        119
  9. 6   a2 2016-01-02 08:01:00.147 2016-01-02 08:01:00.147          0
  10. 7   a2 2016-01-02 18:49:00.123 2016-01-02 08:01:00.147        648
  11. 8   a3 2016-02-01 12:00:00.147 2016-02-01 12:00:00.147          0
  12. 9   a3 2016-02-01 13:59:00.123 2016-02-01 12:00:00.147        119
  13. 10  a3 2016-02-02 08:01:00.147 2016-02-02 08:01:00.147          0
  14. 11  a3 2016-02-02 18:49:00.123 2016-02-02 08:01:00.147        648

答案2

得分: 1

  1. df['datetime_baseline'] = df.groupby(['id', df['datetime'].dt.date])["datetime"].transform('min')
  2. df['timedelta'] = np.round((df['datetime'] - df['datetime_baseline']).dt.seconds / 60)
  4. print(df)
英文:

Try:

  1. df['datetime_baseline'] = df.groupby(['id', df['datetime'].dt.date])["datetime"].transform('min')
  2. df['timedelta'] = np.round((df['datetime'] - df['datetime_baseline']).dt.seconds / 60)
  4. print(df)

Prints:

  1.     id                datetime       datetime_baseline  timedelta
  2. 0   a1 2016-01-01 00:01:00.156 2016-01-01 00:01:00.156        0.0
  3. 1   a1 2016-01-01 12:00:00.425 2016-01-01 00:01:00.156      719.0
  4. 2   a1 2016-01-02 00:59:00.123 2016-01-02 00:59:00.123        0.0
  5. 3   a1 2016-01-02 14:16:00.548 2016-01-02 00:59:00.123      797.0
  6. 4   a2 2016-01-01 12:00:00.147 2016-01-01 12:00:00.147        0.0
  7. 5   a2 2016-01-01 13:59:00.123 2016-01-01 12:00:00.147      119.0
  8. 6   a2 2016-01-02 08:01:00.147 2016-01-02 08:01:00.147        0.0
  9. 7   a2 2016-01-02 18:49:00.123 2016-01-02 08:01:00.147      648.0
  10. 8   a3 2016-02-01 12:00:00.147 2016-02-01 12:00:00.147        0.0
  11. 9   a3 2016-02-01 13:59:00.123 2016-02-01 12:00:00.147      119.0
  12. 10  a3 2016-02-02 08:01:00.147 2016-02-02 08:01:00.147        0.0
  13. 11  a3 2016-02-02 18:49:00.123 2016-02-02 08:01:00.147      648.0

huangapple
  • 本文由 发表于 2023年4月11日 01:14:41
  • 转载请务必保留本文链接：https://go.coder-hub.com/75979159.html
  • date
  • datetime
  • pandas
  • python
  • timedelta
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