I was just writing some code which did date/time manipulations, and I
found that the Python 2.3 datetime module does not supply a number of
fairly basic functions. I understand the reasoning (the datetime
module provides a set of datatypes, and doesn't attempt to get into
the murky waters of date algorithms) but as these things can be quite
tricky to get right, I was wondering if anyone has already implemented
any date algorithms, or alternatively if I'd missed some easy way of
doing what I'm after.
My specific requirements were:
1. Get the last day of the month contatining a given date(time). I
really was surprised to find this one missing, as it seems to me that
the datetime module must know what the last day for each month is, so
exposing it wouldn't have been hard.
2. Add a number of months to a date. This is messy, as there are
options (what is one month after 31st Jan). The trouble is that the
calculation, while simple, is tricky to get right (month is 1-based,
so off-by-1 errors are easy to make, and divmod isn't as useful as
you'd first think).
Other "nice to have" functions which I didn't need for this program,
but which I have found useful in the past, are
- Round a date(time) to a {year,month,day,quarter,week,...} Or
truncate. Or (occasionally) chop to the next higher boundary
(ceiling).
- Calculate the number of {years,months,days,...} between two dates.
(Which is more or less the equivalent of rounding a timedelta).
These latter two aren't very well defined right now, because I have no
immediate use case.
In the absence of anything else, I'll probably write a "date
utilities" module for myself at some point. But I'd hate to be
reinventing the wheel.
Paul.
14  6387 
"Paul Moore" <pa********@atosorigin.com> wrote in message
news:18**************************@posting.google.c om... I was just writing some code which did date/time manipulations, and I
found that the Python 2.3 datetime module does not supply a number of
fairly basic functions. I understand the reasoning (the datetime
module provides a set of datatypes, and doesn't attempt to get into
the murky waters of date algorithms) but as these things can be quite
tricky to get right, I was wondering if anyone has already implemented
any date algorithms, or alternatively if I'd missed some easy way of
doing what I'm after.
My specific requirements were:
1. Get the last day of the month contatining a given date(time). I
really was surprised to find this one missing, as it seems to me that
the datetime module must know what the last day for each month is, so
exposing it wouldn't have been hard.
I'm kind of surprised to see that it's missing, too.
2. Add a number of months to a date. This is messy, as there are
options (what is one month after 31st Jan). The trouble is that the
calculation, while simple, is tricky to get right (month is 1-based,
so off-by-1 errors are easy to make, and divmod isn't as useful as
you'd first think).
That's application dependent. If a bond, for example, has interest
payable on the 30th of the month, you probably want the 30th,
except in February you want the last day of the month. However,
the contract may specify something else. And in no case do you
want the date to suddenly change to the 28th because you went
through February.
Other "nice to have" functions which I didn't need for this program,
but which I have found useful in the past, are
- Round a date(time) to a {year,month,day,quarter,week,...} Or
truncate. Or (occasionally) chop to the next higher boundary
(ceiling).
- Calculate the number of {years,months,days,...} between two dates.
(Which is more or less the equivalent of rounding a timedelta).
These latter two aren't very well defined right now, because I have no
immediate use case.
Most of these don't have well defined, globally useful use cases.
It's heavily application dependent what you want out of these.
In the absence of anything else, I'll probably write a "date
utilities" module for myself at some point. But I'd hate to be
reinventing the wheel.
There's a very well regarded date module out there in the
Vaults of Parnassus. The name escapes me at the moment,
but a bit of spelunking through the Vaults will turn up several
date routines that may do what you want.
John Roth
Paul.
"John Roth" <ne********@jhrothjr.com> writes: "Paul Moore" <pa********@atosorigin.com> wrote in message
news:18**************************@posting.google.c om...
1. Get the last day of the month contatining a given date(time). I
really was surprised to find this one missing, as it seems to me that
the datetime module must know what the last day for each month is, so
exposing it wouldn't have been hard.
I'm kind of surprised to see that it's missing, too.
The best solution I could find was
def month_end(dt):
# Get the next month
y, m = dt.year, dt.month
if m == 12:
y += 1
m = 1
else:
m += 1
# Use replace to cater for both datetime and date types. This
# leaves the time component of a datetime unchanged - it's
# arguable whether this is the right thing.
return dt.replace(year=y, month=m, day=1) - datetime.timedelta(days=1)
It's not hard - but it's mildly tricky (I made a few false starts and
some silly off-by-one errors) and I'd much rather grab it from a
library than make the same mistakes next time I need it.
2. Add a number of months to a date. This is messy, as there are
options (what is one month after 31st Jan). The trouble is that the
calculation, while simple, is tricky to get right (month is 1-based,
so off-by-1 errors are easy to make, and divmod isn't as useful as
you'd first think).
