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[BUG] IntervalIndex.get_loc error for decreasing index #25860

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makbigc opened this issue Mar 24, 2019 · 1 comment

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@makbigc
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commented Mar 24, 2019

Code Sample, a copy-pastable example if possible

In [1]: import pandas as pd                                                     

In [2]: intv1 = pd.IntervalIndex.from_arrays([1, 3, 5], [2, 4, 6])              

In [3]: intv1                                                                   
Out[3]: 
IntervalIndex([(1, 2], (3, 4], (5, 6]],
              closed='right',
              dtype='interval[int64]')

In [4]: intv1.get_loc(intv1[0])                                                 
Out[4]: 0

In [5]: intv2 = pd.IntervalIndex.from_arrays([4, 2, 1], [5, 3, 2])              

In [6]: intv2                                                                   
Out[6]: 
IntervalIndex([(4, 5], (2, 3], (1, 2]],
              closed='right',
              dtype='interval[int64]')

In [7]: intv2.get_loc(intv2[0])                                                 
---------------------------------------------------------------------------
KeyError                                  Traceback (most recent call last)
<ipython-input-7-00c58bb4fb1e> in <module>
----> 1 intv2.get_loc(intv2[0])

~/Code/pandas/pandas/core/indexes/interval.py in get_loc(self, key, method)
    777                 return slice(start, stop)
    778             else:
--> 779                 raise KeyError(original_key)
    780 
    781         else:

KeyError: Interval(4, 5, closed='right')

Problem description

This IntervalIndex is non-overlapping_monotonic. IntervalIndex.get_loc runs in KeyError, even the key is its first element.

After start, stop = self._find_non_overlapping_monotonic_bounds(key), start and stop get the same value. Thereafter, no conditional handles this case.

Expected Outcome

In [5]: intv2 = pd.IntervalIndex.from_arrays([4, 2, 1], [5, 3, 2])

In [6]: intv2
Out[6]:
IntervalIndex([(4, 5], (2, 3], (1, 2]],
closed='right',
dtype='interval[int64]')

In [7]: intv2.get_loc(intv2[0])
Out[7]: 0

Output of pd.show_versions()

INSTALLED VERSIONS

commit: 0198873
python: 3.7.1.final.0
python-bits: 64
OS: Linux
OS-release: 4.18.0-16-generic
machine: x86_64
processor: x86_64
byteorder: little
LC_ALL: None
LANG: en_HK.UTF-8
LOCALE: en_HK.UTF-8

pandas: 0.25.0.dev0+308.g01988735f
pytest: 4.0.2
pip: 18.1
setuptools: 40.6.3
Cython: 0.29.2
numpy: 1.15.4
scipy: 1.1.0
pyarrow: None
xarray: None
IPython: 7.2.0
sphinx: 1.8.2
patsy: 0.5.1
dateutil: 2.7.5
pytz: 2018.7
blosc: None
bottleneck: 1.2.1
tables: 3.4.4
numexpr: 2.6.8
feather: None
matplotlib: 3.0.2
openpyxl: 2.5.12
xlrd: 1.2.0
xlwt: 1.3.0
xlsxwriter: 1.1.2
lxml.etree: 4.2.5
bs4: 4.6.3
html5lib: 1.0.1
sqlalchemy: 1.2.15
pymysql: None
psycopg2: None
jinja2: 2.10
s3fs: None
fastparquet: None
pandas_gbq: None
pandas_datareader: None
gcsfs: None

@jschendel

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commented Mar 25, 2019

Thanks, I can confirm this behavior on master.

For what it's worth, I've been rewriting the IntervalIndex indexing methods as part of #16316 and this issue is not present in my new implementation:

In [2]: intv1 = pd.IntervalIndex.from_arrays([1, 3, 5], [2, 4, 6])

In [3]: intv1.get_loc(intv1[0])
Out[3]: 0

In [4]: intv2 = pd.IntervalIndex.from_arrays([4, 2, 1], [5, 3, 2])

In [5]: intv2.get_loc(intv2[0])
Out[5]: 0

Overall that branch is still a WIP and going somewhat slowly but I should have it ready in time for 0.25.0. Would still take a fix in the meantime but be aware that it might get overwritten shortly thereafter.

@jreback jreback added this to the 0.25.0 milestone Jul 1, 2019

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