BUG: KeyError from unexpected DatetimeIndex partial str indexing

[Alpima-Quant]

• 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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Code Sample, a copy-pastable example

import pandas as pd
df = pd.DataFrame({'a': [0, 1]}, index=[pd.Timestamp('2002-12-30T20'), pd.Timestamp('2003-01-03T20')])
df['T 1.75 1/3'] = [1, 2]

Problem description

When using __setitem__ or __getitem__ with a str key on a pandas DataFrame where the index has ._supports_partial_string_indexing = True, pandas first tries to convert the str key to a slice using index._get_string_slice(key).

It seems to me that we have too loose of an interpretation of what one of these partial string slices looks like (or perhaps when / how one might use them).

In my use case I have some timeseries data for some US treasury bills, where the names are something like f"T {coupon} {date}", e.g. "T 1.75 1/3" and I want to assign a new column to a DataFrame with some data on this instrument.
However df['T 1.75 1/3'] = value - as in the example above - raises "ValueError: cannot set using a slice indexer with a different length than the value"

Expected Output

A new column with label "T 1.75 1/3" and values [1, 2] is assigned to the pandas DataFrame df.

Output of pd.show_versions()

INSTALLED VERSIONS

commit : f2c8480
python : 3.9.2.final.0
python-bits : 64
OS : Linux
OS-release : 4.19.128-microsoft-standard
Version : #1 SMP Tue Jun 23 12:58:10 UTC 2020
machine : x86_64
processor :
byteorder : little
LC_ALL : None
LANG : C.UTF-8
LOCALE : en_US.UTF-8

pandas : 1.2.3
numpy : 1.20.1
pytz : 2021.1
dateutil : 2.8.1
pip : 21.0.1
setuptools : 53.0.0
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 : 7.21.0
pandas_datareader: None
bs4 : None
bottleneck : None
fsspec : None
fastparquet : None
gcsfs : None
matplotlib : None
numexpr : None
odfpy : None
openpyxl : None
pandas_gbq : None
pyarrow : None
pyxlsb : None
s3fs : None
scipy : None
sqlalchemy : None
tables : None
tabulate : None
xarray : None
xlrd : None
xlwt : None
numba : None

[rhshadrach]

Thanks for the report! Doing this on master, one also gets the message:

FutureWarning: Indexing a DataFrame with a datetimelike index using a single string to slice the rows, like frame[string], is deprecated and will be removed in a future version. Use frame.loc[string] instead.

It appears that once this is removed, this issue will be resolved as well.

[jbrockmendel]

Part of this is also that the Timestamp parser (via the dateutil parser) is interpreting 'T 1.75 1/3' as a datetime when it probably shouldn't

>>> pd.Timestamp('T 1.75 1/3')
Timestamp('2003-01-01 00:00:00')

[suzyahyah]

Perhaps can close this issue;

In v2.3.2, the problem reported

import pandas as pd
df = pd.DataFrame({'a': [0, 1]}, index=[pd.Timestamp('2002-12-30T20'), pd.Timestamp('2003-01-03T20')])
df['T 1.75 1/3'] = [1, 2]

has the expected behavior described in the original post,

A new column with label "T 1.75 1/3" and values [1, 2] is assigned to the pandas DataFrame df.

[rhshadrach]

@suzyahyah - I see the same on main. However #40357 (comment) still produces a Timestamp when I believe it should raise.