BUG: `date_range` with `freq="C"` (business days) return value changed on 1.5.0

[adamchainz]

Pandas version checks

• I have checked that this issue has not already been reported.

• I have confirmed this bug exists on the latest version of pandas.

• I have confirmed this bug exists on the main branch of pandas.

Reproducible Example

import datetime as dt
import pandas as pd
import pytz

paris = pytz.timezone("Europe/Paris")
start = dt.datetime.fromisoformat("2022-07-22 22:00:00+00:00").astimezone(paris)
print(pd.date_range(start, periods=1, freq="C"))

Issue Description

On Pandas 1.4.4, the above shows:

DatetimeIndex(['2022-07-25 00:00:00+02:00'], dtype='datetime64[ns, Europe/Paris]', freq='C')

On Pandas 1.5.1, it shows:

DatetimeIndex(['2022-07-23 00:00:00+02:00', '2022-07-25 00:00:00+02:00'], dtype='datetime64[ns, Europe/Paris]', freq='C')

There are no warnings from the previous code.

Expected Behavior

THe same results, unless I have missed something in the release notes (I tried reading all results matching date/time/business, couldn't see anything relevant).

Installed Versions

INSTALLED VERSIONS
------------------
commit           : ca60aab7340d9989d9428e11a51467658190bb6b
python           : 3.10.8.final.0
python-bits      : 64
OS               : Darwin
OS-release       : 21.6.0
Version          : Darwin Kernel Version 21.6.0: Mon Aug 22 20:19:52 PDT 2022; root:xnu-8020.140.49~2/RELEASE_ARM64_T6000
machine          : arm64
processor        : arm
byteorder        : little
LC_ALL           : None
LANG             : en_GB.UTF-8
LOCALE           : en_GB.UTF-8
pandas           : 1.4.4

numpy            : 1.23.4

pytz             : 2022.6

dateutil         : 2.8.2

setuptools       : 65.5.0

pip              : 22.3

Cython           : None

pytest           : None

hypothesis       : None

sphinx           : None

blosc            : None

feather          : None

xlsxwriter       : None

lxml.etree       : None

html5lib         : None

pymysql          : None

psycopg2         : None

jinja2           : None

IPython          : None

pandas_datareader: None

bs4              : None

bottleneck       : None

brotli           : None

fastparquet      : None

fsspec           : None

gcsfs            : None

markupsafe       : None

matplotlib       : None

numba            : None

numexpr          : None

odfpy            : None

openpyxl         : None

pandas_gbq       : None

pyarrow          : None

pyreadstat       : None

pyxlsb           : None

s3fs             : None

scipy            : None

snappy           : None

sqlalchemy       : None

tables           : None

tabulate         : None

xarray           : None

xlrd             : None

xlwt             : None

zstandard        : None

[mroeschke]

Thanks for the report. I'm seeing the same on main.

From a quick look, #48385 touched the related code path most recently but the core logic didn't seem to change. cc @jbrockmendel

This looks to be the root cause of the behavior

(Pdb) start
Timestamp('2022-07-23 00:00:00+0200', tz='Europe/Paris')
(Pdb) offset
<CustomBusinessDay>
(Pdb) start + (0 * offset) # = end
Timestamp('2022-07-25 00:00:00+0200', tz='Europe/Paris')  # date rolled forward

[lohmanndouglas]

@mroeschke I would like to contribute to this issue, but I believe that I will require some support as it is my first contribution to pandas.

I did a git bisect and figure out that the issue was introduced in #47019. More specifically in the lines from 273 to 276
.

I reverted the modification, only in these lines, and it worked. I need a better understanding of what is happing and figure out how to write a test case for it.

One possible solution is just to revert does lines, but I think we are looking for a better solution :)

I will continue studying the code to propose a good solution.

