BUG: the pd.to_timedelta() incorrectly parses the 'M' as minutes.

[IcToxi]

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

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

• (optional) I have confirmed this bug exists on the master branch of pandas.

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Note: Please read this guide detailing how to provide the necessary information for us to reproduce your bug.

Code Sample, a copy-pastable example

>> pd.Timestamp.today() + pd.to_timedelta('1 M')

Problem description

The documentation in version 1.0.5 is described as follows:

Possible values: (‘Y’, ‘M’, ‘W’, ‘D’, ‘days’, ‘day’, ‘hours’, hour’, ‘hr’, ‘h’,...

Since in such a particular order, it should be understood as year, month, and week.

But running my code gives the result of a minute for 'M'. However, 'Y' produces the correct parsing result.

I've seen some discussion there about parameters optimization, and whatever the outcome of the discussion, you shouldn't continue to place this bug here, please either change the function or change the documentation.

And perhaps you might consider separating to_datedelta api to reduce complexity.

Thanks.

Expected Output

Output of pd.show_versions()

INSTALLED VERSIONS

commit : None
python : 3.7.6.final.0
python-bits : 64
OS : Linux
OS-release : 3.10.0-1127.10.1.el7.x86_64
machine : x86_64
processor : x86_64
byteorder : little
LC_ALL : None
LANG : en_US.UTF-8
LOCALE : en_US.UTF-8

pandas : 1.0.5
numpy : 1.18.1

[MJafarMashhadi]

I see that 'M' unit is depricated for the same ambiguity (#16344) and the docs are updated in 095ad41 which will appear in version 1.1.0. But it raises an error for pd.to_timedelta(1, unit='M') but not for pd.to_timedelta('1 M').

[selasley]

From timedeltas.pyx

        if unit == 'M':
            # To parse ISO 8601 string, 'M' should be treated as minute,
            # not month
            unit = 'm'

I'm working on an update to the to_timedelta docs to document this behavior

[IcToxi]

Thanks for the reply.

Sorry, I would like to ask before the issue is closed. Do you no longer consider supporting the delta calculation of the month?

I thought I no longer need to import dateutil every time.

[MJafarMashhadi]

Correct me if I'm wrong; based on the replies on that issue, I guess the largest supported unit is a week now (?). One reason is months can have different lengths, 28 to 31 days; which one should we pick? It's ambiguous!

[jreback]

Timedelta are fixed time representations. You can simply use offsets for any real date begin/end manipulatinos which is why they exist. see https://pandas.pydata.org/pandas-docs/stable/user_guide/timeseries.html#dateoffset-objects

I thought we deprecated M, Y here but maybe not.

I'll repurpose this issue to be a doc one if you'd like to submit a PR to update.

[MJafarMashhadi]

@selasley, you're already working on a PR I guess you might want to comment the word "take" to assign this issue to yourself.
Also, note that Y is deprecated

[selasley]

I submitted PR #34979. Haven't submitted often so I'm not 100% sure I'm doing it correctly. Comments and improvements are welcome.

Using M, y or Y for the units argument in to_datetime raises a ValueError in pandas 1.0.5 and in the 1.1.0 master branch. They are accepted in the arg argument in both versions but their behavior has confused some users. M and Y are no longer listed as accepted units in the docstring for to_timedelta in the 1.1 master branch. The PR is an attempt to document the behavior of M and Y in the arg argument.

>>> pd.__version__
'1.0.5'
>>> pd.to_timedelta('1M')
Timedelta('0 days 00:01:00')
>>> pd.to_timedelta('1y')
Timedelta('365 days 05:49:12')
>>> pd.to_timedelta('1Y')
Timedelta('365 days 05:49:12')
>>> pd.to_timedelta(1, unit='M')
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "/Users/selasley/venvs/py38/lib/python3.8/site-packages/pandas/core/tools/timedeltas.py", line 86, in to_timedelta
    raise ValueError(
ValueError: Units 'M' and 'Y' are no longer supported, as they do not represent unambiguous timedelta values durations.
>>> pd.to_timedelta(1, unit='Y')
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "/Users/selasley/venvs/py38/lib/python3.8/site-packages/pandas/core/tools/timedeltas.py", line 86, in to_timedelta
    raise ValueError(
ValueError: Units 'M' and 'Y' are no longer supported, as they do not represent unambiguous timedelta values durations.

>>> pd.__version__
'1.1.0.dev0+1946.g36b781872'
>>> help(pd.to_timedelta)
...
    unit : str, optional
        Denotes the unit of the arg for numeric `arg`. Defaults to ``"ns"``.
    
        Possible values:
    
        * 'W'
        * 'D' / 'days' / 'day'
        * 'hours' / 'hour' / 'hr' / 'h'
        * 'm' / 'minute' / 'min' / 'minutes' / 'T'
        * 'S' / 'seconds' / 'sec' / 'second'
        * 'ms' / 'milliseconds' / 'millisecond' / 'milli' / 'millis' / 'L'
        * 'us' / 'microseconds' / 'microsecond' / 'micro' / 'micros' / 'U'
        * 'ns' / 'nanoseconds' / 'nano' / 'nanos' / 'nanosecond' / 'N'
...

[MJafarMashhadi]

I'm getting confused too, I believe having Y depreciated means that pd.to_timedelta('1Y') should raise an error. Just like pd.to_timedelta(1, unit='Y') does!

[MarcoGorelli]

closed by #34979