I did look up the help for that function to make this meme but I must have missed that option. in my defense I’ve only been using Python for like 10 years
Python does have a year option that they are not using.
No, it doesn’t:
help(datetime.timedelta)
Help on class timedelta in module datetime:
class timedelta(builtins.object)
| Difference between two datetime values.
|
| timedelta(days=0, seconds=0, microseconds=0, milliseconds=0, minutes=0, hours=0, weeks=0)
|
| All arguments are optional and default to 0.
| Arguments may be integers or floats, and may be positive or negative.
Small correction, it’s every fourth year except every hundreth year except every 400th year. I’m assuming the misconception comes from the last time it was a 100th also being a 1000th (2000) but the next time a leap year will end in 00 is 2400
Ahh right! Thanks for correcting me. Now that you mention it I remember too. It also makes sense, a year is roughly 365.2425 days long. Add 0.25 (one out of four), subtract 0.01 (one out of hundred), add another 0.0025 (2.5 out of thousand which is 1 out of 400)
The python version seems buggy as fuck. Depending on which year you run it it’s off by 1-3 days
Python does have a year option that they are not using. Depending on the application I would use 365 for a year to get a consistent number of days.
I did look up the
helpfor that function to make this meme but I must have missed that option. in my defense I’ve only been using Python for like 10 yearsNo, it doesn’t:
help(datetime.timedelta)
Help on class timedelta in module datetime: class timedelta(builtins.object) | Difference between two datetime values. | | timedelta(days=0, seconds=0, microseconds=0, milliseconds=0, minutes=0, hours=0, weeks=0) | | All arguments are optional and default to 0. | Arguments may be integers or floats, and may be positive or negative.Your right. I was thinking about pd.DateOffset()
That sounds serious, can you give some example values we can test?
look I’m not trying to be a dick or anything, but do you not know about leap years and which years they are?
Yes, and I have no idea…
Leap years are each fourth year, except each hundredth year, except each thousandth fourhundredth year.
1896 leap year
1900 not leap year
1904 leap year
…
1996 leap year
2000 leap year
2004 leap year
…
2096 leap year
2100 not leap year
2104 leap year
Then you just arrange the 10 year window in different positions to overlap 1 to 3 leap years to reveal the three outcomes of the bug.
- / - - - / - - - /
- - / - - - / - - -
- - 0 - - - / - - -
- is a normal year, / is a leap year, 0 is an exceptional non-leap year.
Small correction, it’s every fourth year except every hundreth year except every 400th year. I’m assuming the misconception comes from the last time it was a 100th also being a 1000th (2000) but the next time a leap year will end in 00 is 2400
Ahh right! Thanks for correcting me. Now that you mention it I remember too. It also makes sense, a year is roughly 365.2425 days long. Add 0.25 (one out of four), subtract 0.01 (one out of hundred), add another 0.0025 (2.5 out of thousand which is 1 out of 400)
https://youtu.be/-5wpm-gesOY
edit: just realized it was the python and not ruby example, I was very tired and distracted when I was reading this thread.
Sure, here’s one example for each case:
1 day off: 3650 days before 1907-01-01 is 1897-01-02
2 days off: 3650 days before 2027-01-01 is 2017-01-03
3 days off: 3650 days before 2025-01-01 is 2015-01-04
29 February 2028, 29 February 2032, 29 February 2036…