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convert time

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  convert  time "守株待兔" <1248283536@qq.com> - 2011-09-11 12:46 +0800
    Re: convert  time Steven D'Aprano <steve+comp.lang.python@pearwood.info> - 2011-09-11 15:00 +1000
      Re: convert  time Ben Finney <ben+python@benfinney.id.au> - 2011-09-11 15:30 +1000
      Re: convert time nn <pruebauno@latinmail.com> - 2011-09-12 10:42 -0700

#13110 — convert time

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how can i convert "Dec 11" into  2011-12?

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#13112

守株待兔 wrote:

> how can i convert "Dec 11" into  2011-12?


if my_str == "Dec 11":
    return 1999  # 2011 - 12


Does that help?

But seriously... 2011-12 is not a proper date, so the simplest way is
probably something like this:

def convert(date_str):
    month, short_year = date_str.split()
    if len(short_year) == 4:
        year = short_year
    else:
        year = '20' + short_year
    months = 'Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec'.split()
    month = months.index(month) + 1
    return year + '-' + str(month)

Otherwise see the time and datetime modules:

http://docs.python.org/library/time.html
http://docs.python.org/library/datetime.html


-- 
Steven

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#13116

Steven D'Aprano <steve+comp.lang.python@pearwood.info> writes:

> But seriously... 2011-12 is not a proper date

It's valid by ISO 8601. The standard allows any number of parts to be
dropped, from least to most significant, in order to have a value with
deliberately reduced precision.

   <URL:https://secure.wikimedia.org/wikipedia/en/wiki/ISO_8601#General_principles>

> Otherwise see the time and datetime modules:

    >>> import datetime
    >>> text = "2011-12"
    >>> datetime.datetime.strptime(text, "%Y-%m")
    datetime.datetime(2011, 12, 1, 0, 0)

-- 
 \            “It's not what you pay a man, but what he costs you that |
  `\                                             counts.” —Will Rogers |
_o__)                                                                  |
Ben Finney

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#13193 — Re: convert time

On Sep 11, 1:00 am, Steven D'Aprano <steve
+comp.lang.pyt...@pearwood.info> wrote:
> 守株待兔 wrote:
> > how can i convert "Dec 11" into  2011-12?
>
> if my_str == "Dec 11":
>     return 1999  # 2011 - 12
>
> Does that help?
>
> But seriously... 2011-12 is not a proper date, so the simplest way is
> probably something like this:
>
> def convert(date_str):
>     month, short_year = date_str.split()
>     if len(short_year) == 4:
>         year = short_year
>     else:
>         year = '20' + short_year
>     months = 'Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec'.split()
>     month = months.index(month) + 1
>     return year + '-' + str(month)
>
> Otherwise see the time and datetime modules:
>
> http://docs.python.org/library/time.htmlhttp://docs.python.org/library/datetime.html
>
> --
> Steven

Just a small comment that you can get "months" by doing:

from calendar import month_abbr
months = list(month_abbr)

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