8000 WIP: handling of nan in unique, fixes #2111 by jaimefrio · Pull Request #5487 · numpy/numpy · GitHub
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WIP: handling of nan in unique, fixes #2111 #5487

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WIP: handling of nan in unique, fixes #2111 #5487

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6 changes: 6 additions & 0 deletions numpy/lib/arraysetops.py
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Original file line number Diff line number Diff line change
Expand Up @@ -196,6 +196,12 @@ def unique(ar, return_index=False, return_inverse=False, return_counts=False):
aux = ar
flag = np.concatenate(([True], aux[1:] != aux[:-1]))

# gh-2111: NaNs never compare equal, so they need special handling,
# but they always sort to the end
if issubclass(ar.dtype.type, np.inexact) and np.isnan(aux[-1]):
nanidx = np.searchsorted(aux, np.nan)
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@mattip mattip Sep 23, 2019 •
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what if you replace this with nanidx = np.nonzero(np.isnan(aux))[0] ?

Edit: ... and the next line with flag[nanindex[:-1] + 1] = False

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Well, that creates an intermediate array of the same size as aux, so it's O(n) in both space and time. The solution above is O(log n) in time and O(1) in space, so it will almost certainly have better performance. It could be argued that your proposed code may be more understandable to a beginner, but I don't think np.searchsorted should be considered advanced or obscure, and should be perfectly fine to use, especially in numpy's source code.

So I guess I like my code better... ;-)

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@mattip's solution has the benefit of working for all types. So maybe we want both?

flag[nanidx+1:] = False

if not optional_returns:
ret = aux[flag]
else:
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12 changes: 12 additions & 0 deletions numpy/lib/tests/test_arraysetops.py
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Expand Up @@ -115,6 +115,18 @@ def check_all(a, b, i1, i2, c, dt):
a2, a2_inv = np.unique(a, return_inverse=True)
assert_array_equal(a2_inv, np.zeros(5))

# test for arrays with nans, gh-2111
a = [5, np.nan, 1, 2, 1, 5, np.nan]*10
b = [1, 2, 5, np.nan]
i1 = [2, 3, 0, 1]
i2 = [2, 3, 0, 1, 0, 2, 3]*10
c = np.multiply([2, 1, 2, 2], 10)
for dt in np.typecodes['Float']:
aa = np.array(a, dt)
bb = np.array(b, dt)
check_all(aa, bb, i1, i2, c, dt)


def test_intersect1d(self):
# unique inputs
a = np.array([5, 7, 1, 2])
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