CODEWITHSUNDEEP
pythondatetime
VorX
VorX

Reputation: 455

Datetime strptime in python

I've tried the following code :

import datetime
d = datetime.datetime.strptime("01/27/2012", "%m/%d/%Y")
print(d)

and the output is : 

2012-01-27 00:00:00

I'am using Linux Mint :

test@testsrv ~/pythonvault $ date
Fri Jun 16 21:40:57 EEST 2017

So,the question is why the output of python code returns a date in "%Y/%m/%d" ( 2012-01-27 ) instead of "%m/%d/%Y" format ?

Please note that I'am using python 2.7

Any help would be appreciated.

Upvotes: 27

Views: 96615

Answers (6)

user8234870
user8234870

Reputation:

The datetime.strptime() class method creates a datetime object from a string representing a date and time and a corresponding format string.

  • %a

    Weekday as locale’s abbreviated name.

Sun, Mon, …, Sat (en_US);
So, Mo, …, Sa (de_DE)

  • %A

    Weekday as locale’s full name.

Sunday, Monday, …, Saturday (en_US);
Sonntag, Montag, …, Samstag (de_DE)

  • %w

    Weekday as a decimal number, where 0 is Sunday and 6 is Saturday.

0, 1, …, 6

  • %d

    Day of the month as a zero-padded decimal number.

01, 02, …, 31

  • %b

    Month as locale’s abbreviated name.

Jan, Feb, …, Dec (en_US);
Jan, Feb, …, Dez (de_DE)

  • %B

     Month as locale’s full name.

 January, February, …, December (en_US);
 Januar, Februar, …, Dezember (de_DE)

  • %m

    Month as a zero-padded decimal number.

01, 02, …, 12

  • %y

    Year without century as a zero-padded decimal number.

00, 01, …, 99

  • %Y

    Year with century as a decimal number.

0001, 0002, …, 2013, 2014, …, 9998, 9999

  • %H

    Hour (24-hour clock) as a zero-padded decimal number.

00, 01, …, 23

  • %I

    Hour (12-hour clock) as a zero-padded decimal number.

01, 02, …, 12

  • %p

    Locale’s equivalent of either AM or PM.

AM, PM (en_US);
am, pm (de_DE)

  • %M

    Minute as a zero-padded decimal number.

00, 01, …, 59

  • %S

    Second as a zero-padded decimal number.

00, 01, …, 59

  • %f

    Microsecond as a decimal number, zero-padded to 6 digits.

000000, 000001, …, 999999

  • %z

    UTC offset in the form ±HHMM[SS[.ffffff]] (empty string if the object is naive).

(empty), +0000, -0400, +1030, +063415, -030712.345216

  • %Z

    Time zone name (empty string if the object is naive).

(empty), UTC, GMT

  • %j

    Day of the year as a zero-padded decimal number.

001, 002, …, 366

  • %U

    Week number of the year (Sunday as the first day of the week) as a zero-padded decimal number. All days in a new year preceding the

    first Sunday are considered to be in week 0.

    00, 01, …, 53

  • %W

    Week number of the year (Monday as the first day of the week) as a zero-padded decimal number. All days in a new year preceding the

    first Monday are considered to be in week 0.

    00, 01, …, 53

  • %c

    Locale’s appropriate date and time representation.

Tue Aug 16 21:30:00 1988 (en_US);
Di 16 Aug 21:30:00 1988 (de_DE)

  • %x

    Locale’s appropriate date representation.

08/16/88 (None);
08/16/1988 (en_US);
16.08.1988 (de_DE)

  • %X

    Locale’s appropriate time representation.

21:30:00 (en_US);
21:30:00 (de_DE)

  • %%

    A literal '%' character.

%

for more information please visit official website for python

Example Programs for better understanding

Program 1:

from datetime import datetime

data = "2022-02-09 12:25:22"

dat = datetime.strptime(data, "%Y-%m-%d %H:%M:%S")
print(dat)

Output:

2022-02-09 12:25:22

Program 2:

from datetime import datetime
data = "Thursday, 10 February 2022 11:28:45 am"
dat = datetime.strptime(data, "%A, %d %B %Y %H:%M:%S %p")
print(dat)

Output:

2022-02-10 11:28:45

Upvotes: 1

RedKoder
RedKoder

Reputation: 61

There is a difference between datetime.strp[arse]time() and datetime.strf[ormat]time().

The first one, strptime() allows you to create a date object from a string source, provided you can tell it what format to expect:

strDate = "11-Apr-2019_09:15:42"
dateObj = datetime.strptime(strDate, "%d-%b-%Y_%H:%M%S")

print(dateObj)  # shows: 2019-04-11 09:15:42

The second one, strftime() allows you to export your date object to a string in the format of your choosing:

dateObj = datetime(2019, 4, 11, 9, 19, 25)
strDate = dateObj.strftime("%m/%d/%y %H:%M:%S")

print(strDate)  # shows: 04/11/19 09:19:25

What you're seeing is simply the default string format of a datetime object because you didn't explicitly tell it what format to use.

Checkout http://strftime.org/ for a list of all the different string format options that are availble.

Upvotes: 2

Mahmudul Hasan
Mahmudul Hasan

Reputation: 87

import datetime

str_time= "2018-06-03 08:00:00"
date_date = datetime.datetime.strptime(str_time, "%Y-%m-%d %H:%M:%S")
print date_date

Upvotes: 2

Ravi
Ravi

Reputation: 31397

You need to make sure you provide input accordingly

datetime.strptime(date_string,date_string_format).strftime(convert_to_date_string_format)

To print the date in specified format you need to provide format as below.

import datetime
d =datetime.datetime.strptime("01/27/2012","%m/%d/%Y").strftime('%m/%d/%Y')
print d

Output:

01/27/2012

>>Demo<<

Upvotes: 34

Spencer D
Spencer D

Reputation: 3486

As astutely noted in the comments, you are parsing to a datetime object using the format you specified.

strptime(...) is String Parse Time. You have specified the format for how the string should be interpreted to initialize a Datetime object, but that format is only utilized for initialization. By default, when you go to print that datetime object, you are getting the representation of str(DatetimeObjectInstance) (in your case, str(d)).

If you want a different format, you should use String Format Time (strftime(...))

Upvotes: 3

taras
taras

Reputation: 6914

datetime.strptime(date_string, format) function returns a datetime object corresponding to date_string, parsed according to format. When you print datetime object, it is formatted as a string in ISO 8601 format, YYYY-MM-DDTHH:MM:SS

References:

Upvotes: 5

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