Description
Code Sample, a copy-pastable example if possible
In [2]: idx1 = pd.IntervalIndex.from_breaks(range(5))
...: idx1
...:
Out[2]:
IntervalIndex([(0, 1], (1, 2], (2, 3], (3, 4]]
closed='right',
dtype='interval[int64]')
In [3]: idx2 = pd.Index([2, 3])
...: idx2
...:
Out[3]: Int64Index([2, 3], dtype='int64')
In [4]: idx1.symmetric_difference(idx2)
Out[4]:
IntervalIndex([(0, 1], (3, 4]]
closed='right',
dtype='interval[int64]')
Seems to be working properly between two IntervalIndex
:
In [5]: idx1[:-1].symmetric_difference(idx1[1:])
Out[5]:
IntervalIndex([(0, 1], (3, 4]]
closed='right',
dtype='interval[int64]')
Problem description
Set ops between an IntervalIndex
and non-IntervalIndex
should raise.
Seems to be caused by a typo:
pandas/pandas/core/indexes/interval.py
Line 1167 in 412988e
Using the typo version raises:
In [6]: idx1.symmetric_differnce(idx2)
---------------------------------------------------------------------------
ValueError: can only do set operations between two IntervalIndex objects that are closed on the same side
The typo version also seems to have better perf:
In [11]: idx3 = pd.IntervalIndex.from_breaks(range(10**5))
In [12]: %timeit idx3[:-1].symmetric_difference(idx3[1:])
1.25 s ± 23.8 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
In [13]: %timeit idx3[:-1].symmetric_differnce(idx3[1:])
153 ms ± 6.54 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
Output of pd.show_versions()
pandas: 0.22.0.dev0+213.g412988e
pytest: 3.1.2
pip: 9.0.1
setuptools: 27.2.0
Cython: 0.25.2
numpy: 1.13.1
scipy: 0.19.1
pyarrow: 0.6.0
xarray: 0.9.6
IPython: 6.1.0
sphinx: 1.5.6
patsy: 0.4.1
dateutil: 2.6.0
pytz: 2017.2
blosc: None
bottleneck: None
tables: 3.4.2
numexpr: 2.6.2
feather: 0.4.0
matplotlib: 2.0.2
openpyxl: 2.4.8
xlrd: 1.1.0
xlwt: 1.3.0
xlsxwriter: 0.9.8
lxml: 3.8.0
bs4: None
html5lib: 0.999
sqlalchemy: 1.1.13
pymysql: None
psycopg2: None
jinja2: 2.9.6
s3fs: None
fastparquet: 0.1.0
pandas_gbq: None
pandas_datareader: None