Python pandas unique value ignoring NaN

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I want to use unique in groupby aggregation, but I don't want nan in the unique result.

An example dataframe:

df = pd.DataFrame({'a': [1, 2, 1, 1, np.nan, 3, 3], 'b': [0,0,1,1,1,1,1],
    'c': ['foo', np.nan, 'bar', 'foo', 'baz', 'foo', 'bar']})

       a  b    c
0 1.0000  0  foo
1 2.0000  0  NaN
2 1.0000  1  bar
3 1.0000  1  foo
4    nan  1  baz
5 3.0000  1  foo
6 3.0000  1  bar

And the groupby:

df.groupby('b').agg({'a': ['min', 'max', 'unique'], 'c': ['first', 'last', 'unique']})

Its result is:

       a                             c                      
     min    max           unique first last           unique
b                                                           
0 1.0000 2.0000       [1.0, 2.0]   foo  foo       [foo, nan]
1 1.0000 3.0000  [1.0, nan, 3.0]   bar  bar  [bar, foo, baz]

But I want it without nan:

       a                        c                      
     min    max      unique first last           unique
b                                                           
0 1.0000 2.0000  [1.0, 2.0]   foo  foo            [foo]
1 1.0000 3.0000  [1.0, 3.0]   bar  bar  [bar, foo, baz]

How can I do that? Of course I have several columns to aggregate and every column needs different aggregation functions, so I don't want to do the unique aggregations one-by-one and separately from other aggregations.

pandas unique group-by null python
4 Answers

You can use the below code,

    df.apply(lambda x: len(x.dropna().unique()))
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