Parse date string and change format

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Parse date string and change format

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?

Asked By: Nimmy

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Answer #1:

datetime module could help you with that:

datetime.datetime.strptime(date_string, format1).strftime(format2)

For the specific example you could do

>>> datetime.datetime.strptime('Mon Feb 15 2010', '%a %b %d %Y').strftime('%d/%m/%Y')
'15/02/2010'
>>>
Answered By: SilentGhost

Answer #2:

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
Answered By: llazzaro

Answer #3:

>>> 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'
Answered By: ghostdog74

Answer #4:

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
Answered By: anjaneyulubatta505

Answer #5:

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.

Answered By: thavan

Answer #6:

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'
Answered By: ???S???

Answer #7:

@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
Answered By: Gobryas

Answer #8:

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']
Answered By: nimbous

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