Retrieving a date from a complex string in Python

Question

I'm trying to get a single datetime out of two strings using datetime.strptime.

The time is pretty easy (ex. 8:53PM), so I can do something like:

theTime = datetime.strptime(givenTime, "%I:%M%p")

However, the string has more than just a date, it's a link in a format similar to http://site.com/?year=2011&month=10&day=5&hour=11. I know that I could do something like:

theDate = datetime.strptime(givenURL, "http://site.com/?year=%Y&month=%m&day=%d&hour=%H")

but I don't want to get that hour from the link since it's being retrieved elsewhere. Is there a way to put a dummy symbol (like %x or something) to serve as a flexible space for that last variable?

In the end, I envision having a single line similar to:

theDateTime = datetime.strptime(givenURL + givenTime, ""http://site.com/?year=%Y&month=%m&day=%d&hour=%x%I:%M%p")

(although, obviously, the %x wouldn't be used). Any ideas?

Answer 1

Think that if you would like to simple skip time from the URL you can use split for example the following way:

givenURL = 'http://site.com/?year=2011&month=10&day=5&hour=11'
pattern = "http://site.com/?year=%Y&month=%m&day=%d"
theDate = datetime.strptime(givenURL.split('&hour=')[0], pattern)

So not sure that understood you correctly, but:

givenURL = 'http://site.com/?year=2011&month=10&day=5&hour=11'
datePattern = "http://site.com/?year=%Y&month=%m&day=%d"
timePattern = "&time=%I:%M%p"

theDateTime = datetime.strptime(givenURL.split('&hour=')[0] + '&time=' givenTime, datePattern + timePattern)

Answer 2

import datetime
import re

givenURL  = 'http://site.com/?year=2011&month=10&day=5&hour=11'
givenTime = '08:53PM'

print ' givenURL == ' + givenURL
print 'givenTime == ' + givenTime

regx = re.compile('year=(\d\d\d\d)&month=(\d\d?)&day=(\d\d?)&hour=\d\d?')
print '\nmap(int,regx.search(givenURL).groups()) ==',map(int,regx.search(givenURL).groups())

theDate = datetime.date(*map(int,regx.search(givenURL).groups()))
theTime = datetime.datetime.strptime(givenTime, "%I:%M%p")

print '\ntheDate ==',theDate,type(theDate)
print '\ntheTime ==',theTime,type(theTime)


theDateTime = theTime.replace(theDate.year,theDate.month,theDate.day)
print '\ntheDateTime ==',theDateTime,type(theDateTime)

result

 givenURL == http://site.com/?year=2011&month=10&day=5&hour=11
givenTime == 08:53PM

map(int,regx.search(givenURL).groups()) == [2011, 10, 5]

theDate == 2011-10-05 <type 'datetime.date'>

theTime == 1900-01-01 20:53:00 <type 'datetime.datetime'>

theDateTime == 2011-10-05 20:53:00 <type 'datetime.datetime'>

Edit 1

As strptime() is slow, I improved my code to eliminate it

from datetime import datetime
import re
from time import clock


n = 10000

givenURL  = 'http://site.com/?year=2011&month=10&day=5&hour=11'
givenTime = '08:53AM'

# eyquem
regx = re.compile('year=(\d\d\d\d)&month=(\d\d?)&day=(\d\d?)&hour=\d\d? (\d\d?):(\d\d?)(PM|pm)?')
t0 = clock()
for i in xrange(n):
    given = givenURL + ' ' + givenTime
    mat = regx.search(given)
    grps = map(int,mat.group(1,2,3,4,5))
    if mat.group(6):
        grps[3] += 12 # when it is PM/pm, the hour must be augmented with 12
    theDateTime1 = datetime(*grps)
print clock()-t0,"seconds   eyquem's code"
print theDateTime1


print

# Artsiom Rudzenka
dateandtimePattern = "http://site.com/?year=%Y&month=%m&day=%d&time=%I:%M%p"
t0 = clock()
for i in xrange(n):
    theDateTime2 = datetime.strptime(givenURL.split('&hour=')[0] + '&time=' + givenTime, dateandtimePattern)
print clock()-t0,"seconds   Artsiom's code"
print theDateTime2

print
print theDateTime1 == theDateTime2

result

0.460598763251 seconds   eyquem's code
2011-10-05 08:53:00

2.10386180366 seconds   Artsiom's code
2011-10-05 08:53:00

True

My code is 4.5 times faster. That may be interesting if there are a lot of such transformations to perform

Answer 3

There's no way to do that with the format string. However, if the hour doesn't matter, you can get it from the URL as in your first example and then call theDateTime.replace(hour=hour_from_a_different_source).

That way you don't have to do any additional parsing.