In pandas how can we make the datetime column from this data?
df = pd.DataFrame({'date': ['2020-02-04T22:03:44.846000+00:00']})
print(df)
date
0 2020-02-04T22:03:44.846000+00:00
I am not sure what is the letter "T" here.
pat = '%y-%m-%dT%H:%M%:%SZ'
df['date'] = pd.to_datetime(df['date'],format=pat)
I am not sure what is the correct format here.
Thanks to @Felipe,
I got the answer.
df['date'] = pd.to_datetime(df['date'],infer_datetime_format=True)
df = pd.DataFrame({'date': ['2020-02-04T22:03:44.846000+00:00']})
df['year'] = df['date'].dt.year
print(df)
date year
0 2020-02-04 22:03:44.846000+00:00 2020
You can parse date time string format using this reference: https://docs.python.org/3/library/datetime.html#strftime-and-strptime-behavior
import numpy as np
import pandas as pd
pd.options.display.max_columns = 10
pd.set_option('display.max_colwidth', -1)
df = pd.DataFrame({'date': ['2020-02-04T22:03:44.846000+00:00']})
df['date1'] = pd.to_datetime(df['date'],format='%Y-%m-%dT%H:%M:%S.%f%z')
df['date2'] = pd.to_datetime(df['date'],infer_datetime_format=True)
df['hour'] = df['date1'].dt.hour
print(df)
0 2020-02-04T22:03:44.846000+00:00 2020-02-04 22:03:44.846000+00:00
date2 hour
0 2020-02-04 22:03:44.846000+00:00 22
ISO 8601