Question

我有一个由产品名称和唯一客户电子邮件组成的字典，他们购买了这样的项目：

customer_emails = {
'Backpack':['customer1@gmail.com','customer2@gmail.com','customer3@yahoo.com','customer4@msn.com'], 
'Baseball Bat':['customer1@gmail.com','customer3@yahoo.com','customer5@gmail.com'],
'Gloves':['customer2@gmail.com','customer3@yahoo.com','customer4@msn.com']}

我正在尝试迭代每个键的值，并确定其他键中匹配的电子邮件数量。我将这个字典转换为DataFrame，并使用类似这样的

得到了我想要的单列比较的答案

customers[customers['Baseball Bat'].notna() == True]['Baseball Bat'].isin(customers['Gloves']).sum()

我想要完成的是创建一个基本上看起来像这样的DataFrame，这样我就可以轻松地将它用于相关图表。

             Backpack  Baseball Bat    Gloves
Backpack            4             2         3
Baseball Bat        2             3         1 
Gloves              3             1         3

我认为这样做的方法是迭代customer_emails词典，但我不确定如何选择一个键来将其值与其他所有词进行比较，所以然后存储它。

Answer 1

Start with pd.DataFrame.from_dict:

df = pd.DataFrame.from_dict(customer_emails, orient='index').T

df
              Backpack         Baseball Bat               Gloves
0  customer1@gmail.com  customer1@gmail.com  customer2@gmail.com
1  customer2@gmail.com  customer3@yahoo.com  customer3@yahoo.com
2  customer3@yahoo.com  customer5@gmail.com    customer4@msn.com
3    customer4@msn.com                 None                 None

Now, use stack + get_dummies + sum + dot:

v = df.stack().str.get_dummies().sum(level=1)
v.dot(v.T)

              Backpack  Baseball Bat  Gloves
Backpack             4             2       3
Baseball Bat         2             3       1
Gloves               3             1       3

Alternatively, switch stack with melt for some added performance.

v = (df.melt()
       .set_index('variable')['value']
       .str.get_dummies()
       .sum(level=0)
)
v.dot(v.T)

variable      Backpack  Baseball Bat  Gloves
variable                                    
Backpack             4             2       3
Baseball Bat         2             3       1
Gloves               3             1       3

Answer 2

You can first find all the counts for each product and corresponding emails, then pass the resulting dictionary to pd.DataFrame:

import pandas as pd
emails = {'Baseball Bat': ['customer1@gmail.com', 'customer3@yahoo.com', 'customer5@gmail.com'], 'Backpack': ['customer1@gmail.com', 'customer2@gmail.com', 'customer3@yahoo.com', 'customer4@msn.com'], 'Gloves': ['customer2@gmail.com', 'customer3@yahoo.com', 'customer4@msn.com']}
results = {a:{c:sum(h in j for h in b) for c, j in emails.items()} for a, b in emails.items()}
df = pd.DataFrame(results)

Output:

               Backpack  Baseball Bat  Gloves
Backpack             4             2       3
Baseball Bat         2             3       1
Gloves               3             1       3

Answer 3

使用相同的逻辑创建系列，然后我们使用intersection列表

s=pd.Series(customer_emails)

pd.DataFrame(np.reshape([len(set(x).intersection(set(y)))for x in s for y in s ],(3,3)),index=s.index,columns=s.index)
Out[299]: 
              Backpack  Baseball Bat  Gloves
Backpack             4             2       3
Baseball Bat         2             3       1
Gloves               3             1       3

比较字典的值并返回匹配值的计数

3 个答案: