CODEWITHSUNDEEP
pythondatetime
Rohit-Pandey
Rohit-Pandey

Reputation: 2159

How to convert a string into date-format in python?

I have a string like 23 July 1914 and want to convert it to 23/07/1914 date format. But my code gives error.

from datetime import datetime
datetime_object = datetime.strptime('1 June 2005','%d %m %Y')
print datetime_object

Upvotes: 0

Views: 4213

Answers (5)

void
void

Reputation: 2652

Here is what you should be doing:

from datetime import datetime
datetime_object = datetime.strptime('1 June 2005','%d %B %Y')
s = datetime_object.strftime("%d/%m/%y")
print(s)

Output:

>>> 01/06/05

You see your strptime requires two parameters.

strptime(string[, format])

And the string will be converted to a datetime object according to a format you specify.

There are various formats

%a - abbreviated weekday name %A - full weekday name %b - abbreviated month name %B - full month name %c - preferred date and time representation %C - century number (the year divided by 100, range 00 to 99) %d - day of the month (01 to 31) %D - same as %m/%d/%y %e - day of the month (1 to 31) %g - like %G, but without the century %G - 4-digit year corresponding to the ISO week number (see %V). %h - same as %b %H - hour, using a 24-hour clock (00 to 23)

The above are some examples. Take a look here for formats

Take a goood look at these two!

%b - abbreviated month name %B - full month name

It should be in a similar pattern to the string you provide. Confusing take a look at these examples.

>>> datetime.strptime('1 jul 2009','%d %b %Y')
datetime.datetime(2009, 7, 1, 0, 0)
>>> datetime.strptime('1 Jul 2009','%d %b %Y')
datetime.datetime(2009, 7, 1, 0, 0)
>>> datetime.strptime('jul 21 1996','%b %d %Y')
datetime.datetime(1996, 7, 21, 0, 0)

As you can see based on the format the string is turned into a datetime object. Now take a look!

>>> datetime.strptime('1 July 2009','%d %b %Y')
Traceback (most recent call last):
  File "<pyshell#12>", line 1, in <module>
    datetime.strptime('1 July 2009','%d %b %Y')
  File "/usr/lib/python3.5/_strptime.py", line 510, in _strptime_datetime
    tt, fraction = _strptime(data_string, format)
  File "/usr/lib/python3.5/_strptime.py", line 343, in _strptime
    (data_string, format))
ValueError: time data '1 July 2009' does not match format '%d %b %Y'

Why error because jun or Jun (short form) stands for %b. When you supply a June it gets confused. Now what to do? Changed the format. 

>>> datetime.strptime('1 July 2009','%d %B %Y')
datetime.datetime(2009, 7, 1, 0, 0)

Simple now converting the datetime object is simple enough.

>>> s = datetime.strptime('1 July 2009','%d %B %Y')
>>> s.strftime('%d/%m/%Y')
'01/07/2009

Again the %m is the format for displaying it in months (numbers) read more about them.

Upvotes: 3

Honza Z&#237;ka
Honza Z&#237;ka

Reputation: 483

%d means "Day of the month as a zero-padded decimal number."

%m means "Month as a zero-padded decimal number."

Neither day or month are supplied what you tell it to expect. What you need it %B for month (only if your locale is en_US), and %-d for day.

Upvotes: 0

BPL
BPL

Reputation: 9863

This should work:

from datetime import datetime
print(datetime.strptime('1 June 2005', '%d %B %Y').strftime('%d/%m/%Y'))
print(datetime.strptime('23 July 1914', '%d %B %Y').strftime('%d/%m/%Y'))

For more info you can read about strftime-strptime-behavior

Upvotes: 0

MSeifert
MSeifert

Reputation: 152795

The placeholder for "Month as locale’s full name." would be %B not %m:

>>> from datetime import datetime
>>> datetime_object = datetime.strptime('1 June 2005','%d %B %Y')
>>> print(datetime_object)
2005-06-01 00:00:00

>>> print(datetime_object.strftime("%d/%m/%Y"))
01/06/2005

Upvotes: 0

Jake Conkerton-Darby
Jake Conkerton-Darby

Reputation: 1101

Your error is in the format you are using to strip your string. You use %m as the format specifier for month, but this expects a 0 padded integer representing the month of the year (e.g. 06 for your example). What you want to use is %B, which expects an month of the year written out fully (e.g. June in your example).

For a full explanation of the datetime format specifiers please consult the documentation, and if you have any other issues please check there first.

Upvotes: 7

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