Re: How to find difference in years between two dates?
thebjorn wrote: > For the purpose of finding someone's age I was looking for a way to > find how the difference in years between two dates, so I could do > something like: > > age = (date.today() - born).year > > but that didn't work (the timedelta class doesn't have a year > accessor). > > I looked in the docs and the cookbook, but I couldn't find anything, so > I came up with: > > def age(born): > now = date.today() > birthday = date(now.year, born.month, born.day) Bad luck if the punter was born on 29 Feb and the current year is not a leap year. > return now.year - born.year - (birthday > now and 1 or 0) Holy code bloat, Batman! Try this: return now.year - born.year - (birthday > now) > > i.e. calculate the "raw" years first and subtract one if he hasn't had > his birthday yet this year... It works, but I'd rather use a standard > and generic approach if it exists...? > It's the irregular-size months that cause the problems. If you can work out the months difference, then just floor_div by 12 to get the years difference. Below is some code from the ancient times when everybody and his dog each had their own date class :-) HTH, John 8<--- methods from a date class def months_until(self, to_date): """Return number of months between from_date (self) and to_date. """ from_date = self signum = 1 if from_date > to_date: from_date, to_date = to_date, from_date signum = -1 d1, m1, y1 = from_date.day, from_date.month, from_date.year d2, m2, y2 = to_date.day, to_date.month, to_date.year mdiff = (y2 - y1) * 12 + m2 - m1 if d2 < d1 and (d2 < 28 or d2 != last_day_of_month(y2, m2)): # the test d2 < 28 is not necessary; it is an optimisation # to avoid calling last_day_of_month unnecessarily mdiff = mdiff - 1 return mdiff * signum def years_until(self, to_date): """Return number of years between from_date (self) and to_date. """ md = self.months_until(to_date) if md >= 0: return md // 12 else: # ensure division truncates towards zero return -((-md) // 12) 8<--- module-level functions and constants # days in month _dim = (None, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31) def last_day_of_month(y, m): """Return day (1..31) which is last day of month m in year y """ if m == 2: return 28 + _leap(y) else: if not (1 <= m <= 12): raise DateError, "month not in 1..12" return _dim[m] def _leap(y): if y % 4: return 0 if y % 100: return 1 if y % 400: return 0 return 1 8< -- http://mail.python.org/mailman/listinfo/python-list
Re: How to find difference in years between two dates?
thebjorn wrote:
> For the purpose of finding someone's age I was looking for a way to
> find how the difference in years between two dates, so I could do
> something like:
>
> age = (date.today() - born).year
>
> but that didn't work (the timedelta class doesn't have a year
> accessor).
>
> I looked in the docs and the cookbook, but I couldn't find anything, so
> I came up with:
>
> def age(born):
> now = date.today()
> birthday = date(now.year, born.month, born.day)
> return now.year - born.year - (birthday > now and 1 or 0)
>
> i.e. calculate the "raw" years first and subtract one if he hasn't had
> his birthday yet this year... It works, but I'd rather use a standard
> and generic approach if it exists...?
You may want to have a look at mxDatetime, which has a RelativeDateTime
type that seems to do what you want:
http://www.egenix.com/files/python/mxDateTime.html
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Re: How to find difference in years between two dates?
"thebjorn" <[EMAIL PROTECTED]> wrote: > def age(born): > now = date.today() > birthday = date(now.year, born.month, born.day) > return now.year - born.year - (birthday > now and 1 or 0) I don't get that last line. There's two things in particular that are puzzling me. 1) What does "birthday > now" mean? It sounds like you haven't been born yet. 2) I find the "and 1 or 0" part very confusing. I can't remember all the minor rules about operator precedence, but I'm sure this works out to some clever hack involving boolean short-circuit evaluation to get around the lack of a ternary operator in python. If I need to pull out the reference manual to decipher what an expression means, it's too complicated. Try something like: if birthday > now: return now.year - born.year - 1 else: return now.year - born.year It takes up a little more space, but it's bog easy to understand without scratching your head or diving into the manual to refresh your memory of obscure language details. -- http://mail.python.org/mailman/listinfo/python-list
Re: How to find difference in years between two dates?
