I'm having a bad time with date parsing and formatting today.
Points for somebody who can parse this date format into a datetime.date or datetime.datetime (I'm not too fussy but I'd prefer .date):
5th November 2010
Using dateutil:
In [2]: import dateutil.parser as dparser
In [3]: date = dparser.parse('5th November 2010')
In [4]: date
Out[4]: datetime.datetime(2010, 11, 5, 0, 0)
Unfortunately, strptime has no format characters for "skip an ordinal suffix" -- so, I'd do the skipping first, with a little RE, and then parse the resulting "clear" string. I.e.:
>>> import re
>>> import datetime
>>> ordn = re.compile(r'(?<=\d)(st|nd|rd|th)\b')
>>> def parse(s):
... cleans = ordn.sub('', s)
... dt = datetime.datetime.strptime(cleans, '%d %B %Y')
... return dt.date()
...
>>> parse('5th November 2010')
datetime.date(2010, 11, 5)
Your preference for date vs datetime is no problem of course, that's what the .date() method of datetime objects is for;-).
Third-party extensions like dateutil can be useful if you need to do a lot of "fuzzy" date parsing (or other fancy date-related stuff;-), by the way.
If the ordinal is constant then:
datetime.strptime(s, '%dth %B %Y')
Else:
date_str = '5th November 2010'
modified_date_str = date_str[0:1] + date_str[3:]
datetime.strptime(modified_date_str, '%d %B %Y')
Or like ~unutbu said use dateutil :)
Related
I have a date string with the format 'Mon Feb 15 2010'. I want to change the format to '15/02/2010'. How can I do this?
datetime module could help you with that:
datetime.datetime.strptime(date_string, format1).strftime(format2)
For the specific example you could do
>>> import datetime
>>> datetime.datetime.strptime('Mon Feb 15 2010', '%a %b %d %Y').strftime('%d/%m/%Y')
'15/02/2010'
>>>
You can install the dateutil library. Its parse function can figure out what format a string is in without having to specify the format like you do with datetime.strptime.
from dateutil.parser import parse
dt = parse('Mon Feb 15 2010')
print(dt)
# datetime.datetime(2010, 2, 15, 0, 0)
print(dt.strftime('%d/%m/%Y'))
# 15/02/2010
convert string to datetime object
from datetime import datetime
s = "2016-03-26T09:25:55.000Z"
f = "%Y-%m-%dT%H:%M:%S.%fZ"
out = datetime.strptime(s, f)
print(out)
output:
2016-03-26 09:25:55
>>> from_date="Mon Feb 15 2010"
>>> import time
>>> conv=time.strptime(from_date,"%a %b %d %Y")
>>> time.strftime("%d/%m/%Y",conv)
'15/02/2010'
As this question comes often, here is the simple explanation.
datetime or time module has two important functions.
strftime - creates a string representation of date or time from a datetime or time object.
strptime - creates a datetime or time object from a string.
In both cases, we need a formating string. It is the representation that tells how the date or time is formatted in your string.
Now lets assume we have a date object.
>>> from datetime import datetime
>>> d = datetime(2010, 2, 15)
>>> d
datetime.datetime(2010, 2, 15, 0, 0)
If we want to create a string from this date in the format 'Mon Feb 15 2010'
>>> s = d.strftime('%a %b %d %y')
>>> print s
Mon Feb 15 10
Lets assume we want to convert this s again to a datetime object.
>>> new_date = datetime.strptime(s, '%a %b %d %y')
>>> print new_date
2010-02-15 00:00:00
Refer This document all formatting directives regarding datetime.
#codeling and #user1767754 : The following two lines will work. I saw no one posted the complete solution for the example problem that was asked. Hopefully this is enough explanation.
import datetime
x = datetime.datetime.strptime("Mon Feb 15 2010", "%a %b %d %Y").strftime("%d/%m/%Y")
print(x)
Output:
15/02/2010
You may achieve this using pandas as well:
import pandas as pd
pd.to_datetime('Mon Feb 15 2010', format='%a %b %d %Y').strftime('%d/%m/%Y')
Output:
'15/02/2010'
You may apply pandas approach for different datatypes as:
import pandas as pd
import numpy as np
def reformat_date(date_string, old_format, new_format):
return pd.to_datetime(date_string, format=old_format, errors='ignore').strftime(new_format)
date_string = 'Mon Feb 15 2010'
date_list = ['Mon Feb 15 2010', 'Wed Feb 17 2010']
date_array = np.array(date_list)
date_series = pd.Series(date_list)
old_format = '%a %b %d %Y'
new_format = '%d/%m/%Y'
print(reformat_date(date_string, old_format, new_format))
print(reformat_date(date_list, old_format, new_format).values)
print(reformat_date(date_array, old_format, new_format).values)
print(date_series.apply(lambda x: reformat_date(x, old_format, new_format)).values)
Output:
15/02/2010
['15/02/2010' '17/02/2010']
['15/02/2010' '17/02/2010']
['15/02/2010' '17/02/2010']
Just for the sake of completion: when parsing a date using strptime() and the date contains the name of a day, month, etc, be aware that you have to account for the locale.
