对 dict 值的输出进行排序并使用排序后的值打印 dict

Question

我尝试通过以下代码整理后打印 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从字典中获取排序后的 "value"。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

实例之间不支持

我已经从另一个字典的实例创建了字典。我没有使用 lambda，直接使用 lambda 获得所需的输出。想检查是否可以使用 lambda 对 dict 值进行排序。

Answer 1

感谢以下声明修复了它 "newlist = sorted(alphaCars(mydict), key=itemgetter(1),reverse=True)"。感谢大家的帮助

对 dict 值的输出进行排序并使用排序后的值打印 dict

sort output of dict values and print the dict with sorted values

dictionary

key-value-store

python-3.x