There is another library called arrow
really great to make manipulation on python date.
import arrow
import datetime
a = arrow.get('24052010', 'DMYYYY').date()
print(isinstance(a, datetime.date)) # True
If you are lazy and don't want to fight with string literals, you can just go with the parser
module.
from dateutil import parser
dt = parser.parse("Jun 1 2005 1:33PM")
print(dt.year, dt.month, dt.day,dt.hour, dt.minute, dt.second)
>2005 6 1 13 33 0
Just a side note, as we are trying to match any
string representation, it is 10x slower than strptime
Directly related question:
What if you have
datetime.datetime.strptime("2015-02-24T13:00:00-08:00", "%Y-%B-%dT%H:%M:%S-%H:%M").date()
and you get:
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "/usr/local/lib/python2.7/_strptime.py", line 308, in _strptime
format_regex = _TimeRE_cache.compile(format)
File "/usr/local/lib/python2.7/_strptime.py", line 265, in compile
return re_compile(self.pattern(format), IGNORECASE)
File "/usr/local/lib/python2.7/re.py", line 194, in compile
return _compile(pattern, flags)
File "/usr/local/lib/python2.7/re.py", line 251, in _compile
raise error, v # invalid expression
sre_constants.error: redefinition of group name 'H' as group 7; was group 4
and you tried:
<-24T13:00:00-08:00", "%Y-%B-%dT%HH:%MM:%SS-%HH:%MM").date()
but you still get the traceback above.
Answer:
>>> from dateutil.parser import parse
>>> from datetime import datetime
>>> parse("2015-02-24T13:00:00-08:00")
datetime.datetime(2015, 2, 24, 13, 0, tzinfo=tzoffset(None, -28800))
you have a date string like this, "24052010" and you want date object for this,
from datetime import datetime
cus_date = datetime.strptime("24052010", "%d%m%Y").date()
this cus_date will give you date object.
you can retrieve date string from your date object using this,
cus_date.strftime("%d%m%Y")
For single value the datetime.strptime
method is the fastest
import arrow
from datetime import datetime
import pandas as pd
l = ['24052010']
%timeit _ = list(map(lambda x: datetime.strptime(x, '%d%m%Y').date(), l))
6.86 µs ± 56.5 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
%timeit _ = list(map(lambda x: x.date(), pd.to_datetime(l, format='%d%m%Y')))
305 µs ± 6.32 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
%timeit _ = list(map(lambda x: arrow.get(x, 'DMYYYY').date(), l))
46 µs ± 978 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)
For a list of values the pandas pd.to_datetime
is the fastest
l = ['24052010'] * 1000
%timeit _ = list(map(lambda x: datetime.strptime(x, '%d%m%Y').date(), l))
6.32 ms ± 89.6 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
%timeit _ = list(map(lambda x: x.date(), pd.to_datetime(l, format='%d%m%Y')))
1.76 ms ± 27.3 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
%timeit _ = list(map(lambda x: arrow.get(x, 'DMYYYY').date(), l))
44.5 ms ± 522 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
For ISO8601 datetime format the ciso8601
is a rocket
import ciso8601
l = ['2010-05-24'] * 1000
%timeit _ = list(map(lambda x: ciso8601.parse_datetime(x).date(), l))
241 µs ± 3.24 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
import datetime
datetime.datetime.strptime('24052010', '%d%m%Y').date()
Use time module to convert data.
Code snippet:
import time
tring='20150103040500'
var = int(time.mktime(time.strptime(tring, '%Y%m%d%H%M%S')))
print var
Source: Stackoverflow.com