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Get rid of MultiIndex conversion in IntervalIndex.is_unique #25159

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Get rid of MultiIndex conversion in IntervalIndex.is_unique #25159

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fbce54f
Remove MultiIndex conversion in is_unique
makbigc Feb 5, 2019
986ce9a
Add benchmark
makbigc Mar 21, 2019
fda328b
New IntervalIndex.is_unique
makbigc Mar 24, 2019
b382b17
Add whatsnew entry
makbigc Mar 24, 2019
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14 changes: 13 additions & 1 deletion asv_bench/benchmarks/index_object.py
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Original file line number Diff line number Diff line change
@@ -1,7 +1,7 @@
import numpy as np
import pandas.util.testing as tm
from pandas import (Series, date_range, DatetimeIndex, Index, RangeIndex,
Float64Index)
Float64Index, IntervalIndex)


class SetOperations(object):
Expand Down Expand Up @@ -181,4 +181,16 @@ def time_get_loc(self):
self.ind.get_loc(0)


class IntervalIndexMethod(object):
# GH 24813
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is this where we asv on is_unique?

def setup(self):
N = 10**5
left = np.append(np.arange(N), np.array(0))
right = np.append(np.arange(1, N + 1), np.array(1))
self.intv = IntervalIndex.from_arrays(left, right)

def time_is_unique(self):
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can you add another benchmark with N**3, e.g. s small case

self.intv.is_unique


from .pandas_vb_common import setup # noqa: F401
1 change: 1 addition & 0 deletions doc/source/whatsnew/v0.25.0.rst
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Expand Up @@ -177,6 +177,7 @@ Performance Improvements
- Improved performance of :meth:`pandas.core.groupby.GroupBy.quantile` (:issue:`20405`)
- Improved performance of :meth:`read_csv` by faster tokenizing and faster parsing of small float numbers (:issue:`25784`)
- Improved performance of :meth:`read_csv` by faster parsing of N/A and boolean values (:issue:`25804`)
- Improved performance of :meth:`IntervalIndex.is_unique` by removing conversion to `MultiIndex` (:issue:`24813`)

.. _whatsnew_0250.bug_fixes:

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8 changes: 7 additions & 1 deletion pandas/core/indexes/interval.py
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Expand Up @@ -463,7 +463,13 @@ def is_unique(self):
"""
Return True if the IntervalIndex contains unique elements, else False
"""
return self._multiindex.is_unique
left = self.values.left
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why isnt the answer just: self.left.is_unique & self.right.is_unique

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@jschendel jschendel Mar 24, 2019 •
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That's a little too strict; you don't necessarily need left or right uniqueness, only pairwise uniqueness, as you can have duplicate endpoints on one side as long as the other side also isn't the same, e.g. [Interval(0, 1), Interval(0, 2), Interval(0, 3)] should be unique.

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i think u can simply construct a numpy array and check that then via np.unique by first reshaping

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@makbigc makbigc Apr 7, 2019 •
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I have tried something below

arr = np.array(list(zip(self.values.left, self.values.right)))
np.unique(arr, axis=0)

But np.unique cannot filter out the duplicate np.nan

In [2]: arr = np.array([1, 3, 3, np.nan, np.nan])                               

In [3]: np.unique(arr)                                                          
Out[3]: array([ 1.,  3., nan, nan])

I​'m still looking for other way.

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use pd.unique

right = self.values.right
for i in range(len(self)):
mask = (left[i] == left) & (right[i] == right)
if mask.sum() > 1:
return False
return True

@cache_readonly
@Appender(_interval_shared_docs['is_non_overlapping_monotonic']
Expand Down