Bruno Desthuilliers wrote: > thebjorn wrote: > > For the purpose of finding someone's age I was looking for a way to > > find how the difference in years between two dates, so I could do > > something like: > > > > age = (date.today() - born).year > > > > but that didn't work (the timedelta class doesn't have a year > > accessor). > > > > I looked in the docs and the cookbook, but I couldn't find anything, so > > I came up with: > > > > def age(born): > > now = date.today() > > birthday = date(now.year, born.month, born.day) > > return now.year - born.year - (birthday > now and 1 or 0) > > > > i.e. calculate the "raw" years first and subtract one if he hasn't had > > his birthday yet this year... It works, but I'd rather use a standard > > and generic approach if it exists...? > > You may want to have a look at mxDatetime, which has a RelativeDateTime > type that seems to do what you want: > http://www.egenix.com/files/python/mxDateTime.html > Which pieces of the following seem to be working to you? >>> import mx.DateTime >>> f = mx.DateTime.RelativeDateTimeDiff >>> d = mx.DateTime.Date >>> f(d(2000, 2, 29), d(2001, 2, 28)) >>> f(d(2000, 2, 29), d(2001, 3, 1)) >>> f(d(2001, 1, 31), d(2001, 2, 28)) >>> g = lambda x, y: f(y, x) >>> g(d(2000, 2, 29), d(2001, 2, 28)) >>> g(d(2000, 2, 29), d(2001, 3, 1)) >>> g(d(2001, 1, 31), d(2001, 2, 28)) >>> and going the other way, adding one month to 31 January gives you some date in March which is ludicrous. -- http://mail.python.org/mailman/listinfo/python-list
Re: How to find difference in years between two dates?
John Machin wrote:
> Bruno Desthuilliers wrote:
>
>>thebjorn wrote:
>>
>>>For the purpose of finding someone's age I was looking for a way to
>>>find how the difference in years between two dates, so I could do
>>>something like:
>>>
>>> age = (date.today() - born).year
>>>
>>>but that didn't work (the timedelta class doesn't have a year
>>>accessor).
>>>
>>>I looked in the docs and the cookbook, but I couldn't find anything, so
>>>I came up with:
>>>
>>> def age(born):
>>> now = date.today()
>>> birthday = date(now.year, born.month, born.day)
>>> return now.year - born.year - (birthday > now and 1 or 0)
>>>
>>>i.e. calculate the "raw" years first and subtract one if he hasn't had
>>>his birthday yet this year... It works, but I'd rather use a standard
>>>and generic approach if it exists...?
>>
>>You may want to have a look at mxDatetime, which has a RelativeDateTime
>>type that seems to do what you want:
>>http://www.egenix.com/files/python/mxDateTime.html
>>
>
>
> Which pieces of the following seem to be working to you?
John, it seems you failed to notice the use of "may" and "seems" in my
post. IIRC, both are supposed to strongly suggest a lack of certitude.
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Re: How to find difference in years between two dates?
Bruno Desthuilliers wrote: > John Machin wrote: > > Bruno Desthuilliers wrote: > > > > > Which pieces of the following seem to be working to you? > > John, it seems you failed to notice the use of "may" and "seems" in my > post. IIRC, both are supposed to strongly suggest a lack of certitude. > > I didn't fail to notice that you were seeming. Re-read my question: It's asking you which bits you were seeming. -- http://mail.python.org/mailman/listinfo/python-list
Re: How to find difference in years between two dates?
Roy Smith: > 2) I find the "and 1 or 0" part very confusing. I can't remember all the > minor rules about operator precedence, but I'm sure this works out to some > clever hack involving boolean short-circuit evaluation to get around the > lack of a ternary operator in python. If I need to pull out the reference > manual to decipher what an expression means, it's too complicated. Try > something like: >From the manual, 5.10: >(Note that neither and nor or restrict the value and type they return to False >and True, but rather return the last evaluated argument. This is sometimes >useful, e.g., if s is a string that should be replaced by a default value if >it is empty, the expression s or 'foo' yields the desired value. Because not >has to invent a value anyway, it does not bother to return a value of the same >type as its argument, so e.g., not 'foo' yields False, not ''.)< Then are such things something good to remove from Python 3.0 (making or and and always return True or False), to simplify the language, and make it more clear and reduce the possibility of bugs? Bye, bearophile -- http://mail.python.org/mailman/listinfo/python-list
Re: How to find difference in years between two dates?
