[Python-ideas] datetime.timedelta literals

[Kyle Lahnakoski]

Pål Grønås Drange,

I do like the idea of literals typed with scientific units, but I often
get short variable names mixed up, so I am not sure if I could use them
without a cheat sheet. Formatting datetime is a good example of how
confusing a collection of short names can get: Is month %m or %M?  Is
minute %m or %i?
https://docs.python.org/2/library/datetime.html#strftime-and-strptime-behavior

In case you are thinking, "Kyle, how can you even *think* "%i" means
minutes?!", please see
https://dev.mysql.com/doc/refman/5.7/en/date-and-time-functions.html#function_date-format 
:)

I made a class, with magic methods, to get close to what you want:

(2.5*HOUR - 14*MINUTE + 9300*MILLISECOND).total_seconds()

I used full names for less confusion, but you can do the same with
shorter names:

(2.5*h- 14*min + 9300*ms).total_seconds()


Maybe the Python parser can be made to add an implied multiplication
between a-number-followed-directly-by-a-variable-name. If so, then I
could write:

(2.5HOUR - 14MINUTE + 9300MILLISECOND).total_seconds()

You can define your short, domain specific, suffixes:

(2.5h - 14m + 9300ms).total_seconds()




On 2018-06-02 08:21, Pål Grønås Drange wrote:
> Elevator pitch:
>
> (2.5h - 14min + 9300ms).total_seconds()
> # 8169.3
>
> from datetime import datetime as dt
> start = dt.now()
> end = dt.now()
> (end-start) < 5s
> # True
>
>
> chrono::duration:
>
> In C++ 14 the std::chrono::duration was introduced which corresponds
> somewhat to
> datetime.timedelta.
>
> C++ 14 introduced so-called chrono literals[1], which are literals
> specified as
> [number][h|min|s|ms|us|ns], e.g.
>
> * 2.5h
> * 14min
> * 9300ms
>
> These literals should correspond to
>
> * datetime.timedelta(0, 9000)  # 2.5h = 2.5*3600 = 9000 seconds
> * datetime.timedelta(0, 840)   # 14min = 14*60 = 840 seconds
> * datetime.timedelta(0, 9, 300000)  # 9300ms = 9 seconds + 3*10^5
> microseconds
>
>
> If a literal was interpreted as a datetime.timedelta, the following
> would work
> out of the box:
>
> 2.5h - 14min + 9300ms * 2
>
> which would correspond to
>
> from datetime import timedelta as D
>
> D(hours=2.5) - D(minutes=14) + D(milliseconds=9300) * 2
> # datetime.timedelta(0, 8178, 600000)  # (*2 precedes, so that's to be
> expected)
>
>
> (D(hours=2.5) - D(minutes=14) + D(milliseconds=9300)) * 2
> # datetime.timedelta(0, 16338, 600000)
>
>
>
> Notes:
>
> * C++ uses `min` instead of `m`.  `min` is a keyword in Python.
> * In C++, `1d` means the first day of a month [2].
> * In C++, `1990y` means the year 1990 (in the Proleptic Gregorian
> calendar) [3].
> * C++ have the types signed integers and not floats, so 2.5h would not
> be valid.
> * My apologies if this has been discussed before; my search-fu gave me
> nothing.
>
>
>
> References:
>
>
> [1] std::literals::chrono_literals::operator""min
> http://en.cppreference.com/w/cpp/chrono/operator%22%22min
>
> [2] http://en.cppreference.com/w/cpp/chrono/day
>
> [3] http://en.cppreference.com/w/cpp/chrono/year
>
>
> Best regards,
> Pål Grønås Drange
>
>
>
> _______________________________________________
> Python-ideas mailing list
> Python-ideas at python.org
> https://mail.python.org/mailman/listinfo/python-ideas
> Code of Conduct: http://python.org/psf/codeofconduct/

-------------- next part --------------
An HTML attachment was scrubbed...
URL: <http://mail.python.org/pipermail/python-ideas/attachments/20180604/18af9027/attachment.html>