I have a string column that looks like below:
2018-24 7:10:0
2018-8 12:1:20
2018-44 13:55:19
The 24,8,44 that you see are the day of the year and not the date.
How can I convert this to datetime column in the below format ?
2018-01-24 07:10:00
2018-01-08 12:01:20
2018-02-13 13:55:19
I am unable to find anything related to converting day of the year ?
You need format string '%Y-%j %H:%M:%S'
In[53]:
import datetime as dt
dt.datetime.strptime('2018-44 13:55:19', '%Y-%j %H:%M:%S')
Out[53]: datetime.datetime(2018, 2, 13, 13, 55, 19)
%j is day of year
For pandas:
In[59]:
import pandas as pd
import io
t="""2018-24 7:10:0
2018-8 12:1:20
2018-44 13:55:19"""
df = pd.read_csv(io.StringIO(t), header=None, names=['datetime'])
df
Out[59]:
datetime
0 2018-24 7:10:0
1 2018-8 12:1:20
2 2018-44 13:55:19
Use pd.to_datetime and pass format param:
In[60]:
df['new_datetime'] = pd.to_datetime(df['datetime'], format='%Y-%j %H:%M:%S')
df
Out[60]:
datetime new_datetime
0 2018-24 7:10:0 2018-01-24 07:10:00
1 2018-8 12:1:20 2018-01-08 12:01:20
2 2018-44 13:55:19 2018-02-13 13:55:19
You can use dateutil.relativedelta for sum the day from the first day of years.
example:
from datetime import datetime
from dateutil.relativedelta import relativedelta
datetime.now()+ relativedelta(days=5)
The documentation at strftime.org identifies the %j format specifier as handling day of the year. I don't know whether it's available on all platforms, but my Mac certainly has it.
Use time.strptime to convert from string to datetime. The output below has a newline inserted for reading convenience:
>>> time.strptime('2018-24 7:10:0', '%Y-%j %H:%M:%S')
time.struct_time(tm_year=2018, tm_mon=1, tm_mday=24, tm_hour=7,
tm_min=10, tm_sec=0, tm_wday=2, tm_yday=24, tm_isdst=-1)
The time.strftime formats datetimes, so you can get what you need by applying it to the output of strptime:
>>> time.strftime('%Y-%m-%d %H:%M:%S',
... time.strptime('2018-24 7:10:0', '%Y-%j %H:%M:%S'))
'2018-01-24 07:10:00'
