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BUG: Index.union() drops Index component for duplicate Index elements not sorted #36289

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@phofl

Description

@phofl
  • I have checked that this issue has not already been reported.

  • I have confirmed this bug exists on the latest version of pandas.

  • (optional) I have confirmed this bug exists on the master branch of pandas.


Note: Please read this guide detailing how to provide the necessary information for us to reproduce your bug.

Code Sample, a copy-pastable example

import pandas as pd

idx1 = pd.Index([0, 0, 1])
idx2 = pd.Index([0, 1])

print(idx1.union(idx2))
# Int64Index([0, 0, 1], dtype='int64')
print(idx2.union(idx1))
# Int64Index([0, 0, 1], dtype='int64')

idx1 = pd.Index([1, 0, 0])

print(idx1.union(idx2))
# Int64Index([0, 0, 1], dtype='int64')
print(idx2.union(idx1))
# Int64Index([0, 1], dtype='int64')

Problem description

In the fourth call idx2.union(idx1) one of the 0 is dropped. Meaning union() depends on the call order and on the index order, which seems buggy.

Expected Output

Int64Index([0, 0, 1], dtype='int64') for every print Statement

Output of pd.show_versions()

master

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