John Machin wrote:
> Bruno Desthuilliers wrote:
>
>>John Machin wrote:
>>
>>>Bruno Desthuilliers wrote:
>>
>>>Which pieces of the following seem to be working to you?
>>
>>John, it seems you failed to notice the use of "may" and "seems" in my
>>post. IIRC, both are supposed to strongly suggest a lack of certitude.
>>
>
> I didn't fail to notice that you were seeming. Re-read my question:
> It's asking you which bits you were seeming.
OP problem:
age = (date.today() - born).year
Possible solution:
import mx.DateTime as dt
def age(date):
return dt.Age(dt.today(), date).years
born = dt.Date(1967, 5, 1)
assert age(born) == 39
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Re: How to find difference in years between two dates?
Roy Smith wrote: > "thebjorn" <[EMAIL PROTECTED]> wrote: > > > def age(born): > > now = date.today() > > birthday = date(now.year, born.month, born.day) > > return now.year - born.year - (birthday > now and 1 or 0) > > I don't get that last line. There's two things in particular that are > puzzling me. > > 1) What does "birthday > now" mean? It sounds like you haven't been born > yet. I'm making a (perhaps tenous) semantic distinction between birthdate, the date you were born on, and birthday, an annual event that may or may not have happened yet this year :-) > 2) I find the "and 1 or 0" part very confusing. I can't remember all the > minor rules about operator precedence, but I'm sure this works out to some > clever hack involving boolean short-circuit evaluation to get around the > lack of a ternary operator in python. You're correct :-) the line was originally: return now.year - born.year - (1 if birthday > now else 0) which gave a nice traceback on the production server that didn't have 2.5 on it :-( The and/or short-circuit is a fairly well established (yet divisive) pydiom, and I was going to say something about people really ought to learn a few simple precedence rules, but then I realized the parenthesis aren't needed in the above The parenthesis are needed in a version someone else mentioned: return now.year - born.year - (birthday > now) but I wouldn't write that, just like I wouldn't write 1 + True.. > If I need to pull out the reference manual to decipher what an expression > means, > it's too complicated. Nah, that's a little too restrictive I think. I will agree that the and/or is more cute than functional, at least in this case. Since it could also throw, how about: def yeardiff(a, b): y = a.year - b.year if (a.month, a.day) < (b.month, b.day): # tuple comparison y -= 1 return y def age(born): return yeardiff(date.today(), born) > if birthday > now: >return now.year - born.year - 1 > else: >return now.year - born.year I prefer to hoist the common expression out of the branches so they don't have an opportunity to get out of sync, but I get your point. -- bjorn -- http://mail.python.org/mailman/listinfo/python-list
Re: How to find difference in years between two dates?
John Machin wrote: > thebjorn wrote: [...] > > > > def age(born): > > now = date.today() > > birthday = date(now.year, born.month, born.day) > > Bad luck if the punter was born on 29 Feb and the current year is not a > leap year. Good catch! Thanks! [..] > Holy code bloat, Batman! Try this: > > return now.year - born.year - (birthday > now) yuck :-) [...] > It's the irregular-size months that cause the problems. If you can work > out the months difference, then just floor_div by 12 to get the years > difference. I don't agree that the irregular sized months cause a problem in this case. They do cause a problem if you're asking "when is today + one month?", i.e. there isn't an unambiguous answer to that question in general (e.g. if today was January 31). We're asking a different kind of question though: "has it been at least one month since January 31?", the answer would be no on Feb 29 and yes on Mar 1. > Below is some code from the ancient times when everybody and his dog > each had their own date class :-) [...] Wow. I'm speechless. (any reason you didn't want to use the calendar module?) -- bjorn -- http://mail.python.org/mailman/listinfo/python-list
Re: How to find difference in years between two dates?
