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(month, year):
w1 = calendar.weekday(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 RelativeDateTime 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**********************@g47g2000cwa.googlegroups .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.washington.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.washington.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((2003, 9, 1, 0, 0, 0, 0, 0, 0)) 'Mon Sep 01 00:00:00 2003' time.asctime((2003, 8, 1, 0, 0, 0, 0, 0, 0)) 'Mon Aug 01 00:00:00 2003' time.asctime((2003, 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.weekday(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.weekday(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.weekday(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**********************@g49g2000cwa.googlegroups .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.washington.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((2003, 9, 1, 0, 0, 0, 0, 0, 0)) 'Mon Sep 01 00:00:00 2003' time.asctime((2003, 8, 1, 0, 0, 0, 0, 0, 0)) 'Mon Aug 01 00:00:00 2003' time.asctime((2003, 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((2005, 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((2005, 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.python
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.washington.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

[Peter Hansen]

Carsten Haese wrote:

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

def expiration(year, month):
weekday = calendar.weekday(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.weekday(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.weekday(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.weekday(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.weekday(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((2005,9,1,0,0,0,0,0,0))
print time.asctime((2005,9,1,0,0,0,1,0,0))
print time.asctime((2005,9,1,0,0,0,2,0,0))
print time.asctime((2005,9,1,0,0,0,3,0,0))

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(time.gmtime(time.mktime((2005,9,1,0,0 ,0,0,0,0))))
print time.asctime(time.gmtime(time.mktime((2005,9,1,0,0 ,0,1,0,0))))
print time.asctime(time.gmtime(time.mktime((2005,9,1,0,0 ,0,2,0,0))))
print time.asctime(time.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.weekday(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_of_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.monthcalendar(2005, 9) if y[4]!=0][2] 16 [y[4] for y in calendar.monthcalendar(2003, 6) if y[4]!=0][2] 20 [y[4] for y in calendar.monthcalendar(2006, 2) if y[4]!=0][2]

17

[tryit]

>>> import calendar

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

[John Machin]

Terry Reedy wrote:

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

Some simplification is possible:

[2*m + int(.6*(m+1)) for m in range(15)] [0, 3, 5, 8, 11, 13, 16, 18, 21, 24, 26, 29, 31, 34, 37] [(13*m+3)//5 for m in range(15)] [0, 3, 5, 8, 11, 13, 16, 18, 21, 24, 26, 29, 31, 34, 37]

n = int((n/7.0- n//7)*7.0 + .5)
n %= 7
# 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