Question

我在python中有这种字典列表

[
  {
    "compania": "Fiat",
    "modelo": "2014",    
    "precio": "1000"
  },
  {
    "compania": "Renault",
    "modelo": "2014",    
    "precio": "2000"
  },
  {
    "compania": "Volkwagen",
    "modelo": "2014",    
    "precio": "3000"
  },
  {
    "compania": "Chevrolet",
     "modelo": "2014",    
     "precio": "1000"    
  },
  {
    "compania": "Peugeot",
    "modelo": "2014",    
    "precio": "2000"
  } 
]

我想转换成这种字典列表列表

{   
  "Fiat": {
    "modelo": "2014",    
    "precio": "1000"
  },
  "Renault": {
    "modelo": "2014",    
    "precio": "2000"
  },
  "Volkwagen": {
    "modelo": "2014",    
    "precio": "3000"
  },
  "Chevrolet": {
    "modelo": "2014",    
    "precio": "1000"
  },
  "Peugeot": {
    "modelo": "2014",    
    "precio": "2000"
  }  
}

Answer 1

我们可以使用dict理解

{a.get('compania'): {k: v for k, v in a.items() if k != 'compania'} for a in c}

{'Fiat': {'modelo': '2014', 'precio': '1000'},
 'Renault': {'modelo': '2014', 'precio': '2000'},
 'Volkwagen': {'modelo': '2014', 'precio': '3000'},
 'Chevrolet': {'modelo': '2014', 'precio': '1000'},
 'Peugeot': {'modelo': '2014', 'precio': '2000'}}

其中c是您的原始数据

Answer 2

您可以通过遍历原始列表的元素来创建字典。假设您的列表名为car_list：

d = { x["compania"]: {"modelo": x["modelo"], "precio": x["precio"] } for x in car_list }

Answer 3

假设您的列表名为l，则可以使用简单的迭代并构建新的字典d来完成此操作：

d = {}

for sub in l:
  d[sub.pop('compania')] = sub

这在字典d中产生：

{'Chevrolet': {'modelo': '2014', 'precio': '1000'},
 'Fiat': {'modelo': '2014', 'precio': '1000'},
 'Peugeot': {'modelo': '2014', 'precio': '2000'},
 'Renault': {'modelo': '2014', 'precio': '2000'},
 'Volkwagen': {'modelo': '2014', 'precio': '3000'}}

说明：对于列表（sub）中的每个词典：sub.pop('compania')删除sub中键“ compania”的条目并返回其关联值。然后，我们将更新字典d，使其具有刚返回的键，然后将其与该键相关联，即删除该条目后sub的其余部分。

Answer 4

具有映射器功能，可返回新的自定义字典列表

a=[
  {
    "compania": "Fiat",
    "modelo": "2014",    
    "precio": "1000"
  },
  {
    "compania": "Renault",
    "modelo": "2014",    
    "precio": "2000"
  },
  {
    "compania": "Volkwagen",
    "modelo": "2014",    
    "precio": "3000"
  },
  {
    "compania": "Chevrolet",
     "modelo": "2014",    
     "precio": "1000"    
  },
  {
    "compania": "Peugeot",
    "modelo": "2014",    
    "precio": "2000"
  } 
]  

def mapper(temp): # define a mapper
    new_temp={}
    new_temp[temp['compania']]={k:v for k,v in temp.items() if k!='compania'}
    return new_temp

map(mapper,a) # call map function with mapper and list as argument

Answer 5

result = {}
for d in l:
    # Store the value of the key 'compania' before popping it from the small dictionary d
    compania = d['compania']
    d.pop('compania')
    # Construct new dictionary with key of the compania and value of the small dictionary without the compania key/value pair
    result[compania] = d
print(result)

输出：

{'Chevrolet': {'modelo': '2014', 'precio': '1000'},
 'Fiat': {'modelo': '2014', 'precio': '1000'},
 'Peugeot': {'modelo': '2014', 'precio': '2000'},
 'Renault': {'modelo': '2014', 'precio': '2000'},
 'Volkwagen': {'modelo': '2014', 'precio': '3000'}}

Answer 6

我将分享简单的解决方案：

>>> {d.pop("compania"):d for d in dd}

Answer 7

您可以简单地使用字典update来生成您喜欢的新字典。

from pprint import PrettyPrinter as pp
d={}
for i in l: # 'l' represents your list of dictionary 
    d.update({i['compania']:{"modelo":i['modelo'],"precio":i['precio']}})
print(pp(indent=4).pprint(d))

输出：

{   'Chevrolet': {'modelo': '2014', 'precio': '1000'},
'Fiat': {'modelo': '2014', 'precio': '1000'},
'Peugeot': {'modelo': '2014', 'precio': '2000'},
'Renault': {'modelo': '2014', 'precio': '2000'},
'Volkwagen': {'modelo': '2014', 'precio': '3000'}}

抄录词典列表

7 个答案: