Using pd.apply() to turn each element in a column to a list, grab the 1st element, and turn it into datetime

Question

I have a dataframe with only one row (in the future, it will be more, but I am just using this as an example). I am trying to grab the "SETTLEMENT_DATE" column.

print(previous["SETTLEMENT_DATE"])

0 2021-06-22 00:00:00.0

Name: SETTLEMENT_DATE, dtype: object .

In order to get the date, I did list(previous["SETTLEMENT_DATE"])[0] , which yields: '2021-06-22 00:00:00.0' . Now I turn this into a date format using:

def create_datetime_object(pd_object):

date_time_str = list(pd_object)[0]
return datetime.strptime(date_time_str, "%Y-%m-%d %H:%M:%S.%f")

This code: create_datetime_object(previous["SETTLEMENT_DATE"])

yields: datetime.datetime(2021, 6, 22, 0, 0)

In the future, I will have multiple rows of data, so I wanted to use pd.apply() to apply this function to the entire column. But when I do that, I get:

previous["SETTLEMENT_DATE"] = previous["SETTLEMENT_DATE"].apply(create_datetime_object)

ValueError: time data '2' does not match format '%Y-%m-%d %H:%M:%S.%f

Does anyone know why I am getting this error and how to solve?

Answer 1

这应该工作：

previous["SETTLEMENT_DATE"] = previous.apply(lambda r : create_datetime_object(r.SETTLEMENT_DATE) , axis=1)