That's application dependent.
True. But for "naive" use, a simple definition does. This is in line
with the datetime module's philosophy of not trying to cater for
"advanced" uses, but to provide something useful for straightforward
use. In this particular case, I'd argue that the obvious definition
(same day number N months on) where applicable, plus a well-documented
"reasonable" answer for the edge cases (eg, Jan 31 plus 1 month) is
useful. In practice, I suspect that 99% of cases involve adding a
number of months to either the first or the last of a month.
If a bond, for example, has interest payable on the 30th of the
month, you probably want the 30th, except in February you want the
last day of the month. However, the contract may specify something
else. And in no case do you want the date to suddenly change to the
28th because you went through February.
But that's not so much a case of adding a month, as a more complex
concept, a "repeating date". Nevertheless, I take your point. In the absence of anything else, I'll probably write a "date
utilities" module for myself at some point. But I'd hate to be
reinventing the wheel.
There's a very well regarded date module out there in the
Vaults of Parnassus. The name escapes me at the moment,
but a bit of spelunking through the Vaults will turn up several
date routines that may do what you want.
I guess you're thinking of mxDateTime. I'm aware of this, and agree
that it's pretty comprehensive. I don't really know why I prefer not
to use it - partly it's just a case of reducing dependencies, also
there are already so many date types in my code (COM, cx_Oracle,
datetime) that I am reluctant to add another - there is already far
too much code devoted to converting representations...
Thanks for the comments,
Paul.
--
This signature intentionally left blank
"Paul Moore" <pa********@atosorigin.com> wrote in message
news:br**********@yahoo.co.uk... "John Roth" <ne********@jhrothjr.com> writes:
"Paul Moore" <pa********@atosorigin.com> wrote in message
news:18**************************@posting.google.c om... 1. Get the last day of the month contatining a given date(time). I
really was surprised to find this one missing, as it seems to me that
the datetime module must know what the last day for each month is, so
exposing it wouldn't have been hard.
I'm kind of surprised to see that it's missing, too.
The best solution I could find was
def month_end(dt):
# Get the next month
y, m = dt.year, dt.month
if m == 12:
y += 1
m = 1
else:
m += 1
This looks incomplete. You could use this to get the sequential date
for the first of the next month, but you still have to subtract one day
and then print out the day.
# Use replace to cater for both datetime and date types. This
# leaves the time component of a datetime unchanged - it's
# arguable whether this is the right thing.
return dt.replace(year=y, month=m, day=1) - datetime.timedelta(days=1)
It's not hard - but it's mildly tricky (I made a few false starts and
some silly off-by-one errors) and I'd much rather grab it from a
library than make the same mistakes next time I need it.
As I said, I don't see any obvious reason why they wouldn't accept
a patch, if you want to submit it.
2. Add a number of months to a date. This is messy, as there are
options (what is one month after 31st Jan). The trouble is that the
calculation, while simple, is tricky to get right (month is 1-based,
so off-by-1 errors are easy to make, and divmod isn't as useful as
you'd first think).
That's application dependent.
True. But for "naive" use, a simple definition does. This is in line
with the datetime module's philosophy of not trying to cater for
"advanced" uses, but to provide something useful for straightforward
use. In this particular case, I'd argue that the obvious definition
(same day number N months on) where applicable, plus a well-documented
"reasonable" answer for the edge cases (eg, Jan 31 plus 1 month) is
useful. In practice, I suspect that 99% of cases involve adding a
number of months to either the first or the last of a month.
Except for the edge case I mention below, this is really too simple;
it's just add one to the month and keep the same date. Hardly worth
a method at all, especially if you have the "last day of month" method.
If a bond, for example, has interest payable on the 30th of the
month, you probably want the 30th, except in February you want the
last day of the month. However, the contract may specify something
else. And in no case do you want the date to suddenly change to the
28th because you went through February.
But that's not so much a case of adding a month, as a more complex
concept, a "repeating date". Nevertheless, I take your point.
Yes. When you've got something application dependent, they tend
not to put in "naive" definitions. What's a 'naive' definition for one
person
is simply wrong for another.
John Roth
Thanks for the comments,
Paul.
--
On 15 Sep 2003 08:07:07 -0700, rumours say that pa********@atosorigin.com (Paul Moore) might have written:
[about missing datetime functionality] 1. Get the last day of the month contatining a given date(time). I
really was surprised to find this one missing, as it seems to me that
the datetime module must know what the last day for each month is, so
exposing it wouldn't have been hard.
