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

[Peter Hansen]

Carsten Haese wrote:

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

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

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)]

;)

True, but do you find that more readable? If I saw that in code I was
maintaining I would likely rewrite it, probably to look a lot like the
first one (though likely with a more descriptive name than "table"...
maybe expirationTable ?).

(And, if I were "optimizing ", I would of course dispense with the
dynamic creation of the static table upon every execution of
expiration(), and move it outside the function.)

-Peter

[Paul Rubin]

Peter Hansen <pe***@engcorp. com> writes:

(And, if I were "optimizing ", I would of course dispense with the
dynamic creation of the static table upon every execution of
expiration(), and move it outside the function.)

Replacing it with a tuple might be enough for that.

[John Machin]

Peter Hansen wrote:

Carsten Haese wrote:

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

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

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)]

;)

True, but do you find that more readable? If I saw that in code I was
maintaining I would likely rewrite it, probably to look a lot like the
first one (though likely with a more descriptive name than "table"...
maybe expirationTable ?).

(And, if I were "optimizing ", I would of course dispense with the
dynamic creation of the static table upon every execution of
expiration(), and move it outside the function.)

-Peter

An alternative:

def expiration(year , month):
return 21 - (calendar.weekd ay(year,month,1 ) + 2) % 7

[Paul McGuire]

Donn,

You didn't look closely enough at those results. The OP's point was
that he did not know how to set all the tuple values correctly. Here's
a clearer example, I think:

import time
print time.asctime((2 005,9,1,0,0,0,0 ,0,0))
print time.asctime((2 005,9,1,0,0,0,1 ,0,0))
print time.asctime((2 005,9,1,0,0,0,2 ,0,0))
print time.asctime((2 005,9,1,0,0,0,3 ,0,0))

Prints:
Mon Sep 01 00:00:00 2005
Tue Sep 01 00:00:00 2005
Wed Sep 01 00:00:00 2005
Thu Sep 01 00:00:00 2005

No matter what time zone you're in, Sep 1, 2005 can't be all those days
of the week! :)

Your code works because you use mktime, which appears to ignore the
dayOfWeek element of the input tuple, as shown by the following:

print time.asctime(ti me.gmtime(time. mktime((2005,9, 1,0,0,0,0,0,0)) ))
print time.asctime(ti me.gmtime(time. mktime((2005,9, 1,0,0,0,1,0,0)) ))
print time.asctime(ti me.gmtime(time. mktime((2005,9, 1,0,0,0,2,0,0)) ))
print time.asctime(ti me.gmtime(time. mktime((2005,9, 1,0,0,0,3,0,0)) ))

Prints:
Thu Sep 01 06:00:00 2005
Thu Sep 01 06:00:00 2005
Thu Sep 01 06:00:00 2005
Thu Sep 01 06:00:00 2005

-- Paul

[Peter Hansen]

Paul Rubin wrote:

Peter Hansen <pe***@engcorp. com> writes:

(And, if I were "optimizing ", I would of course dispense with the
dynamic creation of the static table upon every execution of
expiration( ), and move it outside the function.)

Replacing it with a tuple might be enough for that.

You're right, and in fact that would actually be even faster since then
it's a LOAD_CONST instead of a LOAD_GLOBAL.

-Peter

[Jorgen Grahn]

On Fri, 02 Sep 2005 20:53:44 -0400, Peter Hansen <pe***@engcorp. com> wrote:

Carsten Haese wrote:

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

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

.... True, but do you find that more readable? If I saw that in code I was
maintaining I would likely rewrite it, probably to look a lot like the
first one (though likely with a more descriptive name than "table"...
maybe expirationTable ?).

That doesn't explain anything, IMHO. What 'table' really is is a list of
day-of-month candidates for the third Friday.

For some pieces of code, I find that it's better to document what it /does/
than to try to document /how/ it does it. And maybe add a bunch of unit
tests to /demonstrate/ that it seems to work.

The next programmer can then choose to either (a) understand the code or (b)
rip it out and replace it.

I would leave the body alone, but rename the function 'third_friday_o f_month',
and do 'expiration = third_friday_of _month'.

/Jorgen

--
// Jorgen Grahn <jgrahn@ Ph'nglui mglw'nafh Cthulhu
\X/ algonet.se> R'lyeh wgah'nagl fhtagn!

[tryit]

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?

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

[tryit]

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?

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

import calendar
[y[4] for y in calendar.monthc alendar(2005, 9) if y[4]!=0][2] 16 [y[4] for y in calendar.monthc alendar(2003, 6) if y[4]!=0][2] 20 [y[4] for y in calendar.monthc alendar(2006, 2) if y[4]!=0][2]

17

[tryit]

>>> import calendar

[y[4] for y in calendar.monthc alendar(2005, 9) if y[4]!=0][2] 16 [y[4] for y in calendar.monthc alendar(2003, 6) if y[4]!=0][2] 20 [y[4] for y in calendar.monthc alendar(2006, 2) if y[4]!=0][2]

17

[Terry Reedy]

Laguna wrote:

I want to find the expiration date of stock options (3rd Friday of the
month) for an any give month and year.

From year and month (and day=1) get the day of the week (n in [0,6]) of the

first of the month using some version of the the standard formula (see
below) and look up the third friday date in a precalculated 7-element list,
or, with n=0 on Saturday, 3rd Friday is 21-n

Here is a translation of the guts of a 30-year-old Basic program:

def friday3(m,y): # ints
if m <= 2:
m += 12
y -= 1
d = 1
n = d + 2*m + int(.6*(m+1)) + y + y//4 - y//100 + y//400 + 2
n = int((n/7.0- n//7)*7.0 + .5)
# n=0 is Saturday, making 3rd Friday the 21st.
return 21 - n

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

for m,y in ((9,2005), (6,2003), (2,2006)): print friday3(m,y)

....
16
20
17

Terry J. Reedy