Date Time split in python

Question

I have to split a date time which I get from a software in the below format to separate variables (year,month,day,hour, min,sec)

19 Nov 2015  18:45:00.000


        

Answer 1

First pip install python-dateutil then do:

>>> from dateutil.parser import parse
>>> dt = parse('19 Nov 2015  18:45:00.000')
>>> dt.year
2015
>>> dt.month
11
>>> dt.day
19
>>> dt.hour
18
>>> dt.minute
45
>>> dt.second
0

Answer 2

Below solution should work for you:

import datetime

string = "19 Nov 2015  18:45:00.000"
date = datetime.datetime.strptime(string, "%d %b %Y  %H:%M:%S.%f")

print date

Output would be:

2015-11-19 18:45:00

And you can access the desired values with:

>>> date.year
2015
>>> date.month
11
>>> date.day
19
>>> date.hour
18
>>> date.minute
45
>>> date.second
0

You can check datetime's package documentation under section 8.1.7 for srtptime function's usage.

Answer 3

df2['invoice_date']=pd.to_datetime(df2.invoice_date, format='%m/%d/%Y %H:%M')


df2.insert(loc=5, column='month', value=df2.invoice_date.dt.month)
# +1 to make Monday=1.....until Sunday=7
    df2.insert(loc=6, column='day', value=(df2.invoice_date.dt.dayofweek)+1)
    df2.insert(loc=7, column='hour', value=df2.invoice_date.dt.hour)

Answer 4

As an alternative to wim's answer, if you don't want to install a package, you can do it like so:

import datetime

s = "19 Nov 2015  18:45:00.000"
d = datetime.datetime.strptime(s, "%d %b %Y  %H:%M:%S.%f")

print d.year
print d.month
print d.day
print d.hour
print d.minute
print d.second

This outputs:

2015
11
19
18
45
0

---

This utilizes strptime to parse the string.

Answer 5

Use datetime.strptime(your_string, format)

To contsruct the format string, consult the documentation: https://docs.python.org/2/library/datetime.html#strftime-and-strptime-behavior

From there, you can easily get the year, month, etc. (also in the documentation)