有什么直接方法可以在python中逐列检查日期格式(使用datetime)?

问题描述

我在df中有一个日期列.现在，我想通过逐列检查它是否与特定格式匹配.我是逐行完成的，但是要花很多时间才能运行.我想知道是否有一种方法可以立即检查列，该列可能包含空值或日期的其他格式.尽管我尝试按列进行操作(如果有一条不匹配的记录)，但会进入例外.那么，我们该怎么做呢?

I have a date column in a df. Now I want to check if it is matching with a particular format or not by doing column by column. I did it row by row but it takes a lot of time to run. I want to know if there is a way for checking the column at once the column might contain nulls or a different format of a date. Although I tried doing it column-wise if there is one record that is not matching it goes into except. So, how can we do that?

代码:

df=
    Date
0  12-22-2020
1  22-12-2020
3  22122020
4
5  02-22-2020
formatt='%m-%d-%Y'
try:
    datetime = dt.strptime(str(df['Date']), formatt) 
    print(datetime )                                
except ValueError:
    print('error')

这给出:'错误'

总体上是给予.但我希望显示接受的记录.我们该怎么做?

It's giving as a whole. But I want the accepted records to be displayed. How do we do it?

预期输出:

datetime =
 Date
0  12-22-2020
4
5  02-22-2020

推荐答案

您应该将熊猫 to_datetime 函数与 errors ='coerce'一起使用，并保持正确的行转换和行在初始列中为空.

You should use pandas to_datetime function with errors='coerce' and keep lines correctly converted and lines that were null in the initial column.

代码可能是

dates = pd.to_datetime(df['Date'], errors='coerce', format='%m-%d-%Y')
dates = dates[(~dates.isna())|df['Date'].isnull()]

它给出:

0   2020-12-22
4          NaT
5   2020-02-22

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