I would guess that the simple ones were not included just because they
are only a few lines each:
def end_of_month(a_date):
year_inc, month_inc = divmod(a_date.month, 12)
return a_date.__class__(a_date.year+year_inc, 1+month_inc, 1) - \
datetime.timedelta(days=1)
def month_days(a_date):
return end_of_month(a_date).day
Examples: end_of_month(datetime.date.today())
datetime.date(2003, 9, 30) end_of_month(datetime.datetime(1972,2,11))
datetime.datetime(1972, 2, 29, 0, 0) month_days(datetime.date.today())
30 end_of_month(datetime.date(1977,12,15))
datetime.date(1977, 12, 31)
The add_months_to_date seems not that hard too, but I don't have a need
for one so far, so I haven't coded it :) My needs for datetime
intervals are usually of the 'one month start - another month end' kind,
so end_of_month is handy. If you do think these would be useful, feel
free to use them (or your own versions) for a patch.
Date / timedelta rounding / truncating to various units would have to
take account of many different rules, and none would be very generic
IMHO.
--
TZOTZIOY, I speak England very best,
Microsoft Security Alert: the Matrix began as open source.
On Mon, 15 Sep 2003 20:44:08 +0100, rumours say that Paul Moore
<pa********@atosorigin.com> might have written:
[find the end of the month] The best solution I could find was
def month_end(dt):
# Get the next month
y, m = dt.year, dt.month
if m == 12:
y += 1
m = 1
else:
m += 1
# Use replace to cater for both datetime and date types. This
# leaves the time component of a datetime unchanged - it's
# arguable whether this is the right thing.
return dt.replace(year=y, month=m, day=1) - datetime.timedelta(days=1)
I sent my own version without having seen your own --and mine might seem
obfuscated, compared to yours. Only a minor suggestion: don't use
dt.replace, use dt.__class__ instead, since you wouldn't want your
function to have side-effects (that is, don't destroy the actual object
that dt is bound to.)
--
TZOTZIOY, I speak England very best,
Microsoft Security Alert: the Matrix began as open source.
> I was just writing some code which did date/time manipulations, and I found that the Python 2.3 datetime module does not supply a number of
[...]
Your message arrived in the exact time.. :-)
Have a look at this:
http://article.gmane.org/gmane.comp.python.devel/52956
I'll publish it somewhere an announce here once I get the
time to do so.
My specific requirements were:
1. Get the last day of the month contatining a given date(time). I
really was surprised to find this one missing, as it seems to me that
the datetime module must know what the last day for each month is, so
exposing it wouldn't have been hard.
I don't understand this one. The last day of the month containing
a given datetime? A datetime is absolute, how would it be "contained"
in something else?
--
Gustavo Niemeyer http://niemeyer.net
On Tue, 16 Sep 2003 11:28:58 -0400, rumours say that "Tim Peters"
<ti*****@comcast.net> might have written: [Christos TZOTZIOY Georgiou] I sent my own version without having seen your own --and mine might
seem obfuscated, compared to yours. Only a minor suggestion: don't
use dt.replace, use dt.__class__ instead, since you wouldn't want your
function to have side-effects (that is, don't destroy the actual
object that dt is bound to.)
It's OK to use replace: datetime and date (also time and timedelta) objects
are immutable -- their values can't be changed after initialization. In
particular, x.replace() doesn't mutate x.
In some strange, unprecedented and inexplicable way, you are right...
<sigh> ...again :) RTFM-F virus not removed yet.
PS That bang thingie (!) in Ruby is pythonic.
--
TZOTZIOY, I speak England very best,
Microsoft Security Alert: the Matrix began as open source.
On Tue, Sep 16, 2003 at 05:19:50PM +0300, Christos TZOTZIOY Georgiou wrote:
.... I sent my own version without having seen your own --and mine might seem
obfuscated, compared to yours.
you both use the same indirect way of getting to the end of the month.
why not do what datetime.c does, like:
_days_in_month = [
0, # unused; this vector uses 1-based indexing */
31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31
]
def _is_leap_year(year):
return year % 4 == 0 and (year % 100 !=0 or year % 400 == 0)
def days_in_month(year, month):
if month == 2 and _is_leap_year(year):
return 29
else:
return _days_in_month[month]
def end_of_month(d):
date(d.y, d.m, days_in_month(d.y, d.m))
--
groetjes, carel
Christos "TZOTZIOY" Georgiou <tz**@sil-tec.gr> writes: I sent my own version without having seen your own --and mine might seem
obfuscated, compared to yours. Only a minor suggestion: don't use
dt.replace, use dt.__class__ instead, since you wouldn't want your
function to have side-effects (that is, don't destroy the actual object
that dt is bound to.)
dt.replace doesn't have side effects - it creates a new object. Also,
by using replace(), my version works with date, datetime, or
subclasses. Yours does too, but it sets the time component of a
datetime to zero. It's a matter of preference which behaviour is
"better", I guess.
