datedelta

datedelta.datedelta is datetime.timedelta for date arithmetic.

>>> import datetime
>>> import datedelta

>>> datetime.date(2025, 4, 22) + 2 * datedelta.WEEK
datetime.date(2025, 5, 6)

>>> datetime.date(2025, 4, 22) + 3 * datedelta.MONTH
datetime.date(2025, 7, 22)

It accounts for oddities of the Gregorian calendar.

>>> datetime.date(2024, 2, 29) + datedelta.YEAR
datetime.date(2025, 3, 1)

>>> datetime.date(2024, 2, 29) + 4 * datedelta.YEAR
datetime.date(2028, 2, 29)

It's convenient for computing yearly, monthly, or weekly subscriptions periods.

>>> start_date = datetime.date(2024, 1, 30)
>>> for n in range(12):
...     print(repr(start_date + n * datedelta.MONTH))
datetime.date(2024, 1, 30)
datetime.date(2024, 3, 1)
datetime.date(2024, 3, 30)
datetime.date(2024, 4, 30)
datetime.date(2024, 5, 30)
datetime.date(2024, 6, 30)
datetime.date(2024, 7, 30)
datetime.date(2024, 8, 30)
datetime.date(2024, 9, 30)
datetime.date(2024, 10, 30)
datetime.date(2024, 11, 30)
datetime.date(2024, 12, 30)

>>> start_date = datetime.date(2024, 1, 31)
>>> for n in range(12):
...     print(repr(start_date + n * datedelta.MONTH))
datetime.date(2024, 1, 31)
datetime.date(2024, 3, 1)
datetime.date(2024, 3, 31)
datetime.date(2024, 5, 1)
datetime.date(2024, 5, 31)
datetime.date(2024, 7, 1)
datetime.date(2024, 7, 31)
datetime.date(2024, 8, 31)
datetime.date(2024, 10, 1)
datetime.date(2024, 10, 31)
datetime.date(2024, 12, 1)
datetime.date(2024, 12, 31)

It guarantees consistent results on arithmetic operations that it supports.

Behavior

There are two date arithmetic traps in the Gregorian calendar:

Leap years. Problems arise when adding years to a February 29th gives a result in a non-leap year.
Variable number of days in months. Problems arise when adding months to a 29th, 30th or 31st gives a result in a month where that day doesn't exist.

In both cases, datedelta changes the result to the first day of the next month.

This method gives consistent results provided periods are represented by (start date inclusive, end date exclusive) — that's [start date, end date) if you prefer the mathematical notation. This representation of periods is akin to 0-based indexing, which is the convention Python uses.

For example:

If someone subscribes for a year starting on 2020-02-29 inclusive, the end date must be 2021-03-01 exclusive. If it was 2020-02-28 exclusive, that day would be missing from the subscription period.
If someone subscribes for three months starting on 2020-03-31 inclusive, the end date must be 2020-07-01 exclusive. If it was 2020-06-30 exclusive, that day would be missing from the subscription period.

Operations are always performed on years, then months, then days. This order usually provides the expected behavior. It also minimizes loss of precision.

Installation

pip install datedelta

Usage

The most common operations are adding a datedelta to a date and subtracting a datedelta from a date.

Basic intervals

The YEAR, MONTH, WEEK, and DAY constants support common calculations with little code.

>>> import datetime
>>> import datedelta

>>> datetime.date(2022, 1, 1) + datedelta.YEAR
datetime.date(2023, 1, 1)

>>> datetime.date(2023, 1, 1) - datedelta.YEAR
datetime.date(2022, 1, 1)

>>> datetime.date(2024, 2, 29) + datedelta.YEAR
datetime.date(2025, 3, 1)

>>> datetime.date(2025, 3, 1) - datedelta.YEAR
datetime.date(2024, 3, 1)

>>> datetime.date(2022, 1, 1) + datedelta.MONTH
datetime.date(2022, 2, 1)

>>> datetime.date(2022, 2, 1) - datedelta.MONTH
datetime.date(2022, 1, 1)

>>> datetime.date(2022, 1, 31) + datedelta.MONTH
datetime.date(2022, 3, 1)

>>> datetime.date(2022, 3, 1) - datedelta.MONTH
datetime.date(2022, 2, 1)

>>> datetime.date(2022, 1, 1) + datedelta.WEEK
datetime.date(2022, 1, 8)

>>> datetime.date(2022, 1, 1) - datedelta.WEEK
datetime.date(2021, 12, 25)

>>> datetime.date(2022, 1, 1) + datedelta.DAY
datetime.date(2022, 1, 2)

>>> datetime.date(2022, 1, 1) - datedelta.DAY
datetime.date(2021, 12, 31)

datedelta.DAY behaves exactly like datetime.timedelta(1). It's only provided for consistency.

Arbitrary intervals

datedelta objects provide support for arbitrary calculations.