It's mentioned as a footnote in the docs as well.
As an example:
import locale
print(locale.getlocale())
>> ('nl_BE', 'ISO8859-1')
from datetime import datetime
datetime.strptime('6-Mar-2016', '%d-%b-%Y').strftime('%Y-%m-%d')
>> ValueError: time data '6-Mar-2016' does not match format '%d-%b-%Y'
locale.setlocale(locale.LC_ALL, 'en_US')
datetime.strptime('6-Mar-2016', '%d-%b-%Y').strftime('%Y-%m-%d')
>> '2016-03-06'
use datetime library
http://docs.python.org/library/datetime.html look up 9.1.7.
especiall strptime() strftime() Behavior¶
examples
http://pleac.sourceforge.net/pleac_python/datesandtimes.html
If you dont want to define the input date format then, Install dateparser (pip install dateparser) and,
from dateparser import parse
parse("Mon Feb 15 2010").strftime("%d/%m/%Y")
I have a date string with the format 'Mon Feb 15 2010'. I want to change the format to '15/02/2010'. How can I do this?
datetime module could help you with that:
datetime.datetime.strptime(date_string, format1).strftime(format2)
For the specific example you could do
>>> import datetime
>>> datetime.datetime.strptime('Mon Feb 15 2010', '%a %b %d %Y').strftime('%d/%m/%Y')
'15/02/2010'
>>>
You can install the dateutil library. Its parse function can figure out what format a string is in without having to specify the format like you do with datetime.strptime.
from dateutil.parser import parse
dt = parse('Mon Feb 15 2010')
print(dt)
# datetime.datetime(2010, 2, 15, 0, 0)
print(dt.strftime('%d/%m/%Y'))
# 15/02/2010
convert string to datetime object
from datetime import datetime
s = "2016-03-26T09:25:55.000Z"
f = "%Y-%m-%dT%H:%M:%S.%fZ"
out = datetime.strptime(s, f)
print(out)
output:
2016-03-26 09:25:55
>>> from_date="Mon Feb 15 2010"
>>> import time
>>> conv=time.strptime(from_date,"%a %b %d %Y")
>>> time.strftime("%d/%m/%Y",conv)
'15/02/2010'
As this question comes often, here is the simple explanation.
datetime or time module has two important functions.
strftime - creates a string representation of date or time from a datetime or time object.
strptime - creates a datetime or time object from a string.
In both cases, we need a formating string. It is the representation that tells how the date or time is formatted in your string.
Now lets assume we have a date object.
>>> from datetime import datetime
>>> d = datetime(2010, 2, 15)
>>> d
datetime.datetime(2010, 2, 15, 0, 0)
If we want to create a string from this date in the format 'Mon Feb 15 2010'
>>> s = d.strftime('%a %b %d %y')
>>> print s
Mon Feb 15 10
Lets assume we want to convert this s again to a datetime object.
>>> new_date = datetime.strptime(s, '%a %b %d %y')
>>> print new_date
2010-02-15 00:00:00
Refer This document all formatting directives regarding datetime.
#codeling and #user1767754 : The following two lines will work. I saw no one posted the complete solution for the example problem that was asked. Hopefully this is enough explanation.
import datetime
x = datetime.datetime.strptime("Mon Feb 15 2010", "%a %b %d %Y").strftime("%d/%m/%Y")
print(x)
Output:
15/02/2010
You may achieve this using pandas as well:
import pandas as pd
pd.to_datetime('Mon Feb 15 2010', format='%a %b %d %Y').strftime('%d/%m/%Y')
Output:
'15/02/2010'
You may apply pandas approach for different datatypes as:
import pandas as pd
import numpy as np
def reformat_date(date_string, old_format, new_format):
return pd.to_datetime(date_string, format=old_format, errors='ignore').strftime(new_format)
date_string = 'Mon Feb 15 2010'
date_list = ['Mon Feb 15 2010', 'Wed Feb 17 2010']
date_array = np.array(date_list)
date_series = pd.Series(date_list)
old_format = '%a %b %d %Y'
new_format = '%d/%m/%Y'
print(reformat_date(date_string, old_format, new_format))
print(reformat_date(date_list, old_format, new_format).values)
print(reformat_date(date_array, old_format, new_format).values)
print(date_series.apply(lambda x: reformat_date(x, old_format, new_format)).values)
Output:
15/02/2010
['15/02/2010' '17/02/2010']
['15/02/2010' '17/02/2010']
['15/02/2010' '17/02/2010']
Just for the sake of completion: when parsing a date using strptime() and the date contains the name of a day, month, etc, be aware that you have to account for the locale.
