I have dataframe which is using 2 parameters:
data_frame = pd.DataFrame(columns=l_header, data=conv)
With below content:
column_1 date
0 XYZ 04/11/2016 09:13:55 AM
1 QAZ 05/10/2019 11:31:05 PM
2 WSX 12/08/2019 08:00:00 AM
How can I change date column format (it's a string) from:
%m/%d/%Y %I:%M:%S %p
to:
%Y-%m-%d %H:%M:%S
You can use the strptime and strftime methods from the built-in datetime module to use in the lambda function that will go into the brackets of the apply method:
from datetime import datetime
import pandas as pd
df = {"column_1": ["XYZ", "QAZ", "WSX"],
"date": ["04/11/2016 09:13:55 AM", "05/10/2019 11:31:05 PM", "12/08/2019 08:00:00 AM"]}
df = pd.DataFrame(df)
format1 ="%m/%d/%Y %I:%M:%S %p"
format2 = "%Y-%m-%d %H:%M:%S"
df["date"] = df["date"].apply(lambda x: datetime.strptime(x, format1).strftime(format2))
print(df)
Output:
column_1 date
0 XYZ 2016-04-11 09:13:55
1 QAZ 2019-05-10 23:31:05
2 WSX 2019-12-08 08:00:00
UPDATE:
As pointed out by @YevKad in the comments, there is no need to import datetime, as pandas comes with its own datetime section:
import pandas as pd
df = {"column_1": ["XYZ", "QAZ", "WSX"],
"date": ["04/11/2016 09:13:55 AM", "05/10/2019 11:31:05 PM", "12/08/2019 08:00:00 AM"]}
df = pd.DataFrame(df)
format1 ="%m/%d/%Y %I:%M:%S %p"
format2 = "%Y-%m-%d %H:%M:%S"
df['date'] = pd.to_datetime(df['date'], format=format1).dt.strftime(format2)
print(df)
I think this should be the most convenient way?
>>> df
column_1 date
0 XYZ 04/11/2016 09:13:55 AM
1 QAZ 05/10/2019 11:31:05 PM
2 WSX 12/08/2019 08:00:00 AM
>>> df['date']=pd.to_datetime(df['date'], infer_datetime_format='%Y-%m-%d %H:%M:%S')
column_1 date
0 XYZ 2016-04-11 09:13:55
1 QAZ 2019-05-10 23:31:05
2 WSX 2019-12-08 08:00:00
But it's best if you do this during the loading of the .csv file
>>> data_frame = pd.DataFrame(columns=l_header,
parse_dates=['date'],
infer_datetime_format='%Y-%m-%d %H:%M:%S')
