Use uncertainties.ufloat for uncertain durations

pyproject.toml

@@ -9,7 +9,7 @@ readme = "README.md"
 license = { text = "Apache-2" }
 requires-python = ">= 3.9"
 dynamic = ["version"]
-dependencies = ["lark[interegular]", "numpy", "convertdate", "strenum; python_version < '3.11'"]
+dependencies = ["lark[interegular]", "numpy", "convertdate", "strenum; python_version < '3.11'", "uncertainties"]
 authors = [
     { name = "Rebecca Sutton Koeser" },
     { name = "Cole Crawford" },

src/undate/converters/calendars/gregorian.py

@@ -13,8 +13,10 @@ class GregorianDateConverter(BaseCalendarConverter):
     #: calendar
     calendar_name: str = "Gregorian"
 
-    #: known non-leap year
+    #: arbitrary known non-leap year
     NON_LEAP_YEAR: int = 2022
+    #: arbitrary known leap year
+    LEAP_YEAR: int = 2024
 
     def min_month(self) -> int:
         """First month for the Gregorian calendar."""
@@ -38,6 +40,7 @@ def max_day(self, year: int, month: int) -> int:
             _, max_day = monthrange(year, month)
         else:
             # if year and month are unknown, return maximum possible
+            # TODO: should this return a ufloat?
             max_day = 31
 
         return max_day

src/undate/date.py

@@ -1,9 +1,11 @@
 from enum import IntEnum
+from dataclasses import dataclass
 
 # Pre 3.10 requires Union for multiple types, e.g. Union[int, None] instead of int | None
 from typing import Optional, Union
 
 import numpy as np
+from uncertainties import ufloat  # type: ignore
 
 
 class Timedelta(np.ndarray):
@@ -29,6 +31,27 @@ def days(self) -> int:
         return int(self.astype("datetime64[D]").astype("int"))
 
 
+@dataclass
+class Udelta:
+    """An uncertain timedelta, for durations where the number of days is uncertain.
+    Initialize with a list of possible day durations as integers, which are used
+    to calculate a value for duration in :attr:`days` as an
+    instance of :class:`uncertainties.ufloat`.
+    """
+
+    # NOTE: we will probably need other timedelta-like logic here besides days...
+
+    #: number of days, as an instance of :class:`uncertainties.ufloat`
+    days: ufloat
+
+    def __init__(self, *days: int):
+        min_days = min(days)
+        max_days = max(days)
+        half_diff = (max_days - min_days) / 2
+        midpoint = min_days + half_diff
+        self.days = ufloat(midpoint, half_diff)
+
+
 #: timedelta for single day
 ONE_DAY = Timedelta(1)  # ~ equivalent to datetime.timedelta(days=1)
 #: timedelta for a single  year (non-leap year)

src/undate/interval.py

@@ -104,6 +104,11 @@ def duration(self) -> Timedelta:
         elif not self.latest.known_year and not self.earliest.known_year:
             # under what circumstances can we assume that if both years
             # are unknown the dates are in the same year or sequential?
+
+            # TODO: for Gregorian calendars, if this interval spans end
+            # of February we should return a udelta object since the interval
+            # may or may not include February 29
+
             duration = self.latest.earliest - self.earliest.earliest
             # if we get a negative, we've wrapped from end of one year
             # to the beginning of the next;

src/undate/undate.py

@@ -18,7 +18,7 @@
 from typing import Dict, Optional, Union
 
 from undate.converters.base import BaseDateConverter
-from undate.date import ONE_DAY, ONE_MONTH_MAX, Date, DatePrecision, Timedelta
+from undate.date import ONE_DAY, ONE_MONTH_MAX, Date, DatePrecision, Timedelta, Udelta
 
 
 class Calendar(StrEnum):
@@ -420,13 +420,14 @@ def _get_date_part(self, part: str) -> Optional[str]:
         value = self.initial_values.get(part)
         return str(value) if value else None
 
-    def duration(self) -> Timedelta:
+    def duration(self) -> Timedelta | Udelta:
         """What is the duration of this date?
         Calculate based on earliest and latest date within range,
         taking into account the precision of the date even if not all
         parts of the date are known. Note that durations are inclusive
         (i.e., a closed interval)  and include both the earliest and latest
-        date rather than the difference between them."""
+        date rather than the difference between them.  Returns a :class:`undate.date.Timedelta` when
+        possible, and an :class:`undate.date.Udelta` when the duration is uncertain."""
 
