Find day of week from month and year

[Laguna]

Hi Gurus,

I want to find the expiration date of stock options (3rd Friday of the
month) for an any give month and year. I have tried a few tricks with
the functions provided by the built-in module time, but the problem was
that the 9 element tuple need to be populated correctly. Can anyone
help me out on this one?

Thanks a bunch,
Laguna

Requirements:

d0 = expiration(9, 2005) # d0 would be 16
d1 = expiration(6, 2003) # d1 would be 20
d2 = expiration(2, 2006) # d2 would be 17

[Paul Rubin]

"Laguna" <ed*****@yahoo. com> writes:

I want to find the expiration date of stock options (3rd Friday of the
month) for an any give month and year. I have tried a few tricks with
the functions provided by the built-in module time, but the problem was
that the 9 element tuple need to be populated correctly. Can anyone
help me out on this one?
It's probably simplest to use the calendar module:

http://docs.python.org/lib/module-calendar.html

see the weekday function.
d0 = expiration(9, 2005) # d0 would be 16
d1 = expiration(6, 2003) # d1 would be 20
d2 = expiration(2, 2006) # d2 would be 17

# not completely tested
import calendar
def expiration(mont h, year):
w1 = calendar.weekda y(year, month, 1) # weekday of 1st of month
f1d = 1 + (4-w1) % 7 # date of 1st friday
return f1d + 14 # date of 3rd friday

[Robert Kern]

Laguna wrote:

Hi Gurus,

I want to find the expiration date of stock options (3rd Friday of the
month) for an any give month and year. I have tried a few tricks with
the functions provided by the built-in module time, but the problem was
that the 9 element tuple need to be populated correctly. Can anyone
help me out on this one?

mx.DateTime provides a RelativeDateTim e constructor that handles things
like this.

http://www.egenix.com/files/python/mxDateTime.html

--
Robert Kern
rk***@ucsd.edu

"In the fields of hell where the grass grows high
Are the graves of dreams allowed to die."
-- Richard Harter

[Donn Cave]

In article <11************ **********@g47g 2000cwa.googleg roups.com>,
"Laguna" <ed*****@yahoo. com> wrote:

I want to find the expiration date of stock options (3rd Friday of the
month) for an any give month and year. I have tried a few tricks with
the functions provided by the built-in module time, but the problem was
that the 9 element tuple need to be populated correctly. Can anyone
help me out on this one? .... Requirements:

d0 = expiration(9, 2005) # d0 would be 16
d1 = expiration(6, 2003) # d1 would be 20
d2 = expiration(2, 2006) # d2 would be 17

What do you mean by, "the 9 element tuple need to be populated
correctly"? Do you need someone to tell you what values it
needs? What happens if you use (2005, 9, 1, 0, 0, 0, 0, 0, 0),
for example? If you make this tuple with localtime or gmtime,
do you know what the 7th (tm[6]) element of the tuple is?
What tricks did you try, exactly?

Donn Cave, do**@u.washingt on.edu

[Laguna]

> What do you mean by, "the 9 element tuple need to be populated

correctly"? Do you need someone to tell you what values it
needs? What happens if you use (2005, 9, 1, 0, 0, 0, 0, 0, 0),
for example? If you make this tuple with localtime or gmtime,
do you know what the 7th (tm[6]) element of the tuple is?
What tricks did you try, exactly?

Donn Cave, do**@u.washingt on.edu

Thanks for pointing out. tm[6] = weekday, and tm[7] = Julian data, but
I wouldn't know these values when my input values are month and year.

I will try out the more constructive suggestions from Paul and Robert.

Following is what I have tried. As you can tell, the results are wrong!

import time
time.asctime((2 003, 9, 1, 0, 0, 0, 0, 0, 0)) 'Mon Sep 01 00:00:00 2003' time.asctime((2 003, 8, 1, 0, 0, 0, 0, 0, 0)) 'Mon Aug 01 00:00:00 2003' time.asctime((2 003, 7, 1, 0, 0, 0, 0, 0, 0))

'Mon Jul 01 00:00:00 2003'

[Laguna]

Paul,

Thanks for the suggestion on calendar module. Here is my solution and
it works:

def expiration(year , month):
weekday = calendar.weekda y(year, month, 1)
table = [19, 18, 17, 16, 15, 21, 20]
return table[weekday]

Cheers,
Laguna

[Carsten Haese]

On Fri, 2005-09-02 at 16:46, Laguna wrote:

Paul,

Thanks for the suggestion on calendar module. Here is my solution and
it works:

def expiration(year , month):
weekday = calendar.weekda y(year, month, 1)
table = [19, 18, 17, 16, 15, 21, 20]
return table[weekday]

Cheers,
Laguna

This, of course, can be "optimized" into

def expiration(year , month):
return [19,18,17,16,15, 21,20][calendar.weekda y(year,month,1)]

;)

-Carsten

[Laguna]

Thanks for the "hint" :) I may use your solution if this becomes my
bottleneck!

I try to get away from Perl-ish syntax though.

Best,
L

[Donn Cave]

In article <11************ **********@g49g 2000cwa.googleg roups.com>,
"Laguna" <ed*****@yahoo. com> wrote:

What do you mean by, "the 9 element tuple need to be populated
correctly"? Do you need someone to tell you what values it
needs? What happens if you use (2005, 9, 1, 0, 0, 0, 0, 0, 0),
for example? If you make this tuple with localtime or gmtime,
do you know what the 7th (tm[6]) element of the tuple is?
What tricks did you try, exactly?

Donn Cave, do**@u.washingt on.edu

Thanks for pointing out. tm[6] = weekday, and tm[7] = Julian data, but
I wouldn't know these values when my input values are month and year.

I will try out the more constructive suggestions from Paul and Robert.

Following is what I have tried. As you can tell, the results are wrong!

import time
time.asctime((2 003, 9, 1, 0, 0, 0, 0, 0, 0)) 'Mon Sep 01 00:00:00 2003' time.asctime((2 003, 8, 1, 0, 0, 0, 0, 0, 0)) 'Mon Aug 01 00:00:00 2003' time.asctime((2 003, 7, 1, 0, 0, 0, 0, 0, 0))

'Mon Jul 01 00:00:00 2003'

Well, sure, that tm value will certainly not be the
3rd Friday, but it does correctly represent the first
day of the month. With localtime() you can find out
the day of the week, on the first day of the month.
When you know that, the 3rd Friday is simple arithmetic.

Since other followups have already spoon-fed you a
solution (assuming it works, haven't tried), here's an
example of what I mean -

import time
for m in range(1, 13):
c1 = time.mktime((20 05, m, 1, 0, 0, 0, 0, 0, 0))
d1 = time.localtime( c1)[6]
if d1 > 4:
f3 = 26 - d1
else:
f3 = 19 - d1
# f3 = 19 + (d1 // 5) * 7 - d1
c3 = time.mktime((20 05, m, f3, 0, 0, 0, 0, 0, 0))
print time.ctime(c3)

I don't know if you intend to go on to do much more
programming after this, but that's who I normally
assume we're talking to here, programmers. No one
knows everything and misses nothing, certainly not
me, but it's nice when people come to comp.lang.pytho n
and can account for at least the beginning of some
analysis of their problem. When that's missing, it's
hard to know what's really constructive.

Donn Cave, do**@u.washingt on.edu

[Laguna]

Hey Donn,

I don't mean to offend anyone here. I was just saying that the other
solution is better suited for my problem. I truly appreciate your
analysis and suggestions.

BTW, I am not a programmer :( and I like the simplest solution whenever
possible.

Cheers,
L