在具备条件的情况下，将日期时间增加一秒。

huangapple
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英文:

Add one second to datetime with condition

问题

看到下面的示例熊猫数据。

  1. ID	DT	                InOut
  2. 120	2022-12-22-02:12:123	IN
  3. 120	2022-12-23-04:12:456	OUT
  4. 120	2022-12-26-08:11:125	IN
  5. 120	2022-12-30-02:12:126	OUT

首先，我需要将我的日期时间变量更改为仅保留第二部分的两位数（例如：2023-03-03 14:11:43）。然后，我需要仅对InOut等于OUT的日期时间变量添加一秒。

我的dt变量处于datetime64[ns]格式。我无法为日期时间变量创建虚拟数据。

感谢任何帮助！

  1. import pandas as pd
  2. df = pd.DataFrame({'ID': [120, 120, 120, 120], 
  3.                    'InOut': ['IN', 'OUT', 'IN', 'OUT']})
英文:

See the sample panda data below.

  1. ID	DT	                InOut
  2. 120	2022-12-22-02:12:123	IN
  3. 120	2022-12-23-04:12:456	OUT
  4. 120	2022-12-26-08:11:125	IN
  5. 120	2022-12-30-02:12:126	OUT

First I need to change my datetime variable to only 2 digits for the second part (ex: 2023-03-03 14:11:43).Then I need to add one second to my datetime variable only for InOut = OUT.

My dt variable is the in datetime64[ns] format.
I wasn't able to create dummy data for the datetime variable.

Thanks for any help!

  1. import pandas as pd
  2. df = pd.DataFrame({'ID': [120, 120, 120, 120], 
  3.                    'InOut': ['IN', 'OUT', 'IN', 'OUT']})

答案1

得分: 0

你可以使用strftime()函数将你的日期时间变量转换为所需的格式。要将日期时间变量转换为所需的格式，可以执行以下操作：

  1. df['DT'] = df['DT'].apply(lambda x: datetime.strftime(x, '%Y-%m-%d %H:%M:%S'))

如果你想在InOut=Out的情况下将日期时间增加一秒：

  1. df.loc[df['InOut'] == 'OUT', 'DT'] = df.loc[df['InOut'] == 'OUT', 'DT'].apply(lambda x: datetime.strptime(x, '%Y-%m-d %H:%M:%S') + timedelta(seconds=1))
英文:

You can convert your datetime variable to the desired format using the strftime(). To convert datetime variable to desired format you can do:

  1. df['DT'] = df['DT'].apply(lambda x: datetime.strftime(x, '%Y-%m-%d %H:%M:%S'))

and if you want to add a second to datetimes with InOut=Out:

  1. df.loc[df['InOut'] == 'OUT', 'DT'] = df.loc[df['InOut'] == 'OUT', 'DT'].apply(lambda x: datetime.strptime(x, '%Y-%m-%d %H:%M:%S') + timedelta(seconds=1))

答案2

得分: 0

将DT列转换为日期时间格式，并将秒数截断为两位小数，然后使用.loc根据条件添加一秒。

  1. df['DT'] = (pd.to_datetime(df['DT'], format='%Y-%m-%d-%H:%M:%S%f').dt.floor('s')
  2.               .apply(lambda x: x.replace(microsecond=(x.microsecond // 1000) * 1000)))
  3. df.loc[df['InOut'] == 'OUT', 'DT'] += pd.Timedelta(seconds=1)
  1.     ID                  DT InOut
  2. 0  120 2022-12-22 02:12:12    IN
  3. 1  120 2022-12-23 04:12:46   OUT
  4. 2  120 2022-12-26 08:11:12    IN
  5. 3  120 2022-12-30 02:12:13   OUT
英文:

Convert DT column to datetime format and truncate seconds to two(2) digits, after that use .loc to add a second if a condition is met

  1. df['DT'] = (pd.to_datetime(df['DT'], format='%Y-%m-%d-%H:%M:%S%f').dt.floor('s')
  2.               .apply(lambda x: x.replace(microsecond=(x.microsecond // 1000) * 1000)))
  3. df.loc[df['InOut'] == 'OUT', 'DT'] += pd.Timedelta(seconds=1)
  1.     ID                  DT InOut
  2. 0  120 2022-12-22 02:12:12    IN
  3. 1  120 2022-12-23 04:12:46   OUT
  4. 2  120 2022-12-26 08:11:12    IN
  5. 3  120 2022-12-30 02:12:13   OUT

答案3

得分: 0

请将您的 DT 列调整为有效的日期时间格式，使用 pd.to_datetime，然后再加上 1 秒，使用 pd.Timedelta(seconds=1)：

  1. df['DT'] = pd.to_datetime(df['DT'], format='%Y-%m-%d %H:%M:%S', exact=False)
  2. df.loc[df['InOut'].eq('OUT'), 'DT'] += pd.Timedelta(seconds=1)
  1.         ID                  DT InOut
  2.     0  120 2022-12-22 02:12:12    IN
  3.     1  120 2022-12-23 04:12:46   OUT
  4.     2  120 2022-12-26 08:11:12    IN
  5.     3  120 2022-12-30 02:12:13   OUT

这是调整后的数据表。

英文:

Adjust your DT column to a valid datetime format with pd.to_datetime, then - add 1 second with pd.Timedelta(seconds=1):

  1. df['DT'] = pd.to_datetime(df['DT'], format = '%Y-%m-%d-%H:%M:%S', exact=False)
  2. df.loc[df['InOut'].eq('OUT'), 'DT'] += pd.Timedelta(seconds=1)
  1.     ID                  DT InOut
  2. 0  120 2022-12-22 02:12:12    IN
  3. 1  120 2022-12-23 04:12:46   OUT
  4. 2  120 2022-12-26 08:11:12    IN
  5. 3  120 2022-12-30 02:12:13   OUT

答案4

得分: 0

  1. import pandas as pd
  2. import datetime
  3. import numpy as np
  5. df = pd.DataFrame({'ID':[120, 120, 120, 120],
  6.                    'DT':['2022-12-22-02:12:123', 
  7.                          '2022-12-23-04:12:456', 
  8.                          '2022-12-26-08:11:125', 
  9.                          '2022-12-30-02:12:126'],
  10.                    'InOut':['IN', 'OUT', 'IN', 'OUT']
  11.                    })
  13. df.loc[:, ('DT')] = pd.to_datetime(df['DT'].add('0'*5), 
  14.                                    format='%Y-%m-%d-%H:%M:%S%f') # (1)
  16. r = df[df['InOut'].eq('OUT')].copy()
  17. r['DT'] = r['DT'].add(datetime.timedelta(seconds=1))
  19. df.loc[:, ('DT')] = df['DT'].mask(df['InOut'].eq('OUT'), r['DT'])
  21. print(df)
  1.     ID                      DT InOut
  2. 0  120 2022-12-22 02:12:12.300    IN
  3. 1  120 2022-12-23 04:12:46.600   OUT
  4. 2  120 2022-12-26 08:11:12.500    IN
  5. 3  120 2022-12-30 02:12:13.600   OUT

(1) 注意，我将原始的数据框转换为更常规的日期时间表示形式。

英文: 
  1. import pandas as pd
  2. import datetime
  3. import numpy as np
  5. df = pd.DataFrame({'ID':[120, 120, 120, 120],
  6.                    'DT':['2022-12-22-02:12:123', 
  7.                          '2022-12-23-04:12:456', 
  8.                          '2022-12-26-08:11:125', 
  9.                          '2022-12-30-02:12:126'],
  10.                    'InOut':['IN', 'OUT', 'IN', 'OUT']
  11.                    })
  13. df.loc[:, ('DT')] = pd.to_datetime(df['DT'].add('0'*5), 
  14.                                    format='%Y-%m-%d-%H:%M:%S%f') # (1)
  16. r = df[df['InOut'].eq('OUT')].copy()
  17. r['DT'] = r['DT'].add(datetime.timedelta(seconds=1))
  19. df.loc[:, ('DT')] = df['DT'].mask(df['InOut'].eq('OUT'), r['DT'])
  21. print(df)
  1.     ID                      DT InOut
  2. 0  120 2022-12-22 02:12:12.300    IN
  3. 1  120 2022-12-23 04:12:46.600   OUT
  4. 2  120 2022-12-26 08:11:12.500    IN
  5. 3  120 2022-12-30 02:12:13.600   OUT

(1) Note I turned original df into a more conventional datetime presentation

huangapple
  • 本文由 发表于 2023年3月4日 04:26:53
  • 转载请务必保留本文链接：https://go.coder-hub.com/75631594.html
  • datetime
  • pandas
  • python
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