Question

不幸的是，我正在努力执行以下计划：

我刮了一个大字典，用用户名作为键，并用键的Instagram关注者作为值。

看起来像这样：

dict = {
"User1": ['Follower_A', 'Follower_B', 'Follower_C', 'Follower_D', 'Follower_E', 'Follower_F', 'Follower_G'],
"User2": ['Follower_Z', 'Follower_D', 'Follower_V', 'Follower_E' 'Follower_O'],
"User3": ['Follower_P', 'Follower_Q'],
"User4": ['Follower_M', 'Follower_A', 'Follower_Z', 'Follower_P' 'Follower_L'],
.
.
.,
"UserX": ['Follower_Z', 'Follower_D', 'Follower_V', 'Follower_E' 'Follower_O','Follower_Q']}

为了创建一个网络，我需要一个看起来像这样的数据框：

User    Follower
0   User1   Follower_A
1   User1   Follower_B
2   User1   Follower_C
3   User1   Follower_D
4   User1   Follower_E
5   User1   Follower_F
6   User1   Follower_G
7   User2   Follower_Z
8   User2   Follower_D
9   User2   Follower_V
10  User2   Follower_E
11  User2   Follower_O
12  User3   Follower_P
13  User3   Follower_Q

有人有想法以有效的方式创建它吗？我所有的尝试都没有成功...

谢谢！

Answer 1

对具有DataFrame构造函数的元组使用列表理解和扁平化。

也不要使用变量名dict，因为它是python代码保留字。

d = {
"User1": ['Follower_A', 'Follower_B', 'Follower_C', 'Follower_D', 'Follower_E', 'Follower_F', 'Follower_G'],
"User2": ['Follower_Z', 'Follower_D', 'Follower_V', 'Follower_E', 'Follower_O'],
"User3": ['Follower_P', 'Follower_Q'],
"User4": ['Follower_M', 'Follower_A', 'Follower_Z', 'Follower_P', 'Follower_L'],
"UserX": ['Follower_Z', 'Follower_D', 'Follower_V', 'Follower_E', 'Follower_O','Follower_Q']}

df = pd.DataFrame([(oldk, k) for oldk, oldv in d.items() for k in oldv], 
                   columns=['User','Follower'])
print (df)
     User    Follower
0   User1  Follower_A
1   User1  Follower_B
2   User1  Follower_C
3   User1  Follower_D
4   User1  Follower_E
5   User1  Follower_F
6   User1  Follower_G
7   User2  Follower_Z
8   User2  Follower_D
9   User2  Follower_V
10  User2  Follower_E
11  User2  Follower_O
12  User3  Follower_P
13  User3  Follower_Q
14  User4  Follower_M
15  User4  Follower_A
16  User4  Follower_Z
17  User4  Follower_P
18  User4  Follower_L
19  UserX  Follower_Z
20  UserX  Follower_D
21  UserX  Follower_V
22  UserX  Follower_E
23  UserX  Follower_O
24  UserX  Follower_Q

在下一列中使用重复的键为字典的每个值附加行

1 个答案: