CODEWITHSUNDEEP
pythondatetime
Ashy
Ashy

Reputation: 2074

Python time to age

I'm trying to convert a date string into an age.

The string is like: "Mon, 17 Nov 2008 01:45:32 +0200" and I need to work out how many days old it is.

I have sucessfully converted the date using:

>>> time.strptime("Mon, 17 Nov 2008 01:45:32 +0200","%a, %d %b %Y %H:%M:%S +0200")
(2008, 11, 17, 1, 45, 32, 0, 322, -1)

For some reason %z gives me an error for the +0200 but it doesn't matter that much.

I can get the current time using:

>>> time.localtime()
(2009, 2, 3, 19, 55, 32, 1, 34, 0)

but how can I subtract one from the other without going though each item in the list and doing it manually?

Upvotes: 3

Views: 7916

Answers (6)

jfs
jfs

Reputation: 414129

Since Python 3.2, datetime.strptime() returns an aware datetime object if %z directive is provided:

#!/usr/bin/env python3
from datetime import datetime, timezone, timedelta

s = "Mon, 17 Nov 2008 01:45:32 +0200"
birthday = datetime.strptime(s, '%a, %d %b %Y %H:%M:%S %z')
age = (datetime.now(timezone.utc) - birthday) / timedelta(1) # age in days
print("%.0f" % age)

On older Python versions the correct version of @Tony Meyer's answer could be used:

#!/usr/bin/env python
import time
from email.utils import parsedate_tz, mktime_tz

s = "Mon, 17 Nov 2008 01:45:32 +0200"
ts = mktime_tz(parsedate_tz(s))   # seconds since Epoch
age = (time.time() - ts) / 86400  # age in days
print("%.0f" % age)

Both code examples produce the same result.

Upvotes: 0

Georg Schölly
Georg Schölly

Reputation: 126095

You need to use the module datetime and the object datetime.timedelta

from datetime import datetime

t1 = datetime.strptime("Mon, 17 Nov 2008 01:45:32 +0200","%a, %d %b %Y %H:%M:%S +0200")
t2 = datetime.now()

tdelta = t2 - t1 # actually a datetime.timedelta object
print tdelta.days

Upvotes: 16

Tony Meyer
Tony Meyer

Reputation: 10157

If you don't want to use datetime (e.g. if your Python is old and you don't have the module), you can just use the time module.

s = "Mon, 17 Nov 2008 01:45:32 +0200"
import time
import email.utils # Using email.utils means we can handle the timezone.
t = email.utils.parsedate_tz(s) # Gets the time.mktime 9-tuple, plus tz
d = time.time() - time.mktime(t[:9]) + t[9] # Gives the difference in seconds.

Upvotes: 1

Ashy
Ashy

Reputation: 2074

Thanks guys, I ended up with the following:

def getAge( d ):
    """ Calculate age from date """
    delta = datetime.now() - datetime.strptime(d, "%a, %d %b %Y %H:%M:%S +0200")
    return delta.days + delta.seconds / 86400.0 # divide secs into days

Giving:

>>> getAge("Mon, 17 Nov 2008 01:45:32 +0200")
78.801319444444445

Upvotes: 0

Ben Blank
Ben Blank

Reputation: 56572

In Python, datetime objects natively support subtraction:

from datetime import datetime
age = datetime.now() - datetime.strptime(...)
print age.days

The result is a timedelta object.

Upvotes: 5

Alex. S.
Alex. S.

Reputation: 147206

from datetime import datetime, timedelta
datetime.now()
datetime.datetime(2009, 2, 3, 15, 17, 35, 156000)
datetime.now() - datetime(1984, 6, 29 )
datetime.timedelta(8985, 55091, 206000)
datetime.now() - datetime(1984, 6, 29 )
datetime.timedelta(8985, 55094, 198000) # my age...

timedelta(days[, seconds[, microseconds[, milliseconds[, minutes[, hours[, weeks]]]]]]])

Upvotes: 2

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