将具有排序唯一值的嵌套数据帧转换为Python中的嵌套字典

我试图获取一个嵌套的数据帧并将其转换为一个嵌套的字典

这是我的原始数据帧，具有以下唯一值：

输入：

df.head（5）

输出：

    reviewerName                                  title    reviewerRatings
0        Charles       Harry Potter Book Seven News:...                3.0
1      Katherine       Harry Potter Boxed Set, Books...                5.0
2           Lora       Harry Potter and the Sorcerer...                5.0
3           Cait       Harry Potter and the Half-Blo...                5.0
4          Diane       Harry Potter and the Order of...                5.0

                                                       reviewerRatings
    reviewerName                               title
         Charles    Harry Potter Book Seven News:...               3.0
                    Harry Potter and the Half-Blo...               3.5
                    Harry Potter and the Order of...               4.0
       Katherine    Harry Potter Boxed Set, Books...               5.0
                    Harry Potter and the Half-Blo...               2.5
                    Harry Potter and the Order of...               5.0
...
230898 rows x 1 columns

{'reviewerRatings': 
 {
  ('Charles', 'Harry Potter Book Seven News:...'): 3.0, 
  ('Charles', 'Harry Potter and the Half-Blo...'): 3.5, 
  ('Charles', 'Harry Potter and the Order of...'): 4.0,   
  ('Katherine', 'Harry Potter Boxed Set, Books...'): 5.0, 
  ('Katherine', 'Harry Potter and the Half-Blo...'): 2.5, 
  ('Katherine', 'Harry Potter and the Order of...'): 5.0,
 ...}
}

输入：

len（df['reviewerName'].unique（））

输出：

66130

考虑到66130 unqiue值中的每个值都有多个值（即“Charles”将出现3次），我将66130唯一的“reviewerName”作为新嵌套数据框中的键分配给它们，然后使用“title”和“reviewerRatings”分配值作为同一嵌套数据帧中的另一层key:value

{'Charles': 
 {'Harry Potter Book Seven News:...': 3.0, 
  'Harry Potter and the Half-Blo...': 3.5, 
  'Harry Potter and the Order of...': 4.0},   
 'Katherine':
 {'Harry Potter Boxed Set, Books...': 5.0, 
  'Harry Potter and the Half-Blo...': 2.5, 
  'Harry Potter and the Order of...': 5.0},
...}

输入：

df=df.set_索引（['reviewerName'，'title']）。排序_索引（）

输出：

    reviewerName                                  title    reviewerRatings
0        Charles       Harry Potter Book Seven News:...                3.0
1      Katherine       Harry Potter Boxed Set, Books...                5.0
2           Lora       Harry Potter and the Sorcerer...                5.0
3           Cait       Harry Potter and the Half-Blo...                5.0
4          Diane       Harry Potter and the Order of...                5.0

                                                       reviewerRatings
    reviewerName                               title
         Charles    Harry Potter Book Seven News:...               3.0
                    Harry Potter and the Half-Blo...               3.5
                    Harry Potter and the Order of...               4.0
       Katherine    Harry Potter Boxed Set, Books...               5.0
                    Harry Potter and the Half-Blo...               2.5
                    Harry Potter and the Order of...               5.0
...
230898 rows x 1 columns

{'reviewerRatings': 
 {
  ('Charles', 'Harry Potter Book Seven News:...'): 3.0, 
  ('Charles', 'Harry Potter and the Half-Blo...'): 3.5, 
  ('Charles', 'Harry Potter and the Order of...'): 4.0,   
  ('Katherine', 'Harry Potter Boxed Set, Books...'): 5.0, 
  ('Katherine', 'Harry Potter and the Half-Blo...'): 2.5, 
  ('Katherine', 'Harry Potter and the Order of...'): 5.0,
 ...}
}

作为对 ，我尝试将嵌套数据框转换为嵌套字典

上面新的嵌套数据框列索引在第1行（第3列）显示“reviewerRatings”，在第2行（第1列和第2列）显示“reviewerName”和“title”，当我运行下面的

df.to_dict（）

方法时，输出显示

{reviewerRatingsIndexName:{（reviewerName，title）：reviewerRatings}

输入：

df.to_dict（）

输出：

    reviewerName                                  title    reviewerRatings
0        Charles       Harry Potter Book Seven News:...                3.0
1      Katherine       Harry Potter Boxed Set, Books...                5.0
2           Lora       Harry Potter and the Sorcerer...                5.0
3           Cait       Harry Potter and the Half-Blo...                5.0
4          Diane       Harry Potter and the Order of...                5.0

