忽略NaN的Python pandas 的独特价值 [英] Python pandas unique value ignoring NaN

问题描述

我想在groupby聚合中使用unique，但是我不想在unique结果中使用nan.

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

示例数据框:

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

和groupby:

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

其结果是:

       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]

但是我想要没有nan的

       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]

我该怎么做?当然，我有几个要聚合的列，每个列都需要不同的聚合函数，所以我不想与其他聚合一一对应地进行unique聚合.

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.

谢谢！

推荐答案

尝试ffill

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

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

如果Nan是该组的第一个元素，则上述解决方案将失效.从长远来看，@IanS的解决方案更好.

If Nan is the first element of the group then the above solution breaks. @IanS's solution is better in the long run.

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