在Pandas中将DateTime列添加递增的秒数

huangapple
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英文:

Add incrementing seconds to DateTime column Pandas

问题

以下是翻译好的内容：

  1. 我有以下类型的`df`：
  3.         Date_time         Col1
  4.     0 2023-03-04 10:30:00  10
  5.     1 2023-03-04 10:30:00  11
  6.     2 2023-03-04 10:30:00  21
  7.     3 2023-03-04 10:30:00  54
  8.     4 2023-03-04 10:30:00  12 
  9.     5 2023-03-04 10:30:00  13
  10.     6 2023-03-04 10:30:00  21
  11.     ...
  12.     58 2023-03-04 10:30:00  22
  13.     59 2023-03-04 10:30:00  21
  14.     60 2023-03-04 10:31:00  25
  15.     61 2023-03-04 10:31:00  21
  16.     ...
  18. 由于某种原因，`df`中的秒数始终为`0`，但实际上它们必须递增`1`秒。
  20. 我想要将`Date_time`列添加`1`秒的递增，并保持递增直到`59`秒，然后在分钟变化时重置它。请参见下面的期望结果：
  22.         Date_time         Col1
  23.     0 2023-03-04 10:30:00  10
  24.     1 2023-03-04 10:30:01  11
  25.     2 2023-03-04 10:30:02  21
  26.     3 2023-03-04 10:30:03  54
  27.     4 2023-03-04 10:30:04  12 
  28.     5 2023-03-04 10:30:05  13
  29.     6 2023-03-04 10:30:06  21
  30.     ....
  31.     58 2023-03-04 10:30:58  22
  32.     59 2023-03-04 10:30:59  21
  33.     60 2023-03-04 10:31:00  25
  34.     61 2023-03-04 10:31:01  21
  35.     ...

希望这能满足你的需求。

英文:

I have the following type of df:

  1.     Date_time         Col1
  2. 0 2023-03-04 10:30:00  10
  3. 1 2023-03-04 10:30:00  11
  4. 2 2023-03-04 10:30:00  21
  5. 3 2023-03-04 10:30:00  54
  6. 4 2023-03-04 10:30:00  12 
  7. 5 2023-03-04 10:30:00  13
  8. 6 2023-03-04 10:30:00  21
  9. ...
  10. 58 2023-03-04 10:30:00  22
  11. 59 2023-03-04 10:30:00  21
  12. 60 2023-03-04 10:31:00  25
  13. 61 2023-03-04 10:31:00  21
  14. ...

For some reason, the seconds show 0 throughout the df, however in fact they must be incremented by 1 second.

I would like to add 1 second increment to Date_time column and keep incrementing up to 59 seconds, then reset it as the minute changes. Please see below the desired outcome.

  1.     Date_time         Col1
  2. 0 2023-03-04 10:30:00  10
  3. 1 2023-03-04 10:30:01  11
  4. 2 2023-03-04 10:30:02  21
  5. 3 2023-03-04 10:30:03  54
  6. 4 2023-03-04 10:30:04  12 
  7. 5 2023-03-04 10:30:05  13
  8. 6 2023-03-04 10:30:06  21
  9. ....
  10. 58 2023-03-04 10:30:58  22
  11. 59 2023-03-04 10:30:59  21
  12. 60 2023-03-04 10:31:00  25
  13. 61 2023-03-04 10:31:01  21
  14. ...

答案1

得分: 1

使用 groupby.cumcount 和 TimedeltaIndex：

  1. df['Date_time'] = pd.to_datetime(df['Date_time'])
  3. df['Date_time'] += pd.TimedeltaIndex(df.groupby('Date_time').cumcount(), unit='s')

输出结果：

  1.              Date_time  Col1
  2. 0  2023-03-04 10:30:00    10
  3. 1  2023-03-04 10:30:01    11
  4. 2  2023-03-04 10:30:02    21
  5. 3  2023-03-04 10:30:03    54
  6. 4  2023-03-04 10:30:04    12
  7. 5  2023-03-04 10:30:05    13
  8. 6  2023-03-04 10:30:06    21
  9. ...
  10. 58 2023-03-04 10:30:58    22
  11. 59 2023-03-04 10:30:59    21
  12. 60 2023-03-04 10:31:00    25
  13. 61 2023-03-04 10:31:01    21
英文:

Use a groupby.cumcount and TimedeltaIndex:

  1. df['Date_time'] = pd.to_datetime(df['Date_time'])
  3. df['Date_time'] += pd.TimedeltaIndex(df.groupby('Date_time').cumcount(), unit='s')

Output:

  1.              Date_time  Col1
  2. 0  2023-03-04 10:30:00    10
  3. 1  2023-03-04 10:30:01    11
  4. 2  2023-03-04 10:30:02    21
  5. 3  2023-03-04 10:30:03    54
  6. 4  2023-03-04 10:30:04    12
  7. 5  2023-03-04 10:30:05    13
  8. 6  2023-03-04 10:30:06    21
  9. ...
  10. 58 2023-03-04 10:30:58    22
  11. 59 2023-03-04 10:30:59    21
  12. 60 2023-03-04 10:31:00    25
  13. 61 2023-03-04 10:31:01    21

huangapple
  • 本文由 发表于 2023年4月4日 17:23:17
  • 转载请务必保留本文链接：https://go.coder-hub.com/75927670.html
  • datetime
  • pandas
  • python
  • python-datetime
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