Python 抄写字典目录
我有这种python字典列表Python 抄写字典目录,python,list,dictionary,Python,List,Dictionary,我有这种python字典列表 [ { "compania": "Fiat", "modelo": "2014", "precio": "1000" }, { "compania": "Renault", "modelo": "2014", "precio": "2000" }, { "compania": "Volkwagen", "modelo": "2014", "precio
[
{
"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"
}
]
我想转换成这样的字典列表
{
"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"
}
}
您可以通过迭代原始列表中的元素来创建字典。假设您的列表称为car_list:
我们可以使用听写理解
{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是您的原始数据假设您的列表名为l,您可以通过简单的迭代和构建新的字典d来实现这一点: 这将在字典d中生成:
说明:对于该列表中的每个字典,sub: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,使其具有刚返回内容的键,并将该键与删除该条目后的sub剩余部分关联。使用映射器函数返回新的自定义dict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
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'}}
我想分享一个简单的解决方案:
>>> {d.pop("compania"):d for d in dd}
您只需使用dictionary即可生成您喜欢的新词典
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:
你试过什么,到底有什么问题?还要注意,您的输出是一个将字符串映射到字典的字典,而不是一个字典列表。那么您尝试了什么呢?这似乎相当简单,只需从列表的每个元素中取出d['compania',],并用它创建一个字典,其余的键值对就是它的内容。谢谢汉克斯,我在工作,一些关于这些问题的书或论文,我是新来的。你为什么要使用像.pop这样的破坏性操作?有什么优势吗?一般来说,除非您试图删除内容,否则我会错误地选择不删除内容。@juanpa.arrivillaga只是因为pop简洁地返回了我们想要用于新词典的键,并且在执行后,会留下与该键相关的值。我知道你是从哪里来的:不要在理解结构中使用.pop,它有副作用,你应该避免在函数结构中使用副作用。正是因为解决方案使用了副作用,所以在这种特殊情况下它很短。
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)
{'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'}}
>>> {d.pop("compania"):d for d in dd}
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))
{ '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'}}