Paul.
--
This signature intentionally left blank
Carel Fellinger <ca*************@chello.nl> writes: On Tue, Sep 16, 2003 at 05:19:50PM +0300, Christos TZOTZIOY Georgiou wrote:
... I sent my own version without having seen your own --and mine might seem
obfuscated, compared to yours.
you both use the same indirect way of getting to the end of the month.
why not do what datetime.c does, like:
Mainly because that seems even more like reinventing the wheel.
Writing code that already exists is something I dislike doing at the
best of times. Writing *tricky* code that already exists feels even
more unpleasant. (Yes, I know the leap year calculation isn't that
hard - but lots of people have got it wrong in the past...)
I guess that's my real issue. I know the datetime module already has
this information. But persuading it to tell me requires me to jump
through hoops, whereas reimplementing it feels like admitting defeat.
None of it's hard, though. I've spent more time on emails about the
issue than I'd ever need to spend in implementing anything :-)
Paul.
--
This signature intentionally left blank
On Tue, 16 Sep 2003 20:26:01 +0100, rumours say that Paul Moore
<pf******@yahoo.co.uk> might have written:
[I suggesting dt.__class__() instead of dt.replace()] dt.replace doesn't have side effects - it creates a new object. Also,
by using replace(), my version works with date, datetime, or
subclasses. Yours does too, but it sets the time component of a
datetime to zero. It's a matter of preference which behaviour is
"better", I guess.
You are right about dt.replace not having side-effects, just like some
anonymous<wink> major python contributor commented. Also .replace feels
"better" since it doesn't lose the time information.
PS ...but I still like my use of divmod() ;-)
--
TZOTZIOY, I speak England very best,
Microsoft Security Alert: the Matrix began as open source.
> > you both use the same indirect way of getting to the end of the month. why not do what datetime.c does, like:
Mainly because that seems even more like reinventing the wheel.
Writing code that already exists is something I dislike doing at the
[...]
Why not using calendar.monthrange()?
--
Gustavo Niemeyer http://niemeyer.net pa********@atosorigin.com (Paul Moore) wrote in message news:<18**************************@posting.google. com>... I was just writing some code which did date/time manipulations, and I
found that the Python 2.3 datetime module does not supply a number of
fairly basic functions. I understand the reasoning (the datetime
module provides a set of datatypes, and doesn't attempt to get into
the murky waters of date algorithms) but as these things can be quite
tricky to get right, I was wondering if anyone has already implemented
any date algorithms, or alternatively if I'd missed some easy way of
doing what I'm after.
[description of needed functions]
I wrote a module just like that at my last job. Given a date object,
you could find the first or last day of the month, quarter, or year.
There were also functions to answer questions like "What was the date
5 months ago?" or "What date is the first Monday after October 8?"
Unfortunately, I don't have a copy of the source code here.
Try 'mxDateTime' available at http://www.egenix.com/files/python/e...ownload-mxBASE
It most likely has what you are looking for.
Paul Moore wrote: I was just writing some code which did date/time manipulations, and I
found that the Python 2.3 datetime module does not supply a number of
fairly basic functions. I understand the reasoning (the datetime
module provides a set of datatypes, and doesn't attempt to get into
the murky waters of date algorithms) but as these things can be quite
tricky to get right, I was wondering if anyone has already implemented
any date algorithms, or alternatively if I'd missed some easy way of
doing what I'm after.
My specific requirements were:
1. Get the last day of the month contatining a given date(time). I
really was surprised to find this one missing, as it seems to me that
the datetime module must know what the last day for each month is, so
exposing it wouldn't have been hard.
2. Add a number of months to a date. This is messy, as there are
options (what is one month after 31st Jan). The trouble is that the
calculation, while simple, is tricky to get right (month is 1-based,
so off-by-1 errors are easy to make, and divmod isn't as useful as
you'd first think).
Other "nice to have" functions which I didn't need for this program,
but which I have found useful in the past, are
- Round a date(time) to a {year,month,day,quarter,week,...} Or
truncate. Or (occasionally) chop to the next higher boundary
(ceiling).
- Calculate the number of {years,months,days,...} between two dates.
(Which is more or less the equivalent of rounding a timedelta).
These latter two aren't very well defined right now, because I have no
immediate use case.
In the absence of anything else, I'll probably write a "date
utilities" module for myself at some point. But I'd hate to be
reinventing the wheel.
Paul. This thread has been closed and replies have been disabled. Please start a new discussion. Similar topics
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