                                                       reviewerRatings
    reviewerName                               title
         Charles    Harry Potter Book Seven News:...               3.0
                    Harry Potter and the Half-Blo...               3.5
                    Harry Potter and the Order of...               4.0
       Katherine    Harry Potter Boxed Set, Books...               5.0
                    Harry Potter and the Half-Blo...               2.5
                    Harry Potter and the Order of...               5.0
...
230898 rows x 1 columns

{'reviewerRatings': 
 {
  ('Charles', 'Harry Potter Book Seven News:...'): 3.0, 
  ('Charles', 'Harry Potter and the Half-Blo...'): 3.5, 
  ('Charles', 'Harry Potter and the Order of...'): 4.0,   
  ('Katherine', 'Harry Potter Boxed Set, Books...'): 5.0, 
  ('Katherine', 'Harry Potter and the Half-Blo...'): 2.5, 
  ('Katherine', 'Harry Potter and the Order of...'): 5.0,
 ...}
}

但是对于下面我想要的输出，我希望得到的输出是

{reviewerName:{title:reviewerRating}}

，这正是我在嵌套数据框架中排序的方式

{'Charles': 
 {'Harry Potter Book Seven News:...': 3.0, 
  'Harry Potter and the Half-Blo...': 3.5, 
  'Harry Potter and the Order of...': 4.0},   
 'Katherine':
 {'Harry Potter Boxed Set, Books...': 5.0, 
  'Harry Potter and the Half-Blo...': 2.5, 
  'Harry Potter and the Order of...': 5.0},
...}

有没有办法操纵嵌套的数据帧或嵌套的字典，以便在运行

df.to_dict（）

方法时，它将显示

{reviewerName:{title:reviewerRating}}

谢谢

与lambda函数一起用于

字典

每个

审阅者姓名

，然后通过以下方式输出

系列

转换：

---

---

有两种方法。您可以将

groupby

与

一起使用来记录
，或者使用
集合来迭代行。defaultdict
。值得注意的是，后者并不一定效率较低

+ 
从每个
groupby
对象构造一个序列，并将其转换为字典以给出一系列字典值。最后，通过另一个
to_dict
调用将其转换为字典字典

res = df.groupby('reviewerName')\
        .apply(lambda x: x.set_index('title')['reviewerRatings'].to_dict())\
        .to_dict()

定义
dict
对象的
defaultdict
，并逐行迭代数据帧

from collections import defaultdict

res = defaultdict(dict)
for row in df.itertuples(index=False):
    res[row.reviewerName][row.title] = row.reviewerRatings

生成的
defaultdict
不需要转换回常规
dict
，因为
defaultdict
是
dict
的子类

绩效基准
基准测试是建立和数据相关的。您应该使用自己的数据进行测试，以查看哪些数据最有效

# Python 3.6.5, Pandas 0.19.2

from collections import defaultdict
from random import sample

# construct sample dataframe
np.random.seed(0)
n = 10**4  # number of rows
names = np.random.choice(['Charles', 'Lora', 'Katherine', 'Matthew',
                          'Mark', 'Luke', 'John'], n)
books = [f'Book_{i}' for i in sample(range(10**5), n)]
ratings = np.random.randint(0, 6, n)

df = pd.DataFrame({'reviewerName': names, 'title': books, 'reviewerRatings': ratings})

def jez(df):
    return df.groupby('reviewerName')['title','reviewerRatings']\
             .apply(lambda x: dict(x.values))\
             .to_dict()

def jpp1(df):
    return df.groupby('reviewerName')\
             .apply(lambda x: x.set_index('title')['reviewerRatings'].to_dict())\
             .to_dict()

def jpp2(df):
    dd = defaultdict(dict)
    for row in df.itertuples(index=False):
        dd[row.reviewerName][row.title] = row.reviewerRatings
    return dd

%timeit jez(df)   # 33.5 ms per loop
%timeit jpp1(df)  # 17 ms per loop
%timeit jpp2(df)  # 21.1 ms per loop