It's mentioned as a footnote in the docs as well.
As an example:
import locale
print(locale.getlocale())
>> ('nl_BE', 'ISO8859-1')
from datetime import datetime
datetime.strptime('6-Mar-2016', '%d-%b-%Y').strftime('%Y-%m-%d')
>> ValueError: time data '6-Mar-2016' does not match format '%d-%b-%Y'
locale.setlocale(locale.LC_ALL, 'en_US')
datetime.strptime('6-Mar-2016', '%d-%b-%Y').strftime('%Y-%m-%d')
>> '2016-03-06'
use datetime library
http://docs.python.org/library/datetime.html look up 9.1.7.
especiall strptime() strftime() Behavior¶
examples
http://pleac.sourceforge.net/pleac_python/datesandtimes.html
If you dont want to define the input date format then, Install dateparser (pip install dateparser) and,
from dateparser import parse
parse("Mon Feb 15 2010").strftime("%d/%m/%Y")
I know this has been asked in some form or another, but I'm wondering how to convert a date such as:
Saturday, May 18th, 2019
to something like 2019-05-18 so that I can compare it to other dates. I understand dateutil can do so but unfortunately I'm using Pycharm and it won't let me install that package without upgrading Python from 2.7 to 3. Is there a way to do this simply in Python?
You can use datetime with re module as well:
>>> from datetime import datetime
>>> import re
>>> s = 'Saturday, May 18th, 2019'
>>> datetime.strptime(re.sub('(\d+)(st|nd|rd|th)', '\g<1>', s), '%A, %B %d, %Y')
datetime.datetime(2019, 5, 18, 0, 0)
>>>
And print it nicer:
>>> print(datetime.strptime(re.sub('(\d+)(st|nd|rd|th)', '\g<1>', s), '%A, %B %d, %Y'))
2019-05-18 00:00:00
>>>
The regex solution is preferable in my opinion, but this is a solution without using re, for people not familiar with regexps:
from datetime import datetime
s = 'Saturday, May 18th, 2019'
sl = s.split()
sl[2] = sl[2][:-3] # remove 'th,'
datetime.strptime(' '.join(sl), '%A, %B %d %Y')
I am doing Python date comparizon:
Assume I have a date like this: 'Fri Aug 17 12:34:00 2012 +0000'
I am parsing it in the following manner:
dt=datetime.strptime('Fri Aug 17 12:34:00 2012 +0000', '%a %b %d %H:%M:%S %Y +0000')
I could not find on the documentation page how to handle the remaining +0000?
I want to have a more generic solution then hardcoded value.
Perhaps this is quite easy, any hint?
The default datetime module does not handle timezones very well; beyond your current machine timezone and UTC, they are basically not supported.
You'll have to use an external library for that or handle the timezone offset manually.
External library options:
Use dateutil.parser can handle just about any date and or time format you care to throw at it:
from dateutil import parser
dt = parser.parse(s)
The iso8601 library handles only ISO 8601 formats, which include timezone offsets of the same form:
import iso8601
datetimetext, tz = s.rsplit(None, 1) # only grab the timezone portion.
timezone = iso8601.iso8601.parse_timezone('{}:{}'.format(tz[:3], tz[3:]))
dt = datetime.strptime(datetimetext, '%a %b %d %H:%M:%S %Y').replace(tzinfo=timezone)
Demonstration of each approach:
>>> import datetime
>>> s = 'Fri Aug 17 12:34:00 2012 +0000'
>>> import iso8601
>>> timezone = iso8601.iso8601.parse_timezone('{}:{}'.format(tz[:3], tz[3:]))
>>> datetime.datetime.strptime(datetimetext, '%a %b %d %H:%M:%S %Y').replace(tzinfo=timezone)
datetime.datetime(2012, 8, 17, 12, 34, tzinfo=<FixedOffset '+00:00'>)
>>> from dateutil import parser
>>> parser.parse(s)
datetime.datetime(2012, 8, 17, 12, 34, tzinfo=tzutc())
You might want also give Delorean a look. Which is a wrapper around both pytz and dateutil it provides timezone manipulation as well as easy datetime time zone shift.
Here is how I would solve your question with Delorean.
>>> from delorean import parse
>>> parse("2011/01/01 00:00:00 -0700")
Delorean(datetime=2011-01-01 07:00:00+00:00, timezone=UTC)
From there you can return the datetime by simply return the .datetime attribute. If you wan to do some time shifts simply use the Delorean object and do .shift("UTC") etc.