[mroeschke]

Thanks for investigating @lohmanndouglas. Since this is a regression, it would be valid to revert those line if it fixes this issue and doesn't fail any existing test. A test should be included in pandas/tests/indexes/datetimes/test_date_range.py that asserts that the OP code example exhibits the 1.4.4 behavior

[lohmanndouglas]

Thanks @mroeschke

I will do it and open a PR as soon as possible :)

[lohmanndouglas]

take

[Natim]

One possible solution is just to revert does lines

Yes it looks good to me indeed.

[mroeschke]

@lohmanndouglas while git bisect points to that commit, IMO the "root" issue still points to the behavior of #49441 (comment), namely why in CustomBusinessDay._apply why we have

    @apply_wraps
    def _apply(self, other):
        if self.n <= 0: # why does this include 0
            roll = "forward"
        else:
            roll = "backward"

[lohmanndouglas]

@mroeschke I did not open the PR because I would like to get a deep understanding of what is happening. We might have changed the function _to_dt64D(dt) behavior, consequently revealing another bug in the _apply function.

From the following snippet

    paris = pytz.timezone("Europe/Paris")
    start = dt.datetime.fromisoformat("2022-07-22 22:00:00+00:00").astimezone(paris)
    print(pd.date_range(start, periods=1, freq="C"))

I got in _to_dt64D(dt) function:

dt (input)	After the #47019	Before the #47019
2022-07-23 00:00:00+02:00	2022-07-22	2022-07-23

1. Is the new behavior correct? It should ignore the TZ?

Knowing that we have two different execution flows in the _generate_range function (here is the logic and the apply function)

---

FLOW 1: Before the #47019

    if start and not offset.is_on_offset(start):
        # Incompatible types in assignment (expression has type "datetime",
        # variable has type "Optional[Timestamp]")
        start = offset.rollforward(start)  # type: ignore[assignment]

The offset.is_on_offset(start) returns False, and the IF conditional evaluates to True, setting the start to
2022-07-25 00:00:00+02:00

Then, we set the end to end = start + (periods - 1) * offset, which means that we set the end to 2022-07-25 00:00:00+02:00.

Finally, we check:

   cur = start
   if offset.n >= 0:
       while cur <= end:
           yield cur
           if cur == end:
               # GH#24252 avoid overflows by not performing the addition
               # in offset.apply unless we have to
           break
           
           ...

As the cur == end is True we just return the solution. In this flow (before #47019) it does not pass through the apply function.

FLOW 2: After the #47019

After #47019, the offset.is_on_offset(start) (from the second python snippet) evaluates to True, and the IF condition turns to False (does not perform the roll forward).

Then we also set the end to end = start + (periods - 1) * offset, but here we have
start = 2022-07-23 00:00:00+02:00 and end = 2022-07-25 00:00:00+02:00.

Finally, we start the while to find out the periods. As the start is not equal to the end we call the apply.

    cur = start
    if offset.n >= 0:
        while cur <= end:
            yield cur         
            if cur == end:
                break
            [... omitted code]
            next_date = offset._apply(cur)
           [... omitted code]
            cur = next_date

2. Here is another point: Should the start consider that day 23 is Saturday and avoid starting with an invalid custom business day?

The apply function returns the correct value 2022-07-25 00:00:00+02:00. In the second iteration, the cur is
cur == end finishing the while loop. At this point, we have the start (2022-07-23 00:00:00+02:00) and the cur (2022-07-25 00:00:00+02:00) as the result of our function.

---

So far I have this understanding. We need to certify what should be the correct behavior for the _to_dt64D(dt) function.
If the changes from #47019 introduced an undesired behavior, it is simple to undo the modification, otherwise, we may investigate more about the start, it should not start with an invalid day.

DOING -> Studying the is_on_offset that uses the _to_dt64D(dt) function. This function is the root evidence of bad behavior from _to_dt64D.

    def is_on_offset(self, dt: datetime) -> bool:
        if self.normalize and not _is_normalized(dt):
            return False
        day64 = _to_dt64D(dt)
        return np.is_busday(day64, busdaycal=self.calendar)

I do not have a deep understanding of this code, this is my first issue, and I am very glad for any help.