Bruno Desthuilliers wrote: [...] > Possible solution: > > import mx.DateTime as dt > def age(date): > return dt.Age(dt.today(), date).years > born = dt.Date(1967, 5, 1) > assert age(born) == 39 dealbreaker: >>> age(datetime.date(1970,5,2)) Traceback (most recent call last): File "", line 1, in ? File "c:\python24\lib\site-packages\mx\DateTime\DateTime.py", line 842, in RelativeDateTimeDiff diff = date1 - date2 TypeError: unsupported operand type(s) for -: 'DateTime' and 'datetime.date' I'm getting data from a database, and conversions are out of the question for something like this. Otherwise it's a fine library :-) -- bjorn -- http://mail.python.org/mailman/listinfo/python-list
Re: How to find difference in years between two dates?
On 2006-07-26 17:50:43, thebjorn wrote: > I don't agree that the irregular sized months cause a problem in this > case. They do cause a problem if you're asking "when is today + one > month?", i.e. there isn't an unambiguous answer to that question in > general (e.g. if today was January 31). We're asking a different kind > of question though: "has it been at least one month since January 31?", > the answer would be no on Feb 29 and yes on Mar 1. It's still ambiguous. That's why pretty much no library offers months as units of date differences. They offer days (like DateTimeDelta in mxDateTime does). For example, the mxDateTime guys say about their RelativeDateTime (which allows months as differences): """ Note that dates like Date(1999,1,30) + RelativeDateTime(months=+1) are not supported. The package currently interprets these constructions as Date(1999,2,1) + 30, thus giving the 1999-03-02 which may not be what you'd expect. """ I think your original approach (difference of years, adjusted for whether the date of the birthday has already passed in the current year or not) is the unambiguous way to get the age. Other methods based on date differences in days need careful adjustments for odd cases involving leap years. Gerhard -- http://mail.python.org/mailman/listinfo/python-list
Re: How to find difference in years between two dates?
thebjorn wrote: > John Machin wrote: > > thebjorn wrote: > [...] > > > > > > def age(born): > > > now = date.today() > > > birthday = date(now.year, born.month, born.day) > > > > Bad luck if the punter was born on 29 Feb and the current year is not a > > leap year. > > Good catch! Thanks! Easy catch -- happens at least one per inning[s] :-) > > [..] > > Holy code bloat, Batman! Try this: > > > > return now.year - born.year - (birthday > now) > > yuck :-) But this: return now.year - born.year - (birthday > now and 1 or 0) is not yuck??? > > [...] > > It's the irregular-size months that cause the problems. If you can work > > out the months difference, then just floor_div by 12 to get the years > > difference. > > I don't agree that the irregular sized months cause a problem in this > case. They do cause a problem if you're asking "when is today + one > month?", i.e. there isn't an unambiguous answer to that question in > general (e.g. if today was January 31). We're asking a different kind > of question though: "has it been at least one month since January 31?", > the answer would be no on Feb 29 and yes on Mar 1. If a bank were paying you interest on a monthly basis, and you deposited money on Jan 31 and pulled it out on the last day of February, that would count as one month. This is what I call the "today - yesterday == 1" rule. For computing things like duration of employee service, you need the "today - yesterday == 2" rule -- on the assumption that service counts from start of business yesterday to close of business today. So hire date of 1 Feb to fire date of (last day of Feb) would count as one month. > > > Below is some code from the ancient times when everybody and his dog > > each had their own date class :-) > [...] > > Wow. I'm speechless. (any reason you didn't want to use the calendar > module?) Sorry, I don't understand. Why are you speechless? What would I want to use the calendar module for? Apart from the leap() function and the table of days in a month, the calendar module doesn't have any of the functionality that one would expect in a general-purpose date class. -- http://mail.python.org/mailman/listinfo/python-list
Re: How to find difference in years between two dates?
thebjorn wrote:
> Bruno Desthuilliers wrote:
> [...]
