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

[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()

Details

master

[dsaxton]

Agreed this seems like a bug @phofl. This may be a duplicate of #31326.

[phofl]

Similar but not a duplicate. If the indexes both have the same elements the error is at another place than the other example.