script for seconds in given month?

[edfialk]

Hi, does anyone happen to know of a script that would return the
number of seconds in a month if I give it a month and a year?

My python is a little weak, but if anyone could offer some suggestions
I think I could handle it myself, or if anyone happens to know of a
script already written that performs this I would be extremely
grateful.

Thanks!
-Ed

[Jim]

On Apr 16, 12:22 pm, "edfialk" <edfi...@gmail. comwrote:

Hi, does anyone happen to know of a script that would return the
number of seconds in a month if I give it a month and a year?

My python is a little weak, but if anyone could offer some suggestions
I think I could handle it myself, or if anyone happens to know of a
script already written that performs this I would be extremely
grateful.

Probably there are sophisticated answers, but have you tried
something like:

monthDays={'Jan ':31,'Feb':28, ..}
secs=60*60*24*m onthDays[thisMonth]
if (thisMonth=='Fe b'
and isLeap(thisYear )):
secs+=60*60*24
return secs

?

[pythoncurious]

On Apr 16, 6:22 pm, "edfialk" <edfi...@gmail. comwrote:

Hi, does anyone happen to know of a script that would return the
number of seconds in a month if I give it a month and a year?

something like this might work, it should event handle DST correctly.
You could read up on mktime() if you want to make sure.

from time import mktime
def secondsInMonth( year, month):
s1 = mktime((year,mo nth,1,0,0,0,0,0 ,-1))
s2 = mktime((year,mo nth+1,1,0,0,0,0 ,0,-1))
return s2-s1

/Matt

[skip]

Mattfrom time import mktime
Mattdef secondsInMonth( year, month):
Matt s1 = mktime((year,mo nth,1,0,0,0,0,0 ,-1))
Matt s2 = mktime((year,mo nth+1,1,0,0,0,0 ,0,-1))
Matt return s2-s1

Probably won't work if month==12. ;-)

Skip

[Paul McGuire]

On Apr 16, 11:22 am, "edfialk" <edfi...@gmail. comwrote:

Hi, does anyone happen to know of a script that would return the
number of seconds in a month if I give it a month and a year?

My python is a little weak, but if anyone could offer some suggestions
I think I could handle it myself, or if anyone happens to know of a
script already written that performs this I would be extremely
grateful.

Thanks!
-Ed

Do you need to handle leap seconds too? (not a joke)

-- Paul

[edfialk]

Jim: I need years too, basically from 1960-2000. Don't want to
hardcode all those days :)

Matt: Thanks, I will try this out.

Paul: I don't believe we need leap seconds. Leap days definitely.

I'll let you know how Matt's code works. Any other suggestions feel
free to let me know.

Thanks all!
-Ed

[attn.steven.kuo]

On Apr 16, 10:18 am, [e-mail] wrote:

Mattfrom time import mktime
Mattdef secondsInMonth( year, month):
Matt s1 = mktime((year,mo nth,1,0,0,0,0,0 ,-1))
Matt s2 = mktime((year,mo nth+1,1,0,0,0,0 ,0,-1))
Matt return s2-s1

Probably won't work if month==12. ;-)

Skip

Actually, mktime as described in the C Standard does not
restrict members such as tm_mon of struct tm to the
range 0-11 (1-12 for the corresponding entry in the
time tuple in Python). So the underlying struct in
CPython may be normalized and return a perfectly valid
time even if month is 12.

--
Regards,
Steven

[Shane Geiger]

import datetime

def first_day_of_ne xt_month( year, month ):
"""returns the first day of the next month

>>first_day_of_ next_month(2007 ,5)

datetime.dateti me(2007, 6, 1, 0, 0)

>>first_day_of_ next_month(2007 ,12)

datetime.dateti me(2008, 1, 1, 0, 0)
"""
oneday = datetime.timede lta(days=1)
day = datetime.dateti me(year, month, 1)
if day.day == 1:
day += oneday
while day.day != 1:
day += oneday
return day
from time import mktime
def secondsInMonth( year, month):
s1 = mktime((year,mo nth,1,0,0,0,0,0 ,-1))
day = first_day_of_ne xt_month(year, month)
year = day.year
month = day.month
s2 = mktime((year,mo nth+1,1,0,0,0,0 ,0,-1))
return s2-s1
year = 2007
month = 2
print secondsInMonth( year, month)

sk**@pobox.com wrote:

Mattfrom time import mktime
Mattdef secondsInMonth( year, month):
Matt s1 = mktime((year,mo nth,1,0,0,0,0,0 ,-1))
Matt s2 = mktime((year,mo nth+1,1,0,0,0,0 ,0,-1))
Matt return s2-s1

Probably won't work if month==12. ;-)

Skip

--
Shane Geiger
IT Director
National Council on Economic Education
sg*****@ncee.ne t | 402-438-8958 | http://www.ncee.net

Leading the Campaign for Economic and Financial Literacy

[Paul McGuire]

On Apr 16, 12:49 pm, "edfialk" <edfi...@gmail. comwrote:

Jim: I need years too, basically from 1960-2000. Don't want to
hardcode all those days :)

Matt: Thanks, I will try this out.

Paul: I don't believe we need leap seconds. Leap days definitely.

I'll let you know how Matt's code works. Any other suggestions feel
free to let me know.

Thanks all!
-Ed

I googled for "NIST leap second" and found this table online of leap
seconds, if you do in fact need them: http://tf.nist.gov/pubs/bulletin/leapsecond.htm

-- Paul

[Alex Martelli]

edfialk <ed*****@gmail. comwrote:

Hi, does anyone happen to know of a script that would return the
number of seconds in a month if I give it a month and a year?

My python is a little weak, but if anyone could offer some suggestions
I think I could handle it myself, or if anyone happens to know of a
script already written that performs this I would be extremely
grateful.

import calendar

def seconds_in_mont h(month, year):
nomatter, daysinmonth = calendar.monthr ange(year, month)
return daysinmonth * 24 * 60 * 60
Alex