Question

我试图了解如何在使用get_dummies后解决列问题。例如，我们说我有三个分类变量。第一个变量有2个级别。第二个变量有5个级别。第三个变量有2个级别。

df=pd.DataFrame({"a":["Yes","Yes","No","No","No","Yes","Yes"], "b":["a","b","c","d","e","a","c"],"c":["1","2","2","1","2","1","1"]})

我为所有三个变量创建了虚拟变量，以便在python中的sklearn回归中使用它们。

df1 = pd.get_dummies(df,drop_first=True)

现在我要创建两个交互（乘法）：b c，b a

如何在不使用其特定名称的情况下创建每个虚拟变量与另一个变量之间的乘法：

df1['a_yes_b'] = df1['a_Yes']*df1['b_b']
df1['a_yes_c'] = df1['a_Yes']*df1['b_c']
df1['a_yes_d'] = df1['a_Yes']*df1['b_d']
df1['a_yes_e'] = df1['a_Yes']*df1['b_e']

df1['c_2_b'] = df1['c_2']*df1['b_b']
df1['c_2_c'] = df1['c_2']*df1['b_c']
df1['c_2_d'] = df1['c_2']*df1['b_d']
df1['c_2_e'] = df1['c_2']*df1['b_e']

感谢。

Answer 1

您可以使用循环创建新列，过滤列名称可以使用boolean indexing和str.startswith进行过滤：

a = df1.columns[df1.columns.str.startswith('a')]
b = df1.columns[df1.columns.str.startswith('b')]
c = df1.columns[df1.columns.str.startswith('c')]

for col1 in b:
    for col2 in a:
        df1[col2 + '_' + col1.split('_')[1]] = df1[col1].mul(df1[col2])

for col1 in b:
    for col2 in c:
        df1[col2 + '_' + col1.split('_')[1]] = df1[col1].mul(df1[col2])
print (df1)

   a_Yes  b_b  b_c  b_d  b_e  c_2  a_Yes_b  a_Yes_c  a_Yes_d  a_Yes_e  c_2_b  \
0      1    0    0    0    0    0        0        0        0        0      0   
1      1    1    0    0    0    1        1        0        0        0      1   
2      0    0    1    0    0    1        0        0        0        0      0   
3      0    0    0    1    0    0        0        0        0        0      0   
4      0    0    0    0    1    1        0        0        0        0      0   
5      1    0    0    0    0    0        0        0        0        0      0   
6      1    0    1    0    0    0        0        1        0        0      0   

   c_2_c  c_2_d  c_2_e  
0      0      0      0  
1      0      0      0  
2      1      0      0  
3      0      0      0  
4      0      0      1  
5      0      0      0  
6      0      0      0

但如果a和b只有一列（在示例中是，可能在实际数据中），请使用：filter，mul，squeeze和concat：

a = df1.filter(regex='^a')
b = df1.filter(regex='^b')
c = df1.filter(regex='^c')

dfa = b.mul(a.squeeze(), axis=0).rename(columns=lambda x: a.columns[0] + x[1:])
dfc = b.mul(c.squeeze(), axis=0).rename(columns=lambda x: c.columns[0] + x[1:])

df1 = pd.concat([df1, dfa, dfc], axis=1)
print (df1)
   a_Yes  b_b  b_c  b_d  b_e  c_2  a_Yes_b  a_Yes_c  a_Yes_d  a_Yes_e  c_2_b  \
0      1    0    0    0    0    0        0        0        0        0      0   
1      1    1    0    0    0    1        1        0        0        0      1   
2      0    0    1    0    0    1        0        0        0        0      0   
3      0    0    0    1    0    0        0        0        0        0      0   
4      0    0    0    0    1    1        0        0        0        0      0   
5      1    0    0    0    0    0        0        0        0        0      0   
6      1    0    1    0    0    0        0        1        0        0      0   

   c_2_c  c_2_d  c_2_e  
0      0      0      0  
1      0      0      0  
2      1      0      0  
3      0      0      0  
4      0      0      1  
5      0      0      0  
6      0      0      0

Answer 2

您可以将dataframe列转换为numpy数组，然后相应地相乘。这里是你可以找到方法的链接：

Convert Select Columns in Pandas Dataframe to Numpy Array

Answer 3

这可以解决您的问题：

def get_design_with_pair_interaction(data, group_pair):
    """ Get the design matrix with the pairwise interactions
    
    Parameters
    ----------
    data (pandas.DataFrame):
       Pandas data frame with the two variables to build the design matrix of their two main effects and their interaction
    group_pair (iterator):
       List with the name of the two variables (name of the columns) to build the design matrix of their two main effects and their interaction
    
    Returns
    -------
    x_new (pandas.DataFrame):
       Pandas data frame with the design matrix of their two main effects and their interaction
    
    """
    x = pd.get_dummies(data[group_pair])
    interactions_lst = list(
        itertools.combinations(
            x.columns.tolist(),
            2,
        ),
    ) 
    x_new = x.copy()
    for level_1, level_2 in interactions_lst:
        if level_1.split('_')[0] == level_2.split('_')[0]:
            continue
        x_new = pd.concat(
            [
                x_new,
                x[level_1] * x[level_2]
            ],
            axis=1,
        )
        x_new = x_new.rename(
            columns = {
                0: (level_1 + '_' + level_2)
            }
        )
    return x_new

python中虚拟变量之间的交互

3 个答案: