473,375 Members | 1,307 Online
Bytes | Software Development & Data Engineering Community
Post Job

Home Posts Topics Members FAQ

Join Bytes to post your question to a community of 473,375 software developers and data experts.

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

Oct 17 '07 #1
11 2427
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

Oct 17 '07 #2
On Oct 17, 9:59 pm, Harlin Seritt <harlinser...@yahoo.comwrote:
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

Oct 17 '07 #3
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
sg*****@ncee.net | 402-438-8958 | http://www.ncee.net

Leading the Campaign for Economic and Financial Literacy
Oct 17 '07 #4
On Oct 17, 11:56 pm, Shane Geiger <sgei...@ncee.netwrote:
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

Oct 17 '07 #5
Paul Hankin <pa*********@gmail.comwrites:
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
Oct 18 '07 #6
On Oct 18, 8:56 am, Shane Geiger <sgei...@ncee.netwrote:
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']
>>>
Oct 18 '07 #7
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)]

Oct 18 '07 #8
On 18/10/2007 10:33 AM, Ben Finney wrote:
Paul Hankin <pa*********@gmail.comwrites:
>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.
Oct 18 '07 #9
John Machin <sj******@lexicon.netwrites:
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
Oct 18 '07 #10
En Wed, 17 Oct 2007 21:47:50 -0300, Tim Chase
<py*********@tim.thechases.comescribió:
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:

pyimport locale
pylocale.setlocale(locale.LC_ALL, '')
'Spanish_Argentina.1252'
py>
pyimport calendar
pycalendar.month_abbr[12]
'Dic'
pydef prev_months(since, howmany):
.... return [calendar.month_abbr[(since.month-i-2) % 12 + 1] for i in
range(how
many)]
....
pyimport datetime
pyprev_months(datetime.datetime(2005,2,10), 4)
['Ene', 'Dic', 'Nov', 'Oct']
pyprev_months(datetime.datetime(2005,10,17), 3)
['Sep', 'Ago', 'Jul']

--
Gabriel Genellina

Oct 18 '07 #11
On Oct 18, 12:25 am, "Gabriel Genellina" <gagsl-...@yahoo.com.ar>
wrote:
I prefer the calendar module in that case:

pyimport locale
pylocale.setlocale(locale.LC_ALL, '')
'Spanish_Argentina.1252'
py>
pyimport calendar
pycalendar.month_abbr[12]
'Dic'
pydef prev_months(since, howmany):
... return [calendar.month_abbr[(since.month-i-2) % 12 + 1] for i in
range(how
many)]
...
pyimport datetime
pyprev_months(datetime.datetime(2005,2,10), 4)
['Ene', 'Dic', 'Nov', 'Oct']
pyprev_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

Oct 18 '07 #12

This thread has been closed and replies have been disabled. Please start a new discussion.

Similar topics

6
by: Bimo Remus | last post by:
Hi, I am currently taking a C++ class and am having problems with a homework assignment. My problem is that I need to pull the first and last words out of of a character string array which is in...
11
by: Dennis Marks | last post by:
There seems to be a major program with the automatic display of the last modified date. Using the javascript "document.lastModified" sometimes returns the correct date and sometimes 1 Jan 1970...
13
by: SimonC | last post by:
I would like to return data from the last 2 weeks of each given month in Javascript, but in 2 formats. So, the penultimate week (Monday to Sunday) and the last week (Monday to ??) I'm not...
3
by: Melissa | last post by:
I have this table: TblProjectYear ProjectYearID ProjectYearStartDate ProjectYearEndDate The Project Year will always span across December 31; for example 9/1/04 to 6/30/05. How do I build a...
9
by: Robin Tucker | last post by:
Hiya, I need to test "relative dates" in my program, such as "last six months" or "last 3 months" or "in the last week" etc. How can I do this with a DateTime structure? ie. If NodeDate...
5
by: RandyG | last post by:
how do you pull the current months data without having to manually input the date each time?
4
by: sparks | last post by:
We have a new project here, one that I have never tried maybe its easy I don't know yet. We have people that have records dating back over 5 yrs on a sql server. We have to build an access 2003...
3
by: remya1000 | last post by:
i'm using ASP with MSAccess as database. i have two buttons and two textbox in my page. when i press my first button (First month) i need to display the current month in one textbox and last one...
0
by: Harlin Seritt | last post by:
Is there a module that can pull str values for say the last 3 months? Something like: print lastMonths(3) Thanks
1
by: CloudSolutions | last post by:
Introduction: For many beginners and individual users, requiring a credit card and email registration may pose a barrier when starting to use cloud servers. However, some cloud server providers now...
0
by: Faith0G | last post by:
I am starting a new it consulting business and it's been a while since I setup a new website. Is wordpress still the best web based software for hosting a 5 page website? The webpages will be...
0
isladogs
by: isladogs | last post by:
The next Access Europe User Group meeting will be on Wednesday 3 Apr 2024 starting at 18:00 UK time (6PM UTC+1) and finishing by 19:30 (7.30PM). In this session, we are pleased to welcome former...
0
by: ryjfgjl | last post by:
In our work, we often need to import Excel data into databases (such as MySQL, SQL Server, Oracle) for data analysis and processing. Usually, we use database tools like Navicat or the Excel import...
0
by: taylorcarr | last post by:
A Canon printer is a smart device known for being advanced, efficient, and reliable. It is designed for home, office, and hybrid workspace use and can also be used for a variety of purposes. However,...
0
by: aa123db | last post by:
Variable and constants Use var or let for variables and const fror constants. Var foo ='bar'; Let foo ='bar';const baz ='bar'; Functions function $name$ ($parameters$) { } ...
0
by: ryjfgjl | last post by:
If we have dozens or hundreds of excel to import into the database, if we use the excel import function provided by database editors such as navicat, it will be extremely tedious and time-consuming...
0
by: ryjfgjl | last post by:
In our work, we often receive Excel tables with data in the same format. If we want to analyze these data, it can be difficult to analyze them because the data is spread across multiple Excel files...
0
BarryA
by: BarryA | last post by:
What are the essential steps and strategies outlined in the Data Structures and Algorithms (DSA) roadmap for aspiring data scientists? How can individuals effectively utilize this roadmap to progress...

By using Bytes.com and it's services, you agree to our Privacy Policy and Terms of Use.

To disable or enable advertisements and analytics tracking please visit the manage ads & tracking page.