熊猫Groupby列，并获得0的频率

Question

我有一个数据帧，我想按Col1 Col2 Col3分组，得到值列的0频率：df=

Col1 Col2 Col3 Value
Val1 Val2  A    0
Val1 Val2  A    1
Val1 Val2  A    2
Val1 Val2  A    0
Val1 Val2  A    1

Val1 Val2  B    0
Val1 Val2  B    0
Val1 Val2  B    0
Val1 Val2  B    0
Val1 Val2  B    1
...

我如何应用groupby来实现

Col1 Col2 Col3 Fercentage_of_0
Val1 Val2  A       0.2
Val1 Val2  B       0.8
...

非常感谢。

Answer 1

这里有另一种不使用lambda的方法，这对我来说似乎更容易理解：

df['is_zero'] = df['Value'] == 0
df.groupby(['Col1', 'Col2', 'Col3'])['is_zero'].mean()

Answer 2

对数据框使用Groupby，然后对结果数据框应用size（）方法。例如，假设你有一个名为df的createda数据帧包含这些值

df = pd.DataFrame({'Col1': ['Val1','Val1','Val1','Val1','Val1','Val1','Val1','Val1'], 
               'Col2': ['Val2','Val2','Val2','Val2','Val2','Val2','Val2','Val2'],
               'Col3': ['A','A','A','A','B','B','B','B'],
               'Value':[0,1,2,0,0,0,0,1]}) 

然后，可以使用

df.groupby(['Col1','Col2','Col3','Value']).size()
Col1  Col2  Col3  Value
Val1  Val2  A     0        2
                  1        1
                  2        1
            B     0        3
                  1        1
dtype: int64

Answer 3

一个简单的lambda函数可以为您实现这一点。生成一个列表，其中Value==0，获取此列表的len和组中项目的len。你有百分比吗

df = pd.DataFrame({"Col1":["Val1","Val1","Val1","Val1","Val1","Val1","Val1","Val1","Val1","Val1"],"Col2":["Val2","Val2","Val2","Val2","Val2","Val2","Val2","Val2","Val2","Val2"],"Col3":["A","A","A","A","A","B","B","B","B","B"],"Value":[0,1,2,0,1,0,0,0,0,1]})

df.groupby(["Col1","Col2","Col3"]).\
    agg({"Value":lambda x: len([v for v in x if v==0])/len(x)})

输出

                Value
Col1 Col2 Col3       
Val1 Val2 A       0.4
          B       0.8