Python date string to date object

The Question :

370 people think this question is useful

How do I convert a string to a date object in python?

The string would be: "24052010" (corresponding to the format: "%d%m%Y")

I don’t want a datetime.datetime object, but rather a datetime.date.

The Question Comments :

The Answer 1

627 people think this answer is useful

You can use strptime in the datetime package of Python:

>>> import datetime
>>> datetime.datetime.strptime('24052010', "%d%m%Y").date()
datetime.date(2010, 5, 24)

The Answer 2

88 people think this answer is useful
import datetime
datetime.datetime.strptime('24052010', '%d%m%Y').date()

The Answer 3

63 people think this answer is useful

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))

The Answer 4

29 people think this answer is useful

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

The Answer 5

6 people think this answer is useful

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")

The Answer 6

3 people think this answer is useful

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

The Answer 7

2 people think this answer is useful

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)

The Answer 8

1 people think this answer is useful

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

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