>
>>Possible solution:
>>
>>import mx.DateTime as dt
>>def age(date):
>>return dt.Age(dt.today(), date).years
>>born = dt.Date(1967, 5, 1)
>>assert age(born) == 39
>
>
> dealbreaker:
>
age(datetime.date(1970,5,2))
>
(snip traceback)
What about:
age(dt.Date(1970,5,2))
> I'm getting data from a database, and conversions
> are out of the
> question for something like this.
Which conversion ? How do you get the data ? as a datetime object ? as a
(y,m,d) tuple ? as a "y-m-d" string ? Else ?
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Re: How to find difference in years between two dates?
John Machin wrote: > thebjorn wrote: > > John Machin wrote: > > > thebjorn wrote: [...] > > > Holy code bloat, Batman! Try this: > > > > > > return now.year - born.year - (birthday > now) > > > > yuck :-) > > But this: > return now.year - born.year - (birthday > now and 1 or 0) is not yuck??? Correct. > > [...] > > > It's the irregular-size months that cause the problems. If you can work > > > out the months difference, then just floor_div by 12 to get the years > > > difference. > > > > I don't agree that the irregular sized months cause a problem in this > > case. They do cause a problem if you're asking "when is today + one > > month?", i.e. there isn't an unambiguous answer to that question in > > general (e.g. if today was January 31). We're asking a different kind > > of question though: "has it been at least one month since January 31?", > > the answer would be no on Feb 29 and yes on Mar 1. > > If a bank were paying you interest on a monthly basis, and you > deposited money on Jan 31 and pulled it out on the last day of > February, that would count as one month. This is what I call the "today > - yesterday == 1" rule. For computing things like duration of employee > service, you need the "today - yesterday == 2" rule -- on the > assumption that service counts from start of business yesterday to > close of business today. So hire date of 1 Feb to fire date of (last > day of Feb) would count as one month. You give a good argument that the concept of a month is fuzzy, I still don't believe that it makes the concept of a year fuzzy (or that the fuzziness of month needs to be accounted for when computing years). > Sorry, I don't understand. Why are you speechless? What would I want to > use the calendar module for? Apart from the leap() function and the > table of days in a month, the calendar module doesn't have any of the > functionality that one would expect in a general-purpose date class. Well, I thought replacing a 4 line function with 31 lines, 13 of which duplicated functionality in the standard library was overkill... I came up with this yesterday which seems sufficient? def yeardiff(a, b): y = a.year - b.year if (a.month, a.day) < (b.month, b.day): # tuple comparison y -= 1 return y -- bjorn -- http://mail.python.org/mailman/listinfo/python-list
Re: How to find difference in years between two dates?
thebjorn wrote: > John Machin wrote: > > thebjorn wrote: > > > John Machin wrote: > > > > thebjorn wrote: > [...] > > > > Holy code bloat, Batman! Try this: > > > > > > > > return now.year - born.year - (birthday > now) > > > > > > yuck :-) > > > > But this: > > return now.year - born.year - (birthday > now and 1 or 0) is not yuck??? > > Correct. > > > > [...] > > > > It's the irregular-size months that cause the problems. If you can work > > > > out the months difference, then just floor_div by 12 to get the years > > > > difference. > > > > > > I don't agree that the irregular sized months cause a problem in this > > > case. They do cause a problem if you're asking "when is today + one > > > month?", i.e. there isn't an unambiguous answer to that question in > > > general (e.g. if today was January 31). We're asking a different kind > > > of question though: "has it been at least one month since January 31?", > > > the answer would be no on Feb 29 and yes on Mar 1. > > > > If a bank were paying you interest on a monthly basis, and you > > deposited money on Jan 31 and pulled it out on the last day of > > February, that would count as one month. This is what I call the "today > > - yesterday == 1" rule. For computing things like duration of employee > > service, you need the "today - yesterday == 2" rule -- on the > > assumption that service counts from start of business yesterday to > > close of business today. So hire date of 1 Feb to fire date of (last > > day of Feb) would count as one month. > > You give a good argument that the concept of a month is fuzzy Sorry, I can't imagine where you got "fuzzy" from. Perhaps you mean some other word. The concept is capable of being expressed