convert d-m-Y in python

Question

I want to convert date from one format to another.Let's say, I have some date:

February 27, 1820
April 3, 1885

And I want to convert it:

27 February 1820
3 April 1885

I tried but getting error:

dt = "February 12, 1809" 
dt = datetime.strptime('%d %b %Y')

Any kind of help will be greatly appreciated.

[datetime.strptime()](http://docs.python.org/library/datetime.html#strftime-strptime-behavior) is all you need. — Jacob, Jul 26 '11 at 11:03
Are you really dealing with dates in the 19th century? That's going to create trouble when you want to reformat the date back to readable date using strftime(). — Susam Pal, Jul 26 '11 at 11:10
@ Susam Pal: yes, some are before 1900 & some are after 1900 — Liza, Jul 26 '11 at 11:12

score 2 · Answer 1 · edited Jul 26 '11 at 12:41

2

Nor time.strftime nor datetime.datetime.strftime don't accept year values less then 1900. So it looks like there is job for re (although it is a bad style to use it for such tasks):

>>> import re
>>> s = 'February 27, 1820'
>>> month, day, year = re.findall('(\w+) (\d+), (\d+)', s)[0]  
>>> '%s %s %s' % (day, month, year)
'27 February 1820'

edited Jul 26 '11 at 12:41

Liza

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answered Jul 26 '11 at 11:18

Roman Bodnarchuk

29,461
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How to get `27 February 1820` by your solution? – Liza Jul 26 '11 at 11:46
@Liza Simply writing ``'%s %s %s' % (day, month, year)`` – eyquem Jul 26 '11 at 11:54
probabaly, `'%s %s %s' % (month, year, day)`. Anyway, Thanks a lot! – Liza Jul 26 '11 at 12:05
1

@Liza Warning: test the Roman's code on the following string ``'On February 27, 1820 there were 10, 1000, 100000, maybe more shooting stars in the 98, 456347XBC, Hi-Pot galaxy that were seen by 20, 2000 or 200000 persons'`` and you'll see the problem – eyquem Jul 26 '11 at 12:57

score 1 · Answer 2 · answered Jul 26 '11 at 11:07

1

Use datetime.datetime.strptime, or parsedatetime, or dateutil to parse your string to date.

Then use datetime.datetime.strftime to format your datetime object to string using needed format.

answered Jul 26 '11 at 11:07

warvariuc

57,116
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score 1 · Answer 3 · answered Jul 26 '11 at 11:14

You can use strptime / strftime:

from datetime import datetime

date = datetime.strptime('February 27, 1820', '%B %d, %Y')
new_date = date.strftime('%d %b %Y')

However, there's one gotcha. strftime seems to have problems with years < 1900 on some platforms. There's a workaround for it too.

score 1 · Accepted Answer · edited May 23 '17 at 11:55

Yesterday, I learned from Kirk Strauser that strptime()) is much more slower than other solutions: see this file

So my advice is to use another way. For exemple:

import re

ss = '''February 27, 1820
a line
April 3, 1885'''

regx = re.compile('(January|February|March|'
                  'April|May|June'
                  'July|August|September|'
                  'October|November|December)'
                  ' '
                  '(\d|[012]\d|3[01])'
                  ',(?= \d{4})')

print regx.findall(ss)
print
print regx.sub('\\2 \\1',ss)

Edit 1

The speed of the program can be improved using regx.sub(repl,ss) with repl() being a function that doesn't extract the month and day as group(1) and group(2), but by slicing:

import re
from time import clock

ss = '''February 27, 1820
a line
April 3, 1885'''

regx = re.compile('(January|February|March|'
                  'April|May|June'
                  'July|August|September|'
                  'October|November|December)'
                  ' '
                  '(\d|[012]\d|3[01])'
                  ',(?= \d{4})')

print regx.findall(ss)
print


te = clock()
for i in xrange(10000):
    x = regx.sub('\\2 \\1',ss)
print clock()-te
print x
print




regx = re.compile('(?:January|February|March|'
                  'April|May|June'
                  'July|August|September|'
                  'October|November|December)'
                  ' '
                  '(?:\d|[012]\d|3[01]),'
                  '(?= \d{4})')

def repl(mat):
    sp = mat.group().split()
    return sp[1][0:-1] + ' ' + sp[0]

te = clock()
for i in xrange(1):
    y = regx.sub(repl,ss)
print clock()-te
print y

result

[('February', '27'), ('April', '3')]

2.52965614345
27 February 1820
a line
3 April 1885

0.378833622709
27 February 1820
a line
3 April 1885

PS: I also knew that there is a problem of span of time covered by strftime and strptime (not before 1900) , that's why I immediately choosed to treat the problem with a regex. People find regexes too heavy and impressive to resort to them, but I don't understand this trend, because as soon as you master just a little the regexes, you can do plenty of things, with efficiency and speed. Hura for the regex tool.

convert d-m-Y in python

4 Answers4

Edit 1