Python在datetime.now（）和填充日期的系列之间的年份差异？ [英] Python difference in years between a datetime.now() and a Series filled up with dates?

问题描述

我想在数据集中创建一个新列，这是今天和数据集中已经存在的另一列之间的年差。

上面的代码：

  df ['diff_years'] = datetime.today（）-df ['some_date'] 
 df ['diff_years'] 
  

请给我以下输出（例如）：

  1754天11：44：28.971615 
  

，而我必须得到类似的东西（意味着以上年份的输出）：

  4,8 
（或5）
  

感谢您的帮助！

PS：我想避免循环播放该系列，我相信这会给我一个理想的解决方案，但是由于有一个大系列，我想避免这种方式。

解决方案

几天前，我在项目中遇到了同样的问题，现在我已经尝试过这些，

<$从dateutil.relativedelta导入p $ p> 从日期时间导入日期开始的relativedelta

现在= date.today（）
some_date =日期（df ['some_date']）

rdelta = relativedelta（现在为some_date） ）
print（'diff in years-'，rdelta.years）
print（'remaining months-'，rdelta.months）
print（'remaining days-'，rdelta.days）


应该打印出年份差异

I would like to create a new column in my dataset, which is a difference in years between today and a another column already in the dataset, filled up with dates.

the code above:

df['diff_years'] = datetime.today() - df['some_date']
df['diff_years']

give me the following output (exemple):

1754 days 11:44:28.971615

and i have to get something like (meaning the output above in years):

4,8 
(or 5)

I appreciate any help!

PS.: i would like to avoid looping the series, path i believe would give me a desired solution, but due having a big series i would like to avoid this way.

解决方案

Before some days i was facing same issue in my project now i had tried with these ,

from dateutil.relativedelta import relativedelta
from datetime import date
now = date.today()
some_date = date(df['some_date'])

rdelta = relativedelta(now, some_date)
print('diff in years - ', rdelta.years)
print('remaining months - ', rdelta.months)
print('remaining days - ', rdelta.days)

It should print difference in years

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