Re: Pull Last 3 Months
On Oct 18, 12:25 am, "Gabriel Genellina" <[EMAIL PROTECTED]> wrote: > I prefer the calendar module in that case: > > py> import locale > py> locale.setlocale(locale.LC_ALL, '') > 'Spanish_Argentina.1252' > py> > py> import calendar > py> calendar.month_abbr[12] > 'Dic' > py> def prev_months(since, howmany): > ... return [calendar.month_abbr[(since.month-i-2) % 12 + 1] for i in > range(how > many)] > ... > py> import datetime > py> prev_months(datetime.datetime(2005,2,10), 4) > ['Ene', 'Dic', 'Nov', 'Oct'] > py> prev_months(datetime.datetime(2005,10,17), 3) > ['Sep', 'Ago', 'Jul'] Ah, you beat me to it. I was going to point out that if you're going to be using month strings, you should use calendar, since it can also use the correct locale. Plus, it offers a ridiculously simple solution to this problem compared to all the others. Hyuga -- http://mail.python.org/mailman/listinfo/python-list
Re: Pull Last 3 Months
En Wed, 17 Oct 2007 21:47:50 -0300, Tim Chase <[EMAIL PROTECTED]> escribió: > In the event that you need them in whatever your locale is, you > can use the '%b' formatting to produce them: I prefer the calendar module in that case: py> import locale py> locale.setlocale(locale.LC_ALL, '') 'Spanish_Argentina.1252' py> py> import calendar py> calendar.month_abbr[12] 'Dic' py> def prev_months(since, howmany): ... return [calendar.month_abbr[(since.month-i-2) % 12 + 1] for i in range(how many)] ... py> import datetime py> prev_months(datetime.datetime(2005,2,10), 4) ['Ene', 'Dic', 'Nov', 'Oct'] py> prev_months(datetime.datetime(2005,10,17), 3) ['Sep', 'Ago', 'Jul'] -- Gabriel Genellina -- http://mail.python.org/mailman/listinfo/python-list
Re: Pull Last 3 Months
John Machin <[EMAIL PROTECTED]> writes: > It's a bit hard to see how anybody could imagine that in the expression > [months[(month - i - 1) % 12] for i in range(n)] > the number 12 referred to anything but the number of months in a year. Exactly, that's what people *will* assume. But what if they're wrong, and you're using the number 12 for some other semantic purpose? If the programmer has encountered this type of betrayed assumption before, they'll never be entirely sure that a bare '12' in the code means what they think it means. And the code doesn't say anything about why the number was used, so they're left to guess. Of course, in such a trivial example, it is almost unthinkable that the number 12 would mean anything else; but the entire point of the principle of not using magic numbers is that you don't have to wonder about when that line is crossed. Better to be explicit about it, in every case, IMO. -- \ "Money is always to be found when men are to be sent to the | `\ frontiers to be destroyed: when the object is to preserve them, | _o__) it is no longer so." -- Voltaire, _Dictionnaire Philosophique_ | Ben Finney -- http://mail.python.org/mailman/listinfo/python-list
Re: Pull Last 3 Months
On 18/10/2007 10:33 AM, Ben Finney wrote: > Paul Hankin <[EMAIL PROTECTED]> writes: > >> import datetime >> >> months = 'Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec'.split() >> >> def last_months(n): >> month = datetime.date.today().month >> return [months[(month - i - 1) % 12] for i in range(n)] >> >> print last_months(3) > > Heck you don't even need the magic number 12 in there. > > import datetime > > months = 'Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec'.split() Heck if you really want to be anal, you could even guard against a typo (or one of those spaces actually being '\xA0' [seen it happen]) by adding in here: MONTHS_IN_YEAR = 12 assert len(months) == MONTHS_IN_YEAR > def last_months(n): > month = datetime.date.today().month > return [months[(month - i - 1) % len(months) > for i in range(n)] > > In general I try to avoid magic numbers: always be explicit about the > semantic purpose of the number, either by binding a meaningful name to > it and only using that reference thereafter, or showing how that value > is derived. > It's a bit hard to see how anybody could imagine that in the expression [months[(month - i - 1) % 12] for i in range(n)] the number 12 referred to anything but the number of months in a year. -- http://mail.python.org/mailman/listinfo/python-list
Re: Pull Last 3 Months
On Oct 18, 8:56 am, Shane Geiger <[EMAIL PROTECTED]> wrote:
> A simpler way, imho:
>
> import datetime
> m = {
> 1:'Jan',2:'Feb',3:'Mar',4:'Apr',5:'May',6:'Jun',7:'Jul',8:'Aug',9:'Sep',10:'Oct',11:'Nov',12:'Dec'}
>
> month = datetime.date.today().month
> if month == 1:
> ans = [m[11], m[12], m[1]]
> elif month == 2:
> ans = [m[11], m[12], m[1]]
> else:
> ans = [m[month-2], m[month-1], m[month]]
> print ans
>
1. Why use a dict?
2. The if-elif-else caper doesn't scale well; suppose the OP want to
"pull" the previous 6 months. The % operator is your friend.
Try this:
>>> m = ['Jan', 'Feb', 'Mar', 'Apr', 'May', 'Jun', 'Jul', 'Aug', 'Sep', 'Oct',
>>> 'Nov', 'Dec']
>>> for mo in range(1, 13):
...print mo, [m[(mo - x - 2) % 12] for x in range(3)]
...
1 ['Dec', 'Nov', 'Oct']
2 ['Jan', 'Dec', 'Nov']
3 ['Feb', 'Jan', 'Dec']
4 ['Mar', 'Feb', 'Jan']
5 ['Apr', 'Mar', 'Feb']
6 ['May', 'Apr', 'Mar']
7 ['Jun', 'May', 'Apr']
8 ['Jul', 'Jun', 'May']
9 ['Aug', 'Jul', 'Jun']
10 ['Sep', 'Aug', 'Jul']
11 ['Oct', 'Sep', 'Aug']
12 ['Nov', 'Oct', 'Sep']
>>> for mo in range(1, 13):
...print mo, [m[(mo - x - 2) % 12] for x in range(6)]
...
1 ['Dec', 'Nov', 'Oct', 'Sep', 'Aug', 'Jul']
2 ['Jan', 'Dec', 'Nov', 'Oct', 'Sep', 'Aug']
...snip...
11 ['Oct', 'Sep', 'Aug', 'Jul', 'Jun', 'May']
12 ['Nov', 'Oct', 'Sep', 'Aug', 'Jul', 'Jun']
>>>
--
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Re: Pull Last 3 Months
Paul Hankin <[EMAIL PROTECTED]> writes: > import datetime > > months = 'Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec'.split() > > def last_months(n): > month = datetime.date.today().month > return [months[(month - i - 1) % 12] for i in range(n)] > > print last_months(3) Heck you don't even need the magic number 12 in there. import datetime months = 'Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec'.split() def last_months(n): month = datetime.date.today().month return [months[(month - i - 1) % len(months) for i in range(n)] In general I try to avoid magic numbers: always be explicit about the semantic purpose of the number, either by binding a meaningful name to it and only using that reference thereafter, or showing how that value is derived. -- \ "I hope some animal never bores a hole in my head and lays its | `\ eggs in my brain, because later you might think you're having a | _o__) good idea but it's just eggs hatching." -- Jack Handey | Ben Finney -- http://mail.python.org/mailman/listinfo/python-list
Re: Pull Last 3 Months
> It looks like you copied the month 2 case from the month 1 case
> because you forgot to edit it afterwards. Anyway, a bit of modulo-12
> arithmetic avoids special cases, and allows the number of months to be
> generalised:
nice...
> import datetime
>
> months = 'Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec'.split()
>
> def last_months(n):
> month = datetime.date.today().month
> return [months[(month - i - 1) % 12] for i in range(n)]
>
> print last_months(3)
In the event that you need them in whatever your locale is, you
can use the '%b' formatting to produce them:
import datetime
month_map = dict(
(n, datetime.date(2000,n+1,1).strftime('%b'))
for n in xrange(12)
)
The function can then be written as either a generator:
def last_months(months):
this_month = datetime.date.today().month - 1
for i in xrange(months):
yield month_map[(this_month-i) % 12]
or as a function returning a list/tuple:
def last_months(months):
this_month = datetime.date.today().month - 1
return [month_map[(this_month - i) % 12]
for i in xrange(months)]
--
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Re: Pull Last 3 Months
On Oct 17, 11:56 pm, Shane Geiger <[EMAIL PROTECTED]> wrote:
> A simpler way, imho:
>
> import datetime
> m = {
> 1:'Jan',2:'Feb',3:'Mar',4:'Apr',5:'May',6:'Jun',7:'Jul',8:'Aug',9:'Sep',10:'Oct',11:'Nov',12:'Dec'}
>
> month = datetime.date.today().month
> if month == 1:
> ans = [m[11], m[12], m[1]]
> elif month == 2:
> ans = [m[11], m[12], m[1]]
> else:
> ans = [m[month-2], m[month-1], m[month]]
> print ans
It looks like you copied the month 2 case from the month 1 case
because you forgot to edit it afterwards. Anyway, a bit of modulo-12
arithmetic avoids special cases, and allows the number of months to be
generalised:
import datetime
months = 'Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec'.split()
def last_months(n):
month = datetime.date.today().month
return [months[(month - i - 1) % 12] for i in range(n)]
print last_months(3)
--
Paul Hankin
--
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Re: Pull Last 3 Months
A simpler way, imho:
import datetime
m = {
1:'Jan',2:'Feb',3:'Mar',4:'Apr',5:'May',6:'Jun',7:'Jul',8:'Aug',9:'Sep',10:'Oct',11:'Nov',12:'Dec'
}
month = datetime.date.today().month
if month == 1:
ans = [m[11], m[12], m[1]]
elif month == 2:
ans = [m[11], m[12], m[1]]
else:
ans = [m[month-2], m[month-1], m[month]]
print ans
Tim Chase wrote:
>> Is there a module that can pull str values for say the last 3 months?