precisely. > I still > don't believe that it makes the concept of a year fuzzy (or that the > fuzziness of month needs to be accounted for when computing years). The point is to ensure that (say) 24 months of employment and 2 years of employment are determined in a consistent fashion. > > > Sorry, I don't understand. Why are you speechless? What would I want to > > use the calendar module for? Apart from the leap() function and the > > table of days in a month, the calendar module doesn't have any of the > > functionality that one would expect in a general-purpose date class. > > Well, I thought replacing a 4 line function with 31 lines, 13 of which > duplicated functionality in the standard library was overkill. I think you missed the point that the lines I quoted were straight out of a self-contained library that existed (in C as well as Python) way before the datetime module was a gleam in Fred & the timbot's eyes. Even if I had noticed a leap year function in the calendar module, I would probably not have used it. The Python version of the module was just a stopgap while I fiddled with getting a C extension going. The C leap year function doesn't have any of that modulo stuff in it. > I came > up with this yesterday which seems sufficient? > > def yeardiff(a, b): > y = a.year - b.year > if (a.month, a.day) < (b.month, b.day): # tuple comparison > y -= 1 > return y At least it doesn't blow up when b is leapyear-02-29. It just gives the wrong answer when a is nonleapyear-02-28. E.g. it gives 0 years difference from 1992-02-29 to 1993-02-28 instead of 1. -- http://mail.python.org/mailman/listinfo/python-list
Re: How to find difference in years between two dates?
John Machin wrote: > thebjorn wrote: [...] > > You give a good argument that the concept of a month is fuzzy > > Sorry, I can't imagine where you got "fuzzy" from. Perhaps you mean > some other word. The concept is capable of being expressed precisely. and the second to last date in January plus a month is..? > > > Sorry, I don't understand. Why are you speechless? What would I want to > > > use the calendar module for? Apart from the leap() function and the > > > table of days in a month, the calendar module doesn't have any of the > > > functionality that one would expect in a general-purpose date class. > > > > Well, I thought replacing a 4 line function with 31 lines, 13 of which > > duplicated functionality in the standard library was overkill. > > I think you missed the point that the lines I quoted were straight out > of a self-contained library that existed (in C as well as Python) way > before the datetime module was a gleam in Fred & the timbot's eyes. I'm guessing you missed that I was looking for a method in the stdlib to do this, and failing that an idiomatic solution... > Even if I had noticed a leap year function in the calendar module, I > would probably not have used it. The Python version of the module was > just a stopgap while I fiddled with getting a C extension going. The C > leap year function doesn't have any of that modulo stuff in it. I'm sure it doesn't. You might want to look at the assembly your C compiler produces for a modulo-power-of-2 operation... > > I came up with this yesterday which seems sufficient? > > > > def yeardiff(a, b): > > y = a.year - b.year > > if (a.month, a.day) < (b.month, b.day): # tuple comparison > > y -= 1 > > return y > > At least it doesn't blow up when b is leapyear-02-29. It just gives the > wrong answer when a is nonleapyear-02-28. E.g. it gives 0 years > difference from 1992-02-29 to 1993-02-28 instead of 1. I believe you're mistaken (but feel free to correct me), my grandmother is born on Feb 29 and a quick call to my dad verified that they celebrated it the day after the 28th (unless Mar 1 was a Monday ;-). http://timeanddate.com/date/duration.html also seem to agree with my current understanding, just as a datapoint if nothing else. -- bjorn -- http://mail.python.org/mailman/listinfo/python-list
Re: How to find difference in years between two dates?
Bruno Desthuilliers wrote: > Which conversion ? How do you get the data ? as a datetime object ? as a > (y,m,d) tuple ? as a "y-m-d" string ? Else ? All input routines, whether they're from a web-form, database, command line, or anywhere else, only produce objects from the datetime module for calendar data. That way the program logic doesn't have to guess which format it's getting... I suppose I could do something like: def age(born): mx_born = mx.DateTime.Date(born.year, born.month, born.day) ... -- bjorn -- http://mail.python.org/mailman/listinfo/python-list
Re: How to find difference in years between two dates?