Use the dateutil.parser:
>>> import dateutil.parser
>>> dateutil.parser.parse('Fri Aug 17 12:34:00 2012 +0000')
>>> datetime.datetime(2012, 8, 17, 12, 34, tzinfo=tzutc())
I have a date string with the format 'Mon Feb 15 2010'. I want to change the format to '15/02/2010'. How can I do this?
datetime module could help you with that:
datetime.datetime.strptime(date_string, format1).strftime(format2)
For the specific example you could do
>>> import datetime
>>> datetime.datetime.strptime('Mon Feb 15 2010', '%a %b %d %Y').strftime('%d/%m/%Y')
'15/02/2010'
>>>
You can install the dateutil library. Its parse function can figure out what format a string is in without having to specify the format like you do with datetime.strptime.
from dateutil.parser import parse
dt = parse('Mon Feb 15 2010')
print(dt)
# datetime.datetime(2010, 2, 15, 0, 0)
print(dt.strftime('%d/%m/%Y'))
# 15/02/2010
convert string to datetime object
from datetime import datetime
s = "2016-03-26T09:25:55.000Z"
f = "%Y-%m-%dT%H:%M:%S.%fZ"
out = datetime.strptime(s, f)
print(out)
output:
2016-03-26 09:25:55
>>> from_date="Mon Feb 15 2010"
>>> import time
>>> conv=time.strptime(from_date,"%a %b %d %Y")
>>> time.strftime("%d/%m/%Y",conv)
'15/02/2010'
As this question comes often, here is the simple explanation.
datetime or time module has two important functions.
strftime - creates a string representation of date or time from a datetime or time object.
strptime - creates a datetime or time object from a string.
In both cases, we need a formating string. It is the representation that tells how the date or time is formatted in your string.
Now lets assume we have a date object.
>>> from datetime import datetime
>>> d = datetime(2010, 2, 15)
>>> d
datetime.datetime(2010, 2, 15, 0, 0)
If we want to create a string from this date in the format 'Mon Feb 15 2010'
>>> s = d.strftime('%a %b %d %y')
>>> print s
Mon Feb 15 10
Lets assume we want to convert this s again to a datetime object.
>>> new_date = datetime.strptime(s, '%a %b %d %y')
>>> print new_date
2010-02-15 00:00:00
Refer This document all formatting directives regarding datetime.
#codeling and #user1767754 : The following two lines will work. I saw no one posted the complete solution for the example problem that was asked. Hopefully this is enough explanation.
import datetime
x = datetime.datetime.strptime("Mon Feb 15 2010", "%a %b %d %Y").strftime("%d/%m/%Y")
print(x)
Output:
15/02/2010
You may achieve this using pandas as well:
import pandas as pd
pd.to_datetime('Mon Feb 15 2010', format='%a %b %d %Y').strftime('%d/%m/%Y')
Output:
'15/02/2010'
You may apply pandas approach for different datatypes as:
import pandas as pd
import numpy as np
def reformat_date(date_string, old_format, new_format):
return pd.to_datetime(date_string, format=old_format, errors='ignore').strftime(new_format)
date_string = 'Mon Feb 15 2010'
date_list = ['Mon Feb 15 2010', 'Wed Feb 17 2010']
date_array = np.array(date_list)
date_series = pd.Series(date_list)
old_format = '%a %b %d %Y'
new_format = '%d/%m/%Y'
print(reformat_date(date_string, old_format, new_format))
print(reformat_date(date_list, old_format, new_format).values)
print(reformat_date(date_array, old_format, new_format).values)
print(date_series.apply(lambda x: reformat_date(x, old_format, new_format)).values)
Output:
15/02/2010
['15/02/2010' '17/02/2010']
['15/02/2010' '17/02/2010']
['15/02/2010' '17/02/2010']
Just for the sake of completion: when parsing a date using strptime() and the date contains the name of a day, month, etc, be aware that you have to account for the locale.
It's mentioned as a footnote in the docs as well.
As an example:
import locale
print(locale.getlocale())
>> ('nl_BE', 'ISO8859-1')
from datetime import datetime
datetime.strptime('6-Mar-2016', '%d-%b-%Y').strftime('%Y-%m-%d')
>> ValueError: time data '6-Mar-2016' does not match format '%d-%b-%Y'
locale.setlocale(locale.LC_ALL, 'en_US')
datetime.strptime('6-Mar-2016', '%d-%b-%Y').strftime('%Y-%m-%d')
>> '2016-03-06'
use datetime library
http://docs.python.org/library/datetime.html look up 9.1.7.
especiall strptime() strftime() Behavior¶
examples
http://pleac.sourceforge.net/pleac_python/datesandtimes.html
If you dont want to define the input date format then, Install dateparser (pip install dateparser) and,
from dateparser import parse
parse("Mon Feb 15 2010").strftime("%d/%m/%Y")