大熊猫在每一组中最长的最后一段

Question

我有以下潘达斯的数据：

import pandas as pd

    df = pd.DataFrame(
    
        [
            ("bird", '2022-01',"Falconiformes"),
            ("bird", '2022-02',"Falconiformes"),
            ("bird", '2022-03',"Falconiformes"),
            ("bird", '2022-04',"Falconiformes"),
            ("bird", '2022-05',"Falconiformes"),
            ("bird", '2022-06',"Falconiformes"),
            ("bird", '2022-07',"Falconiformes"),
            ("bird", '2022-08',"Falconiformes"),
            ("bird", '2022-09',"Psittaciformes"),
            ("bird", '2022-10',"Psittaciformes"),
            ("bird", '2022-11',"Psittaciformes"),
            ("bird", '2022-12',"Psittaciformes"),
            ("mammal", '2022-01',"Falconiformes"),
            ("mammal", '2022-02',"Falconiformes"),
            ("mammal",'2022-03',"Falconiformes"),
            ("mammal", '2022-04',"Falconiformes"),
            ("mammal",'2022-05',"Falconiformes"),
            ("mammal", '2022-06',"Psittaciformes"),
            ("mammal", '2022-07',"Falconiformes"),
            ("mammal", '2022-08',"Falconiformes"),
            ("mammal", '2022-09',"Falconiformes"),
            ("mammal", '2022-10',"Falconiformes"),
            ("mammal", '2022-11',"Falconiformes"),
            ("mammal", '2022-12',"Falconiformes"),
        
    
        ],
    
        columns=("animal", "date", "attribute"),
    )

现在事情变得越来越复杂了。对于每一种动物，我想要该组中最新的连续值序列的计数。

结果应该是

result = pd.DataFrame(
    [   ("bird", 'Psittaciformes' ,4),
        ("mammal", 'Falconiformes' ,6),
    ],
    columns=("animal", "attribute", "count"),
)
result

我认为可以用迭代组或类似的方法来编程。我要找的是个独角兽。这应该是可能的，是吗？

Answer 1

可以使用groupby.agg和自定义函数计算count

(df.groupby('animal', as_index=False)
   .agg(attribute=('attribute', 'last'),
        count=('attribute', lambda s: s.eq(s.iloc[-1])[::-1].cummin().sum())
       )
)

产出：

   animal       attribute  count
0    bird  Psittaciformes      4
1  mammal   Falconiformes      6

功能：

s.eq(s.iloc[-1])   # identify values equal to last one
[::-1]             # inverse Series
.cummin()          # set all values False after the first False
.sum()             # count the True

Answer 2

另一种解决办法是：

df_out = df.groupby("animal", as_index=False).apply(
    lambda x: x.groupby((x.attribute != x.attribute.shift()).cumsum())
    .agg(
        animal=("animal", "first"),
        attribute=("attribute", "first"),
        count=("animal", "count"),
    )
    .iloc[-1]
)

print(df_out)

指纹：

   animal       attribute  count
0    bird  Psittaciformes      4
1  mammal   Falconiformes      6