thebjorn wrote: > John Machin wrote: > > thebjorn wrote: > [...] > > > You give a good argument that the concept of a month is fuzzy > > > > Sorry, I can't imagine where you got "fuzzy" from. Perhaps you mean > > some other word. The concept is capable of being expressed precisely. > > and the second to last date in January plus a month is..? The last day in February, according to me and also according to the website that you mention later, e.g. """ >From date: Thursday, January 30, 1992 Added 1 month Resulting date: Saturday, February 29, 1992 """ The precise rule is that if the original day is later than the last day of the target month, clip it to that last day. > > > Even if I had noticed a leap year function in the calendar module, I > > would probably not have used it. The Python version of the module was > > just a stopgap while I fiddled with getting a C extension going. The C > > leap year function doesn't have any of that modulo stuff in it. > > I'm sure it doesn't. You might want to look at the assembly your C > compiler produces for a modulo-power-of-2 operation... I don't need to. AFAIR just about every compiler I've used (BDS C on a Z80 is a likely exception) has generated "& mask" (with extra mucking about if the first operand is signed). Leap year calc involves % 100 and % 400. Factoring out the powers of two leaves 25. Now tell me how I'm managing that without a modulo operation. > > > > I came up with this yesterday which seems sufficient? > > > > > > def yeardiff(a, b): > > > y = a.year - b.year > > > if (a.month, a.day) < (b.month, b.day): # tuple comparison > > > y -= 1 > > > return y > > > > At least it doesn't blow up when b is leapyear-02-29. It just gives the > > wrong answer when a is nonleapyear-02-28. E.g. it gives 0 years > > difference from 1992-02-29 to 1993-02-28 instead of 1. > > I believe you're mistaken (but feel free to correct me), my grandmother > is born on Feb 29 and a quick call to my dad verified that they > celebrated it the day after the 28th (unless Mar 1 was a Monday ;-). > http://timeanddate.com/date/duration.html also seem to agree with my > current understanding, just as a datapoint if nothing else. hmm ... let's look at more than 1 data point from that website: easy cases Jan 1993 to Feb 1993: Jan 10 to Feb 9: 30d (excluding last day) 1m 0d (including last day) Jan 10 to Feb 10: 1m 0d (excluding last day) 1m 1d (including last day) Jan 10 to Feb 11: 1m 1d (excluding last day) 1m 2d (including last day) looks reasonable, consistent, no anomalies; note how 1m 0d (excluding) is "predicted" from the right and below. Now try from Jan 31 1993: Jan 31 to Feb 27: 27d (ex) 28d (in) Jan 31 to Feb 28: 28d (ex) 1m 1d (in) Jan 31 to Mar 01: 1m 1d (ex) 1m 2d (in) So 1 day short of 1m 1d is not 1m 0 d??? I'd call this unreasonable, inconsistent, anomalous -- especially when on the same website you do 1993-01-31 plus 1 month, it gives you 1993-02-28 (as I'd expect). -- http://mail.python.org/mailman/listinfo/python-list
Re: How to find difference in years between two dates?
thebjorn wrote:
> Bruno Desthuilliers wrote:
>> Which conversion ? How do you get the data ? as a datetime object ? as a
>> (y,m,d) tuple ? as a "y-m-d" string ? Else ?
>
> All input routines, whether they're from a web-form, database, command
> line, or anywhere else, only produce objects from the datetime module
> for calendar data.
Seems quite sensible.
> That way the program logic doesn't have to guess
> which format it's getting... I suppose I could do something like:
>
>def age(born):
> mx_born = mx.DateTime.Date(born.year, born.month, born.day)
> ...
Yes, I was thinking of something like this.
> -- bjorn
>
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Re: How to find difference in years between two dates?
