Pull Last 3 Months

[Shane Geiger]

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
sgeiger at ncee.net  |  402-438-8958  |  http://www.ncee.net

Leading the Campaign for Economic and Financial Literacy

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