如何进行比较（1、2、3 个因素的组合）

Question

您能帮我自动计算“我们的”客户在 1、2、3 个因素的组合中所占的份额吗？

我有一个客户和功能数据集。所有的客户都有标签：

• 1 - “我们的”
• 0 - 'not_ours'

clients ours car        house         boat     plane         bike
client1 1     1         0             1         1             1
client2 0     0         0             1         1             1
client3 1     0         0             0         1             1
client4 1     1         0             1         1             1
client5 0     0         0             1         1             1
client6 1     0         0             0         1             1
clientN 0     0         1             0         1             1

我想做 3 个实验：

1. 了解我们在每个 1 因子值中的数量份额。理想的结果：

    factor_value    1                 1                 0                 0
    calculation     quantity of ours  share of ours (%) quantity of ours  share of ours (%)
    car             2                 100%              2                 40%
    house           0                 0%                4                 67%
    boat            2                 50%               2                 67%
    plane           4                 67%               0                 0%
    bike            4                 67%               0                 0%
   

   我们的份额 = 我们在因子价值中的份额。例如。 car' value = 0。比我们的要多 40%，因为 5 个客户没有汽车，其中有 2 个是我们的。

2. 相同的计算，但检查每个因素的 2 个组合：

    car + house
    car + boat
    car  + plane
    car + bike
    house + boat
    house + plane
    house + bike
    boat + plane
    boat + bike
   

3. 考虑 3 个因素的所有可能组合：

    car + house + boat
    car + house + plane
    car + house + bike
    car + boat + plane
    car + boat + bike
    car + plane + bike
   

Answer 1

以下是步骤（以 3 个因素为例）：

#create container for 3 factors combinations
xgb3 = pd.DataFrame([('i', 'j', 'k')], columns = ['factor1', 'factor2', 'factor3'], index=[0])

#take the previous  table with combinations of 2 factors: (res3)
for i in range(len(res3)) :
    a = res3.iloc[i:i + 1,:]['factor1'].values[0]
    b = res3.iloc[i:i + 1,:]['factor2'].values[0]
#add the third factor
    for j in df.iloc[:,5:].columns.values:
#sort - to drop duplicates (e.g. a,c,b and a,b,c)
        to_sort = sorted([a, b, j])
        new_row3 = {'factor1':to_sort[0], 'factor2':to_sort[1], 'factor3':to_sort[2]}
        xgb3 = xgb3.append(new_row3, ignore_index=True)
        
xgb3 = xgb3.drop_duplicates()
#additional drop of duplicates inside the row (e.g. a,a,b). All the items in the row must be unique
xgb3 = xgb3[(xgb3['factor1'] != xgb3['factor3']) & 
            (xgb3['factor2'] != xgb3['factor3']) & 
            (xgb3['factor1'] != xgb3['factor2'])].reset_index(drop=True)


#create container for results
result3 = pd.DataFrame([('','','',0,0,.1,0,0,.1)], columns = ['factor1','factor2','factor3','quantity_inside_combination', 'quantity_of_ours_inside_combination', '%_of_ours_inside_combination', 'quantity_outside_combination', 'quantity_of_ours_outside_combination', '%_of_ours_outside_combination'], index=[0])

for i in range(1, len(xgb3)): #range begins with 1 to skip the first row with technical information
    f = xgb3.iloc[i:i+1,:]['factor1'].values[0]
    s = xgb3.iloc[i:i+1,:]['factor2'].values[0]
    x = xgb3.iloc[i:i+1,:]['factor3'].values[0]
    
    m = df['stl'][(df[f] == 1) & (df[s] == 1) & (df[x] == 1)].count()
    n = df['stl'][(df[f] == 1) & (df[s] == 1) & (df[x] == 1)].sum()
    o = df['stl'][(df[f] == 0) & (df[s] == 0) & (df[x] == 0)].count()
    p = df['stl'][(df[f] == 0) & (df[s] == 0) & (df[x] == 0)].sum()
    
    new_row3 = {'factor1':f, 
               'factor2':s,
                'factor3': x,
               'quantity_inside_combination': m,
               'quantity_of_ours_inside_combination': n,
               '%_of_ours_inside_combination': 1. * n / m,
               'quantity_outside_combination': o,
               'quantity_of_ours_outside_combination': p,
               '%_of_ours_outside_combination': 1. * p / o
              }
    result3 = result3.append(new_row3, ignore_index=True)