python - 减去熊猫中的日期时间

问题描述

我有数据框是这个

In[1]: df1
Out[1]
   Loan Date Negotiation
   2019-03-31
   2019-03-31
   2019-03-31

as Loan Date Negotiation           datetime64[ns]

所以我想让函数从中减去 2 天。如果一个月的最后一天是星期天，我会从中减去 2 天。

从上面的数据框中，2019-03-31 是星期日，

我试过但失败了，这是

 def subtractingDate(dateTime):
     dateTimestamp = pd.Timestamp(dateTime)
     newDate = dateTimestamp - pd.Timedelta("2 days")
     return newDate

 dfMARET.loc[dfMARET["Loan Date Negotiation"].dt.dayofweek == 6, "New Date"] = subtractingDate(dfMARET["Loan Date Negotiation"])

*note: 6 is for sunday

So the error is


TypeError                                 Traceback (most recent call last)
<ipython-input-9-cc2a3348e6ce> in <module>
     16 # a = subtractingDate(dfMARET["Loan Date Negotiation"])
     17 # a
---> 18 dfMARET.loc[dfMARET["Loan Date Negotiation"].dt.dayofweek == 6, "New Date"] = subtractingDate(dfMARET["Loan Date Negotiation"])
     19 dfMARET
     20 

<ipython-input-9-cc2a3348e6ce> in subtractingDate(dateTime)
     10 
     11 def subtractingDate(dateTime):
---> 12     dateTimestamp = pd.Timestamp(dateTime)
     13     newDate = dateTimestamp - pd.Timedelta(days = 2)
     14     return newDate

pandas\_libs\tslibs\timestamps.pyx in pandas._libs.tslibs.timestamps.Timestamp.__new__()

pandas\_libs\tslibs\conversion.pyx in pandas._libs.tslibs.conversion.convert_to_tsobject()```

所以我的期望是

 Loan Date Negotiation
 2019-03-29
 2019-03-29
 2019-03-29

大熊猫的解决方案？

谢谢

标签： pythonpandas

如果需要先将 datetime64 转换为 Timestamp，可以使用：

df['Date'] = [pd.Timestamp(x) for x in df['Date']]

并使用timedelta()：

from datetime import datetime, timedelta
dt = pd.Timestamp(2019,3,31)
new_dt = dt-timedelta(days=2)

new_dt
> datetime.datetime(2019, 3, 29, 0, 0)

new_dt.strftime('%Y-%m-%d')
> '2019-03-29'

或在整个列Date上：

df['New_Date'] = df['Date']-timedelta(days=2)

编辑：完整示例：

import numpy as np
from datetime import datetime, timedelta

df1 = pd.DataFrame([[np.datetime64(datetime(2019, 3, 31))],
                    [np.datetime64(datetime(2019, 3, 27))],
                    [np.datetime64(datetime(2019, 3, 24))]], 
                   columns=['Loan Date Negotiation'])
df1

df1['Loan Date Negotiation'].dtype
> dtype('<M8[ns]')

M8[ns]是的特定类型datetime64[ns]，因此进一步处理应该没有区别。

如果您只想在这一天是星期天的情况下减去两天，您可以使用np.where()：

df1['New_Date'] = np.where(df1['Loan Date Negotiation'].dt.dayofweek==6, 
         df1['Loan Date Negotiation']-timedelta(days=2), 
         df1['Loan Date Negotiation'])

索引 0 和 2 的日期是星期天，减去了 2 天。未触及索引 1 处的日期。

通过对每个时间戳的列表理解进行替代

df1['New_Date'] = [x-timedelta(days=2) if x.weekday()==6 else x for x in df1['Loan Date Negotiation']]

python - 减去熊猫中的日期时间

问题描述

解决方案

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