ordinal -> (year, month, day), considering 01-Jan-0001 as day 1.
(n)
| 86 | assert _DI100Y == 25 * _DI4Y - 1 |
| 87 | |
| 88 | def _ord2ymd(n): |
| 89 | "ordinal -> (year, month, day), considering 01-Jan-0001 as day 1." |
| 90 | |
| 91 | # n is a 1-based index, starting at 1-Jan-1. The pattern of leap years |
| 92 | # repeats exactly every 400 years. The basic strategy is to find the |
| 93 | # closest 400-year boundary at or before n, then work with the offset |
| 94 | # from that boundary to n. Life is much clearer if we subtract 1 from |
| 95 | # n first -- then the values of n at 400-year boundaries are exactly |
| 96 | # those divisible by _DI400Y: |
| 97 | # |
| 98 | # D M Y n n-1 |
| 99 | # -- --- ---- ---------- ---------------- |
| 100 | # 31 Dec -400 -_DI400Y -_DI400Y -1 |
| 101 | # 1 Jan -399 -_DI400Y +1 -_DI400Y 400-year boundary |
| 102 | # ... |
| 103 | # 30 Dec 000 -1 -2 |
| 104 | # 31 Dec 000 0 -1 |
| 105 | # 1 Jan 001 1 0 400-year boundary |
| 106 | # 2 Jan 001 2 1 |
| 107 | # 3 Jan 001 3 2 |
| 108 | # ... |
| 109 | # 31 Dec 400 _DI400Y _DI400Y -1 |
| 110 | # 1 Jan 401 _DI400Y +1 _DI400Y 400-year boundary |
| 111 | n -= 1 |
| 112 | n400, n = divmod(n, _DI400Y) |
| 113 | year = n400 * 400 + 1 # ..., -399, 1, 401, ... |
| 114 | |
| 115 | # Now n is the (non-negative) offset, in days, from January 1 of year, to |
| 116 | # the desired date. Now compute how many 100-year cycles precede n. |
| 117 | # Note that it's possible for n100 to equal 4! In that case 4 full |
| 118 | # 100-year cycles precede the desired day, which implies the desired |
| 119 | # day is December 31 at the end of a 400-year cycle. |
| 120 | n100, n = divmod(n, _DI100Y) |
| 121 | |
| 122 | # Now compute how many 4-year cycles precede it. |
| 123 | n4, n = divmod(n, _DI4Y) |
| 124 | |
| 125 | # And now how many single years. Again n1 can be 4, and again meaning |
| 126 | # that the desired day is December 31 at the end of the 4-year cycle. |
| 127 | n1, n = divmod(n, 365) |
| 128 | |
| 129 | year += n100 * 100 + n4 * 4 + n1 |
| 130 | if n1 == 4 or n100 == 4: |
| 131 | assert n == 0 |
| 132 | return year-1, 12, 31 |
| 133 | |
| 134 | # Now the year is correct, and n is the offset from January 1. We find |
| 135 | # the month via an estimate that's either exact or one too large. |
| 136 | leapyear = n1 == 3 and (n4 != 24 or n100 == 3) |
| 137 | assert leapyear == _is_leap(year) |
| 138 | month = (n + 50) >> 5 |
| 139 | preceding = _DAYS_BEFORE_MONTH[month] + (month > 2 and leapyear) |
| 140 | if preceding > n: # estimate is too large |
| 141 | month -= 1 |
| 142 | preceding -= _DAYS_IN_MONTH[month] + (month == 2 and leapyear) |
| 143 | n -= preceding |
| 144 | assert 0 <= n < _days_in_month(year, month) |
| 145 |
no test coverage detected