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
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