对dict值的输出进行排序并使用排序值打印dict

Question

在通过以下代码整理后，我尝试打印 dict 键。效果很好

import pprint
from operator import itemgetter

cars = [
    {'Name':'Ka','Year Introduced':1996,'Production of the current model':2014,'Generation':'3rd Generation','Vehicle Information':'Developed by Ford Brazil as a super mini car'},
    {'Name':'Fiesta','Year Introduced':1976,'Production of the current model':2017,'Generation':'7th Generation (Mark VIII)','Vehicle Information':'Ford\'s long running subcompact line based on global B-car Platform'},
    {'Name':'Focus','Year Introduced':1998,'Production of the current model':2018,'Generation':'3rd Generation (Mark III)','Vehicle Information':'Ford\'s Compact car based on global C-car platform'},
    {'Name':'Mondeo','Year Introduced':1992,'Production of the current model':2012,'Generation':'2nd Generation (Fusion)','Vehicle Information':'Mid sized passenger sedan with "One-Ford" design based on CD4 platform.'},
    {'Name':'Taurus','Year Introduced':1986,'Production of the current model':2009,'Generation':'6th Generation','Vehicle Information':'Full sized car based on D3 platform'},
    {'Name':'Fiesta ST','Year Introduced':2013,'Production of the current model':2013,'Generation':'1st Generation (6th Generation)','Vehicle Information':'Fiesta\'s high performance factory tune'},
    {'Name':'Focus RS','Year Introduced':2015,'Production of the current model':2015,'Generation':'1st Generation (3rd Generation)','Vehicle Information':'Special high performance Focus developed by SVT'},
    {'Name':'Mustang','Year Introduced':1964,'Production of the current model':2014,'Generation':'6th Generation','Vehicle Information':'Ford\'s long running pony/muscle car'},
    {'Name':'GT','Year Introduced':2004,'Production of the current model':2016,'Generation':'2nd Generation','Vehicle Information':'Ford\'s limited production super car inspired by the legendary race car GT40'}
    ]

# Creates a new dictionary with the name value as the key, and the dictionary as the value   
def newDict(carlist):
    mynewDict ={}
    for car in carlist:
        mynewDict[car['Name'],car['Year Introduced']] = car

    return mynewDict


def alphaCars(mycardict):
    carlist = list(mycardict.keys())
    carlist.sort() 
    #print (carlist)
    return carlist


mydict = newDict(cars)

#pprint.pprint (alphaCars(mydict))



for i in sorted (newDict(cars).keys()):  
    print(i) 

我尝试了同样的方法来从 dict 中获取排序后的“值”。尝试了堆栈溢出中提到的单行代码，但我无法这样做。得到下面提到的错误

错误：'dict' 和 'dict' 的实例之间不支持

我从另一个 dict 的实例创建了 dict。我没有使用 lambda，使用 lambda 获得所需的输出很简单。想检查是否可以在没有 lambda 的情况下对 dict 值进行排序。

Answer 1

感谢以下语句修复它 "newlist = sorted(alphaCars(mydict), key=itemgetter(1),reverse=True)" 。谢谢大家的帮助