x in pd.MultiIndex.drop(x)

[toobaz]

Code Sample, a copy-pastable example if possible

In [2]: 2 in pd.MultiIndex.from_product([[1, 2], [3, 4]]).drop(2)
Out[2]: True

Problem description

#2770 was actually more complex than what I understood when I suggested to close it. While remove_unused_levels() did solve the problem of just resetting the .levels, the above is still buggy. Ironically, the above is even tested.

Expected Output

False

Output of pd.show_versions()

INSTALLED VERSIONS

commit: 8433562
python: 3.5.3.final.0
python-bits: 64
OS: Linux
OS-release: 4.9.0-4-amd64
machine: x86_64
processor:
byteorder: little
LC_ALL: None
LANG: it_IT.UTF-8
LOCALE: it_IT.UTF-8

pandas: 0.23.0.dev0+10.g84335621a
pytest: 3.2.3
pip: 9.0.1
setuptools: 36.7.0
Cython: 0.25.2
numpy: 1.12.1
scipy: 0.19.0
pyarrow: None
xarray: None
IPython: 6.2.1
sphinx: 1.5.6
patsy: 0.4.1
dateutil: 2.6.1
pytz: 2017.2
blosc: None
bottleneck: 1.2.0dev
tables: 3.3.0
numexpr: 2.6.1
feather: 0.3.1
matplotlib: 2.0.0
openpyxl: 2.3.0
xlrd: 1.0.0
xlwt: 1.3.0
xlsxwriter: 0.9.6
lxml: 4.1.1
bs4: 4.5.3
html5lib: 0.999999999
sqlalchemy: 1.0.15
pymysql: None
psycopg2: None
jinja2: 2.10
s3fs: None
fastparquet: None
pandas_gbq: None
pandas_datareader: 0.2.1

[jreback]

this is a defined behavior. why should this not be True? The point is we don't perform this optimization as we don't really know when to do it.

[toobaz]

It should not be True... because it's not in the index: being in the .levels is an implementation detail.

And while I agree about not optimizing implicitly, then we must check both that 2 is in .levels[0] and that the relative code is in the .labels[0]. Or more simply: 2 in a_multiindex should just be implemented as (2,) in a_multiindex, which already returns the correct result.

[jreback]

ok I suppose this is inconsistent. .get_loc should also KeyError (correct now).

[toobaz]

Right!