timedelta additions to pendulum.DateTime

[andreivnegrean]

• I am on the latest Pendulum version.

• I have searched the issues of this repo and believe that this is not a duplicate.

• OS version and name: macOS 13.2.1 (22D68)

• Pendulum version: 3.0.0

Issue

Hi,

It seems like pendulum.DateTime additions of timedelta don't give the same result as in the case of pendulum.Duration additions when crossing a DST timezone boundary.

Examples:

assert DateTime(year=2024, month=11, day=3, hour=1, tzinfo=Timezone("America/Los_Angeles")) + timedelta(days=30) == DateTime(year=2024, month=12, day=3, hour=1, tzinfo=Timezone("America/Los_Angeles"))

throws the following:

AssertionError: assert DateTime(2024, 12, 3, 0, 0, 0, tzinfo=Timezone('America/Los_Angeles')) == DateTime(2024, 12, 3, 1, 0, 0, tzinfo=Timezone('America/Los_Angeles'))

the same assertion passes if adding a Duration instead of timedelta:

assert DateTime(year=2024, month=11, day=3, hour=1, tzinfo=Timezone("America/Los_Angeles")) + Duration(months=1) == DateTime(year=2024, month=12, day=3, hour=1, tzinfo=Timezone("America/Los_Angeles"))

The issue is also reproducible when adding weeks:

assert DateTime(year=2024, month=11, day=3, hour=1, tzinfo=Timezone("America/Los_Angeles")) + timedelta(weeks=1) == DateTime(year=2024, month=11, day=10, hour=1, tzinfo=Timezone("America/Los_Angeles"))

[ariebovenberg]

I suspect this may be because months in Duration is counted on the calendar (1am one month later is still 1am), while days in timedelta considers days to always be 24 hours (1am + 30×24h may not result in 1am again due to DST transitions).

[ericfrederich]

I came across this issue as well. This bug is more than 1 year old, I thought about creating a new issue but instead will paste the new bug template here...

• I am on the latest Pendulum version.
• I have searched the issues of this repo and believe that this is not a duplicate.

• OS version and name: Ubuntu 24.04 (via WSL)
• Pendulum version: 3.1.0

Issue

Given two datetimes d1 and d2 I want to calculate a datetime double such that d2 is the midpoint between d1 and double.

I calculated it 4 different ways but one of them is off by more than 3 days.

double1 = (d1 + (delta * 2)).in_tz("US/Eastern")
double2 = (d2 + delta).in_tz("US/Eastern")
double3 = d1.add(minutes=delta.total_minutes() * 2).in_tz("US/Eastern")
double4 = d2.add(minutes=delta.total_minutes()).in_tz("US/Eastern")

Full Code

You can play around with the input datetimes. I've seen it be off by weird amounts... 3 days, 1 day, 1, hour. I've even seen it match depending on the datetimes. This example has a difference of -3 days -1 hour

#!/usr/bin/env python3

import pendulum

birthday, birthday_tz, event, event_tz = "2025-01-01 12:00:00", "US/Central", "2025-02-08 12:00:00", "US/Eastern"
d1 = pendulum.parse(birthday, tz=birthday_tz)
d2 = pendulum.parse(event, tz=event_tz)
delta = d2 - d1

# calculate the same point in time using different methods
# only double2 is different from the others
double1 = (d1 + (delta * 2)).in_tz("US/Eastern")
double2 = (d2 + delta).in_tz("US/Eastern")
double3 = d1.add(minutes=delta.total_minutes() * 2).in_tz("US/Eastern")
double4 = d2.add(minutes=delta.total_minutes()).in_tz("US/Eastern")

for double in (double1, double2, double3, double4):
    print(f"{double.to_day_datetime_string()} {double.timezone_name}")

print("Equality matrix:")
for i, d_i in enumerate([double1, double2, double3, double4]):
    print("    ", end="")
    for j, d_j in enumerate([double1, double2, double3, double4]):
        if i == j:
            print(".", end="")
        elif d_i == d_j:
            print("=", end="")
        else:
            print("X", end="")
    print()

off_by = double2 - double1
print(f"Off by {off_by.in_words()} ({off_by.in_seconds()} seconds)")

Output

Tue, Mar 18, 2025 12:00 PM US/Eastern
Sat, Mar 15, 2025 11:00 AM US/Eastern
Tue, Mar 18, 2025 12:00 PM US/Eastern
Tue, Mar 18, 2025 12:00 PM US/Eastern
Equality matrix:
    .X==
    X.XX
    =X.=
    =X=.
Off by -3 days -1 hour (-262800 seconds)

[ericfrederich]

I have found that if I multiply the duration object by 1 I get the correct result.

(d1 + (delta * 2)).in_tz("US/Eastern") != (d2 + delta).in_tz("US/Eastern")

(d1 + (delta * 2)).in_tz("US/Eastern") == (d2 + (delta * 1)).in_tz("US/Eastern")