John Machin wrote: > Jan 31 to Feb 27: 27d (ex) 28d (in) > Jan 31 to Feb 28: 28d (ex) 1m 1d (in) > Jan 31 to Mar 01: 1m 1d (ex) 1m 2d (in) > So 1 day short of 1m 1d is not 1m 0 d??? Exactly. Just as a person born on 1999-3-1 isn't a year old on 2000-2-29. Perfectly regular, consistent and reasonable. > I'd call this unreasonable, inconsistent, anomalous -- especially > when on the same website you do 1993-01-31 plus 1 month, it > gives you 1993-02-28 (as I'd expect). You're entitled to your opinion. -- bjorn -- http://mail.python.org/mailman/listinfo/python-list
Re: How to find difference in years between two dates?
thebjorn wrote: > John Machin wrote: > > Jan 31 to Feb 27: 27d (ex) 28d (in) > > Jan 31 to Feb 28: 28d (ex) 1m 1d (in) > > Jan 31 to Mar 01: 1m 1d (ex) 1m 2d (in) > > So 1 day short of 1m 1d is not 1m 0 d??? > > Exactly. Just as a person born on 1999-3-1 isn't a year old on > 2000-2-29. Perfectly regular, consistent and reasonable. Is a person born on 1993-08-01 a year old on 1994--07-31? I don't understand. The examples that I showed went from the last day of a month to the last day of another month. You justify a manifest inconsistency in the results by reference to an example that goes from the first day of a month to the last day of a month?? > > > I'd call this unreasonable, inconsistent, anomalous -- especially > > when on the same website you do 1993-01-31 plus 1 month, it > > gives you 1993-02-28 (as I'd expect). > > You're entitled to your opinion. And you to yours :-) -- http://mail.python.org/mailman/listinfo/python-list
Re: How to find difference in years between two dates?
John Machin wrote:
> I don't understand. The examples that I showed went from the last day
> of a month to the last day of another month. [...]
Q1: is ((date-4days)+4days) == date?
Q2: is (((date-4days)+1month)+4days) == date+1month?
Ok, let's use Python'ish syntax (including numbering the days from 0
upwards, and backwards from -1, the last day of the month), you want
the last day of a month plus a month be the last day of the next
month. Simplistically, something like:
month[-1] + 1 month == (month+1)[-1] {last-to-last}
but that's obviously not the entire rule you want, unless 4-30 + 1
month == 5-31? So you would also like to have:
month[i] + 1 month = (month+1)[i] {lock-step}
we'd like yesterday to be a day ago? So for suitable i:
month[i] - 1 day == month[i-1] {yesterday-1}
month[0] - 1 day == (month-1)[-1] {yesterday-2}
which leads to a natural definition for when tomorrow is:
month[i] + 1 day == month[i+1] {tomorrow-1}
month[-1] + 1 day == (month+1)[0]{tomorrow-2}
So far so good. Now let's count backwards:
month[-1] - 1 day == month[-2]
month[-2] - 1 day == month[-3]
month[-3] - 1 day == month[-4]
etc.
In other words, if you insist that the last day of the month is a well
defined concept and you want a day ago to be yesterday then month[-4],
the forth-to-last day of the month, is necessarily also well
defined. Having a well defined month[i], I'll apply your rules for
adding a month:
month[-4] + 1 month == (month+1)[-4]
but you don't like this, because that means that e.g.:
april[-4] + 1 month == may[-4]
april[27] + 1 month == may[28]
which in addition to {lock-step}:
april[27] + 1 month == may[27]
either gives an inconsistent, ill-formed, or FUZZY system (although
I would call it "regular" ;-)
My approach is simpler since it doesn't define addition, only
subtraction on valid dates, so if I switch to representing dates as
(month, day):
(a, b) - (c, d) := a - c iff b < d {subtract}
else a - c - 1
{subtract} is irregular but well defined for all valid dates (it will
always give you an answer, and it's always the same answer ;-) :
(2,29) - (1,31) == 0
(3,1) - (1,31) == 2
I can add "day addition" and still be ok:
(m,d) + 1 day := (m,d+1){tomorrow-1}
(m,-1) + 1 day := (m+1,0) {tomorrow-2}
(m,d) - 1 day := (m,d-1) {yesterday-1}
(m,0) - 1 day := (m-1,-1) {yesterday-2}
Now my system is well-formed and consitent, even though it is
irregular, and it will answer yes to Q1 above. I can't see a way of
adding month addition to this and stay consistent without enumerating
special cases for every month, so Q2 can't even be asked in my system.
> > You're entitled to your opinion.
>
> And you to yours :-)
Ok, I've explained why I hold mine... You care to do the same?
-- bjorn
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