         # if precision is a single day, duration is one day
         # no matter when it is or what else is known
@@ -437,20 +438,48 @@ def duration(self) -> Timedelta:
         # calculate month duration within a single year (not min/max)
         if self.precision == DatePrecision.MONTH:
             latest = self.latest
+            # if year is unknown, calculate month duration in
+            # leap year and non-leap year, in case length varies
             if not self.known_year:
-                # if year is unknown, calculate month duration in
-                # a single year
-                latest = Date(self.earliest.year, self.latest.month, self.latest.day)
+                # TODO: should leap-year specific logic shift to the calendars,
+                # since it works differently depending on the calendar?
+                possible_years = [
+                    self.calendar_converter.LEAP_YEAR,
+                    self.calendar_converter.NON_LEAP_YEAR,
+                ]
+            # TODO: what about partially known years like 191X ?
+            else:
+                # otherwise, get possible durations for all possible months
+                # for a known year
+                possible_years = [self.earliest.year]
+
+            # for every possible month and year, get max days for that month,
+            possible_max_days = set()
+            # appease mypy, which says month values could be None here
+            if self.earliest.month is not None and self.latest.month is not None:
+                for possible_month in range(self.earliest.month, self.latest.month + 1):
+                    for year in possible_years:
+                        possible_max_days.add(
+                            self.calendar_converter.max_day(year, possible_month)
+                        )
+
+                # if there is more than one possible value for month length,
+                # whether due to leap year / non-leap year or ambiguous month,
+                # return a uncertain delta
+                if len(possible_max_days) > 1:
+                    return Udelta(*possible_max_days)
+
+            # otherwise, calculate timedelta normally
+            max_day = list(possible_max_days)[0]
+            latest = Date(self.earliest.year, self.earliest.month, max_day)
 
-                # latest = datetime.date(
-                #     self.earliest.year, self.latest.month, self.latest.day
-                # )
             delta = latest - self.earliest + ONE_DAY
             # month duration can't ever be more than 31 days
             # (could we ever know if it's smaller?)
 
             # if granularity == month but not known month, duration = 31
             if delta.astype(int) > 31:
+                # FIXME: this depends on calendar!
                 return ONE_MONTH_MAX
             return delta
 

tests/test_date.py

@@ -1,5 +1,7 @@
 import numpy as np
-from undate.date import ONE_YEAR, Date, DatePrecision, Timedelta
+from uncertainties import ufloat
+
+from undate.date import ONE_YEAR, Date, DatePrecision, Timedelta, Udelta
 
 
 class TestDatePrecision:
@@ -77,3 +79,15 @@ def test_init_from_np_timedelta64(self):
 
     def test_days(self):
         assert Timedelta(10).days == 10
+
+
+class TestUdelta:
+    def test_init(self):
+        # february in an unknown year in Gregorian calendar could be 28 or 29 days
+        february_days = ufloat(28.5, 0.5)  # 28 or 29
+        udelt = Udelta(28, 29)
+        # two ufloat values don't actually compare as equal, due to the variance
+        assert udelt != february_days
+        # so inspect the expected values
+        assert udelt.days.nominal_value == 28.5
+        assert udelt.days.std_dev == 0.5

tests/test_undate.py

@@ -4,7 +4,7 @@
 
 from undate import Undate, UndateInterval, Calendar
 from undate.converters.base import BaseCalendarConverter
-from undate.date import DatePrecision, Timedelta
+from undate.date import DatePrecision, Timedelta, Udelta
 
 
 class TestUndate:
@@ -383,10 +383,26 @@ def test_partiallyknown_duration(self):
         # month in unknown year
         assert Undate(month=6).duration().days == 30
         # partially known month
-        assert Undate(year=1900, month="1X").duration().days == 31
-        # what about february?
-        # could vary with leap years, but assume non-leapyear
-        assert Undate(month=2).duration().days == 28
+        # 1X = October, November, or December = 30 or 31 days
+        # should return a Udelta object
+        unknown_month_duration = Undate(year=1900, month="1X").duration()
+        assert isinstance(unknown_month_duration, Udelta)
+        assert unknown_month_duration.days.nominal_value == 30.5
+        assert unknown_month_duration.days.std_dev == 0.5
+
+        # completely unknown month should also return a Udelta object
+        unknown_month_duration = Undate(year=1900, month="XX").duration()
+        assert isinstance(unknown_month_duration, Udelta)
+        # possible range is 28 to 31 days
+        assert unknown_month_duration.days.nominal_value == 29.5
+        assert unknown_month_duration.days.std_dev == 1.5
+
+        # the number of days in feburary of an unknow year is uncertain, since
+        # it could vary with leap years; either 28 or 29 days
+        feb_duration = Undate(month=2).duration()
+        assert isinstance(feb_duration, Udelta)
+        assert feb_duration.days.nominal_value == 28.5
+        assert feb_duration.days.std_dev == 0.5
 
     def test_known_year(self):
         assert Undate(2022).known_year is True