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BUG: IntervalIndex is_non_overlapping_monotonic is incorrect when endpoints are closed on both sides #16588

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BUG: IntervalIndex is_non_overlapping_monotonic is incorrect when endpoints are closed on both sides #16588

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4989ceb
Fix IntervalIndex is_non_overlapping_monotonic bug where closed='both…
alysivji Jun 2, 2017
bdf1146
Clean up code / add comments + additional test cases. Update WhatsNew…
alysivji Jun 10, 2017
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1 change: 1 addition & 0 deletions .gitignore
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Expand Up @@ -21,6 +21,7 @@
.ipynb_checkpoints
.tags
.cache/
.vscode

# Compiled source #
###################
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8 changes: 6 additions & 2 deletions pandas/core/indexes/interval.py
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Expand Up @@ -556,8 +556,12 @@ def is_non_overlapping_monotonic(self):
# must be increasing (e.g., [0, 1), [1, 2), [2, 3), ... )
# or decreasing (e.g., [-1, 0), [-2, -1), [-3, -2), ...)
# we already require left <= right
return ((self.right[:-1] <= self.left[1:]).all() or
(self.left[:-1] >= self.right[1:]).all())
if self.closed == 'both':
return ((self.right[:-1] < self.left[1:]).all() or
(self.left[:-1] > self.right[1:]).all())
else:
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don't use the else here

add a comment on these cases (left/right).

neighter case? (short circuit to False)

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Working on this now. Why would the neither case short circuit to False? Can't we have intervals [(0, 1), (1, 2)] where it is non-overlapping and also cases where it is overlapping [(0, 1), (.5, 3)]

Thanks!

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this is checking monotonicity as well. so you can do the same for neither.

return ((self.right[:-1] <= self.left[1:]).all() or
(self.left[:-1] >= self.right[1:]).all())

@Appender(_index_shared_docs['_convert_scalar_indexer'])
def _convert_scalar_indexer(self, key, kind=None):
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13 changes: 11 additions & 2 deletions pandas/tests/indexes/test_interval.py
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Expand Up @@ -371,10 +371,19 @@ def slice_locs_cases(self, breaks):
assert index.slice_locs(1, 1) == (1, 1)
assert index.slice_locs(1, 2) == (1, 2)

index = IntervalIndex.from_breaks([0, 1, 2], closed='both')
assert index.slice_locs(1, 1) == (0, 2)
index = IntervalIndex.from_tuples(
[(0, 1), (2, 3), (4, 5)], closed='both')
assert index.slice_locs(1, 1) == (0, 1)
assert index.slice_locs(1, 2) == (0, 2)

def test_is_non_overlapping_monotonic(self):
index = IntervalIndex.from_tuples(
[(0, 1), (2, 3), (4, 5)], closed='both')
assert index.is_non_overlapping_monotonic == True
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use is True


index = IntervalIndex.from_breaks(range(4), closed='both')
assert index.is_non_overlapping_monotonic == False
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use is False


def test_slice_locs_int64(self):
self.slice_locs_cases([0, 1, 2])

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