>> Something like:
>>
>> print lastMonths(3)
>>
>> ['Sep', 'Aug', 'Jul']
>>
>
> I don't think there's anything inbuilt. It's slightly
> frustrating that timedelta doesn't accept a "months" parameter
> when it would be rather helpful (but how many days are in a
> month-delta is something that changes from month-to-month).
>
> It's somewhat inelegant, but this works for me:
>
>import datetime
>
>def last_months(months):
> assert months > 0
> d = datetime.date.today()
> m = d.strftime('%b')
> yield m
> while months > 1:
>d -= datetime.timedelta(days=28)
>m2 = d.strftime('%b')
>if m2 <> m:
> m = m2
> months -= 1
> yield m
>
>print list(last_months(3))
>for month in last_months(24): print month
>
> The alternative would likely be to do something like subtract one
> from the current month, and if it drops below 1, decrement the
> year and reset the month to 12. Equally fuzzy:
>
>def lastN(months):
> assert months > 0
> d = datetime.date.today()
> for _ in xrange(months):
>yield d.strftime('%b')
>y,m = d.year, d.month
>if m > 1:
> m -= 1
>else:
> m = 12
> y -= 1
>d = datetime.date(y,m,1)
>
> Use whichever you prefer.
>
> -tkc
>
>
>
>
>
>
--
Shane Geiger
IT Director
National Council on Economic Education
[EMAIL PROTECTED] | 402-438-8958 | http://www.ncee.net
Leading the Campaign for Economic and Financial Literacy
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Re: Pull Last 3 Months
On Oct 17, 9:59 pm, Harlin Seritt <[EMAIL PROTECTED]> wrote:
> Is there a module that can pull str values for say the last 3 months?
> Something like:
>
> print lastMonths(3)
>
> ['Sep', 'Aug', 'Jul']
You should take a look at the 'datetime' module.
You can get the current month:
datetime.datetime.now().month
And a list of all abbreviated month names:
[datetime.datetime(1900, i + 1, 1).strftime('%b') for i in range(12)]
>From there, it shouldn't be too tricky to construct the list of months
you want.
--
Paul Hankin
--
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Re: Pull Last 3 Months
> Is there a module that can pull str values for say the last 3 months?
> Something like:
>
> print lastMonths(3)
>
> ['Sep', 'Aug', 'Jul']
I don't think there's anything inbuilt. It's slightly
frustrating that timedelta doesn't accept a "months" parameter
when it would be rather helpful (but how many days are in a
month-delta is something that changes from month-to-month).
It's somewhat inelegant, but this works for me:
import datetime
def last_months(months):
assert months > 0
d = datetime.date.today()
m = d.strftime('%b')
yield m
while months > 1:
d -= datetime.timedelta(days=28)
m2 = d.strftime('%b')
if m2 <> m:
m = m2
months -= 1
yield m
print list(last_months(3))
for month in last_months(24): print month
The alternative would likely be to do something like subtract one
from the current month, and if it drops below 1, decrement the
year and reset the month to 12. Equally fuzzy:
def lastN(months):
assert months > 0
d = datetime.date.today()
for _ in xrange(months):
yield d.strftime('%b')
y,m = d.year, d.month
if m > 1:
m -= 1
else:
m = 12
y -= 1
d = datetime.date(y,m,1)
Use whichever you prefer.
-tkc
--
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Pull Last 3 Months
Is there a module that can pull str values for say the last 3 months? Something like: print lastMonths(3) ['Sep', 'Aug', 'Jul'] Thanks -- http://mail.python.org/mailman/listinfo/python-list
Pull Last 3 Months
Is there a module that can pull str values for say the last 3 months? Something like: print lastMonths(3) ['Sep', 'Aug', 'Jul'] Thanks -- http://mail.python.org/mailman/listinfo/python-list
