减去熊猫中的日期时间

Question

我的数据框是这个

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

as Loan Date Negotiation           datetime64[ns]

所以我想使函数减去2天。如果该月的最后一天是星期日，我将从中减去2天。

根据上面的数据框，2019年3月31日是星期日

我已经尝试过，但是失败了，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

Answer 1

如果需要先将datetime64转换为Timestamp，则可以使用：

ports:
- 9050:7050

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

timedelta()

或者在整个列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' 上显示：

Date

编辑：完整示例：

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

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]') 是M8[ns]的特定类型，因此在进一步处理中应该没有区别。

如果只想在星期天减去两天，则可以使用datetime64[ns]：

np.where()

索引0和2的日期为星期日，并减去2天。未触及索引1的日期。

在每个单独的时间戳记上通过列表理解的替代方式

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

Answer 2

您接近了！您只需要更改传递给Timedelta()的内容即可。

这是一个例子：

import pandas as pd

ts = pd.Timestamp(2017, 1, 1, 12)
days = pd.Timedelta(days=2)
print(days)
print(ts)
print(ts - days)

输出：

2 days 00:00:00
2017-01-01 12:00:00
2016-12-30 12:00:00