DEV: Error in `main` branch when subtracting a valid numpy timedelta from a Series of timestamps

[Dr-Irv]

import pandas as pd
import numpy as np
s1 = pd.Series(data=pd.date_range("1/1/2020", "2/1/2020"))
td = np.timedelta64(1, "M")
s1-td

With version 2.1.0.dev0+1159.gc126eeb392 and numpy 1.24.3, get the following error:

  File "C:\Code\pandas_dev\pandas\pandas\core\ops\common.py", line 85, in new_method
    return method(self, other)
  File "C:\Code\pandas_dev\pandas\pandas\core\arraylike.py", line 194, in __sub__
    return self._arith_method(other, operator.sub)
  File "C:\Code\pandas_dev\pandas\pandas\core\series.py", line 5735, in _arith_method
    return base.IndexOpsMixin._arith_method(self, other, op)
  File "C:\Code\pandas_dev\pandas\pandas\core\base.py", line 1342, in _arith_method
    rvalues = ops.maybe_prepare_scalar_for_op(rvalues, lvalues.shape)
  File "C:\Code\pandas_dev\pandas\pandas\core\ops\array_ops.py", line 575, in maybe_prepare_scalar_for_op
    return Timedelta(obj)
  File "timedeltas.pyx", line 1846, in pandas._libs.tslibs.timedeltas.Timedelta.__new__
ValueError:  cannot construct a Timedelta from a unit M

Tagging @mcgeestocks since last commit related to this code made by him and @mroeschke since he approved the PR #53421

[mcgeestocks]

I believe this is the new desired behavior. You shouldn't be able to construct timedelta64 from ambiguous time units.

ie. m for month can either be 30 or 31 days.
see #52806 for more details.

[Dr-Irv]

I believe this is the new desired behavior. You shouldn't be able to construct timedelta64 from ambiguous time units.

ie. m for month can either be 30 or 31 days. see #52806 for more details.

While that's true for pandas, numpy is allowing it, so we need to make sure we are either issuing a relevant message when converting a np.timedelta64 with that unit in an arithmetic operation, or convert it to something.

[mcgeestocks]

take

[mcgeestocks]

per discussion going to update to a improved err msg.