Question

我已经编写了一个代码来修改日期，以便在2/29/2008减少1天。代码成功执行并向我显示输出但不更新数据帧。

TO

我写了下面的代码，它给了我输出但它没有更新数据帧

Name    Maint Start Date
Yasin   02/29/08
Susy    01/04/15
James   04/14/06

然后我将其修改为，但所有没有闰年数据的字段都变为空白。

spend_sw[spend_sw['Maint Start Date'] == '02/29/2008'].apply(lambda x: pd.datetime(x['Maint Start Date'].year , x['Maint Start Date'].month, x['Maint Start Date'].day - 1), axis=1)

您能告诉我如何修改现有数据帧的值。

Answer 1

我认为您可以先转换Maint Start Date to_datetime列中的所有值，然后使用ix进行过滤，并在to_timedelta创建的一天减去Timedelta：< / p>

print (spend_sw)
    Name Maint Start Date
0  Yasin         02/29/08
1   Susy         01/04/15
2  James         04/14/06

spend_sw['Maint Start Date'] = pd.to_datetime(spend_sw['Maint Start Date'])

print (spend_sw.ix[spend_sw['Maint Start Date'] == '02/29/08', 'Maint Start Date'])
0   2008-02-29
Name: Maint Start Date, dtype: datetime64[ns]

spend_sw.ix[spend_sw['Maint Start Date'] == '02/29/08', 'Maint Start Date'] = \
spend_sw.ix[spend_sw['Maint Start Date'] == '02/29/08', 'Maint Start Date'] - 
pd.to_timedelta(1, unit='d') 
print (spend_sw)
    Name Maint Start Date
0  Yasin       2008-02-28
1   Susy       2015-01-04
2  James       2006-04-14

mask和offset的另一种解决方案：

spend_sw['Maint Start Date'] = pd.to_datetime(spend_sw['Maint Start Date'])

date = pd.to_datetime('02/29/08')
date1 = date - pd.offsets.Day(1)
#mask by condition
ma = spend_sw['Maint Start Date'] == date
spend_sw['Maint Start Date'] = spend_sw['Maint Start Date'].mask(ma, date1)
print (spend_sw)
    Name Maint Start Date
0  Yasin       2008-02-28
1   Susy       2015-01-04
2  James       2006-04-14

使用修改后的值更新python中的数据帧

1 个答案: