Multiple NaN from np.unique with masked arrays

[hdkire]

The new behavior of np.unique seems inconsistent with masked arrays.
Release notes for 1.2.1 https://numpy.org/devdocs/release/1.21.0-notes.html#np-unique-now-returns-single-nan

Reproducing code example:

import numpy as np
a1 = np.ma.masked_array([1.0, 2.0, np.nan, np.nan],
                        [False, True, False, False])
unique_a1 = np.unique(a1)
print('a1', a1)
# ... a1 [1.0 -- nan nan]
print('unique_a1', unique_a1)
# ... unique_a1 [1.0 -- nan]
# Ok, single NaN.
assert np.count_nonzero(np.isnan(unique_a1)) == 1

a2 = np.ma.masked_array([1.0, 2.0, np.nan, np.nan, 3],
                        [False, True, False, False, True])
unique_a2 = np.unique(a2)
print('a2', a2)
# ... a2 [1.0 -- nan nan --]
print('unique_a2', unique_a2)
# ... unique_a2 [1.0 -- nan nan --]
# Fail, 2 NaNs.
assert np.count_nonzero(np.isnan(unique_a2)) == 1

NumPy/Python version information:

1.21.0 3.9.5 (default, May 11 2021, 08:20:37)
[GCC 10.3.0]

[hdkire]

import numpy as np
a3 = np.ma.masked_array([np.nan, np.nan, 2.0, np.nan, np.nan],
                        [True, False, False, True, False])
unique_a3 = np.unique(a3)
print('a3', a3)
# ... a3 [-- nan 2.0 -- nan]
print('unique_a3', np.unique(a3))
# ... unique_a3 [2.0 --]
# Fail, 0 NaNs.
assert np.count_nonzero(np.isnan(unique_a3)) == 1

[rossbar]

Related to the post-merge discussion in #19301

[KiranHipparagi]

Can I work on this issues?

[KiranHipparagi]

As of Numpy version 1.21.0, np.unique now returns single NaN:

a = np.array([8, 1, np.nan, 3, np.inf, np.nan, -np.inf, -2, np.nan, 3]) np.unique(a)
array([-inf, -2., 1., 3., 8., inf, nan])

[Dinkarkumar]

Is the issue still open ? Can i work on this ?