>>> import datetime
>>> import datedelta

>>> datetime.date(2022, 3, 23) + datedelta.datedelta(years=1, months=1, days=-1)
datetime.date(2023, 4, 22)

>>> datetime.date(2022, 3, 23) - datedelta.datedelta(years=-1, months=-1, days=1)
datetime.date(2023, 4, 22)

>>> datetime.date(2024, 2, 29) + datedelta.datedelta(years=2)
datetime.date(2026, 3, 1)

>>> datetime.date(2024, 2, 29) - datedelta.datedelta(years=2)
datetime.date(2022, 3, 1)

>>> datetime.date(2024, 2, 29) + datedelta.datedelta(years=2, days=-1)
datetime.date(2026, 2, 28)

>>> datetime.date(2024, 2, 29) - datedelta.datedelta(years=2, days=1)
datetime.date(2022, 2, 28)

>>> datetime.date(2024, 2, 29) + datedelta.datedelta(years=2, months=6)
datetime.date(2026, 9, 1)

>>> datetime.date(2024, 2, 29) - datedelta.datedelta(years=2, months=-6)
datetime.date(2022, 9, 1)

>>> datetime.date(2024, 2, 29) + datedelta.datedelta(years=4)
datetime.date(2028, 2, 29)

>>> datetime.date(2024, 2, 29) - datedelta.datedelta(years=4)
datetime.date(2020, 2, 29)

>>> datetime.date(2024, 2, 29) + datedelta.datedelta(years=4, days=1)
datetime.date(2028, 3, 1)

>>> datetime.date(2024, 2, 29) - datedelta.datedelta(years=4, days=-1)
datetime.date(2020, 3, 1)

>>> datetime.date(2024, 2, 29) + datedelta.datedelta(years=4, months=6)
datetime.date(2028, 8, 29)

>>> datetime.date(2024, 2, 29) - datedelta.datedelta(years=4, months=-6)
datetime.date(2020, 8, 29)

These results are mathematically consistent, as explained in "Behavior" above.

Other operations

datedelta instances can be added, subtracted, and multiplied with an integer. However, there are some restrictions on addition and subtraction.

Adding then subtracting a given datedelta to a date doesn't always return the original date. In order to prevent bugs caused by this behavior, when the result of adding or subtracting two datedelta isn't well defined, a ValueError is raised.

>>> import datedelta

>>> datedelta.YEAR + datedelta.YEAR
datedelta.datedelta(years=2)

>>> 3 * datedelta.YEAR
datedelta.datedelta(years=3)

>>> datedelta.YEAR - datedelta.DAY
datedelta.datedelta(years=1, days=-1)

>>> datedelta.YEAR - datedelta.YEAR
Traceback (most recent call last):
    ...
ValueError: cannot subtract datedeltas with same signs

>>> datedelta.datedelta(months=6) + datedelta.datedelta(months=-3)
Traceback (most recent call last):
    ...
ValueError: cannot add datedeltas with opposite signs

Limitations

Additions involving datedelta are neither associative nor commutative in general.

Here are two examples where adding a datedelta then subtracting it doesn't return the original value:

>>> import datetime
>>> import datedelta

>>> datetime.date(2024, 2, 29) + datedelta.datedelta(years=1)
datetime.date(2025, 3, 1)

>>> datetime.date(2025, 3, 1) - datedelta.datedelta(years=1)
datetime.date(2024, 3, 1)

>>> datetime.date(2024, 1, 31) + datedelta.datedelta(months=1)
datetime.date(2024, 3, 1)

>>> datetime.date(2024, 3, 1) - datedelta.datedelta(months=1)
datetime.date(2024, 2, 1)

Here are two examples where adding two datedelta gives a different result depending on the order of operations:

>>> import datetime
>>> import datedelta

>>> datetime.date(2024, 2, 29) + datedelta.datedelta(months=6) + datedelta.datedelta(years=1)
datetime.date(2025, 8, 29)

>>> datetime.date(2024, 2, 29) + datedelta.datedelta(years=1) + datedelta.datedelta(months=6)
datetime.date(2025, 9, 1)

>>> datetime.date(2024, 1, 31) + datedelta.datedelta(months=2) + datedelta.datedelta(months=5)
datetime.date(2024, 8, 31)

>>> datetime.date(2024, 1, 31) + datedelta.datedelta(months=5) + datedelta.datedelta(months=2)
datetime.date(2024, 9, 1)

To avoid problems, you should always start from the same reference date and add a single datedelta. Don't chain additions or subtractions.

To minimize the risk of incorrect results, datedelta only implements operations that have unambiguous semantics:

Adding a datedelta to a date
Subtracting a datedelta from a date
Adding a datedelta to a datedelta when components have the same sign
Subtracting a datedelta from a datedelta when components have opposite signs

(PEP 20 says: "In the face of ambiguity, refuse the temptation to guess.")

Alternatives

datedelta.datedelta is smarter than datetime.timedelta because it knows about years and months in addition to days.

Compared to pendulum.Duration or dateutil.relativedelta.relativedelta, datedelta.datedelta has a few benefits:

It handles non-existing results in a mathematically consistent fashion: it adjusts to the first day of the next month while pendulum and dateutil adjust to the last day of the current month.
It provides an API designed to prevent programming mistakes: it requires keyword arguments, rejects operations expressing incorrect business logic, and omits error-prone features of dateutil like the "replace" behavior or explicit control of leap days.
It has very small footprint, by virtue of providing of very small subset of the features found in pendulum or dateutil. That makes it a good choice if you're otherwise happy with the standard library's datetime module.

Changelog

1.4

Update supported Python versions.

1.3

Add WEEK constant.

1.2

Optimize hashing and pickling.

1.1

Add YEAR, MONTH, and DAY constants.

1.0

Initial stable release.

aaugustin/datedelta

aaugustin

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