how to find the unique non nan values in a numpy array?

As previous answers have already stated, numpy can't count nans directly, because it can't compare nans. numpy.ma.count_masked is your friend. For example, like this:

>>> import numpy.ma as ma
>>> a = np.array([ 0.,  1., np.nan, np.nan,  4.])
>>> a
np.array([ 0.,  1., nan, nan,  4.])
>>> a_masked = ma.masked_invalid(a)
>>> a_masked
masked_array(data=[0.0, 1.0, --, --, 4.0],
             mask=[False, False,  True,  True, False],
       fill_value=1e+20)
>>> ma.count_masked(a_masked)
2

I'd suggest using pandas. I think it's a direct replacement, but pandas keeps the original order unlike numpy.

import numpy as np
import pandas as pd

my_array1=np.array([5,4,2,2,4,np.nan,np.nan,6])

np.unique(my_array1)
# array([ 2.,  4.,  5.,  6., nan, nan])

pd.unique(my_array1)
# array([ 5.,  4.,  2., nan,  6.]) 

I'm using numpy 1.17.4 and pandas 0.25.3. Hope this helps!

You can use np.unique to find unique values in combination with isnan to filter the NaN values:

In [22]:

my_array1=np.array([5,4,2,2,4,np.nan,np.nan,6])
np.unique(my_array1[~np.isnan(my_array1)])
Out[22]:
array([ 2.,  4.,  5.,  6.])

as to why you get multiple NaN values it's because NaN values cannot be compared normally:

In [23]:

np.nan == np.nan
Out[23]:
False

so you have to use isnan to perform the correct comparison

using set:

In [24]:

set(my_array1[~np.isnan(my_array1)])
Out[24]:
{2.0, 4.0, 5.0, 6.0}

You can call len on any of the above to get a size:

In [26]:

len(np.unique(my_array1[~np.isnan(my_array1)]))
Out[26]:
4