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.

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)

import datetime
datetime.datetime.strptime('24052010', '%d%m%Y').date()

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))

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

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)

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

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

string “24052010”
In a very manual way you could just go like this:-
first split the string as (yyyy-mm-dd) format so you could get a tuple something like this (2010, 5, 24), then simply convert this tuple to a date format something like 2010-05-24.

you could run this code on a list of string object similar to above and convert the entire list of tuples object to date object by simply unpacking(*tuple) check the code below.

import datetime
#for single string simply use:-

my_str= "24052010"
date_tup = (int(my_str[4:]),int(my_str[2:4]),int(my_str[:2]))
print(datetime.datetime(*date_tup))

output: 2012-01-01 00:00:00 

# for a list of string date objects you could use below code.
date_list = []
str_date = ["24052010", "25082011", "25122011","01012012"]

for items in str_date:
    date_list.append((int(items[4:]),int(items[2:4]),int(items[:2])))

for dates in date_list:
    # unpack all tuple objects and convert to date
    print(datetime.datetime(*dates))

output:
2010-05-24 00:00:00
2011-08-25 00:00:00
2011-12-25 00:00:00
2012-01-01 00:00:00