Python：我如何排序字典裏面的列表

Question

我想按class1的值排序score_list。

我可以按字母順序對score_list進行排序，只需對鍵進行排序，但我無法弄清楚如何從最大到最小的數值排序列表。

此外，我想score_list只存儲最後3個輸入，如果可能，我想找到每個學生最後3個分數的平均值。

enter code here: 




score_list = [] 
class1 = {} 
what_class = input ("what class would you like?") 
if what_class == ("1") : 
    print("class1 results") 
    student_name =input("name?") 
    score = input("score?") 
    choose = input("do you want to entre another student? y,n") 
    while choose == "y": 
     student_name =input("name?") 
     score = input("score?") 
     score_list.append(score) 
     choose = input("do you want to entre another score? y,n") 
     class1.setdefault(student_name,score_list) 
     if choose == "n": 
      sorted_names = sorted(class1.keys()) 
      print(class1) 

class1_items = class1.items() 
print (type(class1_items)) 

for i in class1_items: 
    print (i) 

，如果有人能幫助我，我會非常感激它
感謝你的幫助

[]的目標更好看：1

enter code here 
import heapq 
from collections import deque 
print("alphabetical(type:A),highest to lowest(type:B),Average(type:C)") 
student_name = input("ENTER NAME") 
result_type = input ("WHICH TYPE OF RESULTS WOULD YOU LIKE?") 
which_class = input ("which class?") 

if result_type == ("A") or result_type == ("a"): 
    print("alphabetical order(type:A)") 


if result_type == ("B") or result_type == ("b"): 
    print("results in highest to lowest.") 
    score_list = deque(maxlen=3) 
    scores = input("Please enter your score.") 
    choose = input("do you want to entre another score? y,n") 
    while choose != "n": 
     scores = input("Please enter your score.") 
     choose = input("do you want to entre another score? y,n") 
     score_list.append(scores) 
     if choose != "y" : 
      print(score_list) 
      print((student_name)+"'s:"+"lowest to largest"+str(heapq.nsmallest(3, score_list))) 
      print((student_name)+"'s:"+"largest to lowest"+str(heapq.nlargest(3, score_list))) 

if result_type == ("C") or result_type == ("c"): 
    print("results in average order.") 
    score_list = deque(maxlen=3) 
    scores = input("Please enter your score.") 
    choose = input("do you want to entre another score? y,n") 
    while choose != "n": 
     scores = input("Please enter the scores.") 
     choose = input("do you want to entre another score? y,n") 
     score_list.append(int(scores)) 
     #print(len(score_list)) 
     #print(sum(score_list)) 
     total = (int(sum(score_list))) 
     length = (int(len(score_list))) 
     average= (total)/(length) 
     if choose != "y" : 
      print(student_name+"'s "+"average score:"+str(average)) 
      #print("average: " +str(average)) 

這是怎麼了我改進了它，我仍然無法弄清楚按字母順序排序的部分。

the objectives to help you know what i am trying to do

Answer 1

---

更新，你應該能夠只粘貼並運行下面的代碼 但儘管如此，你需要通過你的自我，這意味着包裹輸入迴路，連接接口/函數調用

import json 
class Quize(): 
    def __init__(self): 
     self.db = {} 

    def createClass(self, class_name): 
     self.db[class_name] = {} 

    def createUser(self, class_name, user_name): 
     self.db[class_name][user_name] = [] 

    def addScore(self, class_name, user_name, score): 
     self.db[class_name][user_name].append(score) 
     self.db[class_name][user_name] = self.db[class_name][user_name][-3:] 

    def printClass(self, class_name, order = None): 
     if not order: 
      print json.dumps(self.db[class_name], indent=4) 
     else: 
      to_sort = self.db[class_name] 
      print 'order', order 
      if order == 'by highest': 
       print sorted([ [max(to_sort[name]), name] for name in to_sort ], cmp = lambda x, y : y[0] - x[0]) 
      if order == 'by lowest': 
       print sorted([ [min(to_sort[name]), name] for name in to_sort ], cmp = lambda x, y : x[0] - y[0]) 
      if order == 'by average': 
       print sorted([ [sum(to_sort[name])/len(to_sort[name]), name] for name in to_sort ], 
           cmp = lambda x, y : y[0] - x[0]) 
      if order == 'by name': 
       print sorted([ [to_sort[name], name] for name in to_sort ], cmp = lambda x, y : x[1] > y[1]) 
    def printUser(self, class_name, user_name): 
     print json.dumps(self.db[class_name][user_name], indent=4) 

    def dump(self): 
     print json.dumps(self.db, indent = 4) 

ins = Quize() 
ins.createClass('0302') 
ins.createUser('0302', 'jack') 
ins.createUser('0302', 'tom') 

ins.addScore('0302', 'jack', 99) 
ins.addScore('0302', 'jack', 92) 
ins.addScore('0302', 'tom', 99) 
ins.addScore('0302', 'tom', 92) 
ins.addScore('0302', 'tom', 19) 
ins.addScore('0302', 'tom', 29) 
ins.addScore('0302', 'tom', 66) 

ins.printClass('0302', 'by highest') 
ins.printClass('0302', 'by lowest') 
ins.printClass('0302', 'by average') 
ins.printClass('0302', 'by name') 

ins.printUser('0302', 'tom') 
ins.printUser('0302', 'jack') 
ins.printClass('0302') 

ins.addScore('0302', 'jack', 11) 
ins.addScore('0302', 'jack', 37) 
ins.addScore('0302', 'tom', 80) 
ins.addScore('0302', 'tom', 72) 

ins.printClass('0302', 'by highest') 
ins.printClass('0302', 'by lowest') 
ins.printClass('0302', 'by average') 
ins.printClass('0302', 'by name') 

ins.printUser('0302', 'tom') 
ins.printUser('0302', 'jack') 
ins.printClass('0302') 

Answer 2

這裏有很多問題，但我會介紹幾個。您需要確保將輸入轉換爲int（），否則您將對字符串進行數學計算，這會導致問題。其次，也是最重要的是，字典沒有排序，所以你永遠不會得到相同的最後3個條目。

您可以使用列表理解來獲取等級並對它們進行排序，這是else之後的第一件事：因爲輸入順序是不相關的。對於接下來的兩個，你不能使用字典值，你必須參考你用來跟蹤輸入等級的列表。

grade_list = [] 
class1 = {} 
if input('what class would you like? ') == '1': 
    while True: 
      student_name = input('name? ') 
      score = int(input('score? ')) # must convert input to int 
      grade_list.append(score) 
      class1[student_name] = score 
      if input('do you want to enter another student? ') == 'y': 
       pass 
      else: 
       # fetch grades 
       grades = [v for k, v in class1.items()] 
       # print highest to lowest 
       print(sorted(grades, reverse=True)) 
       # print last 3 grades 
       print(grade_list[-3:]) 
       # print avg of last 3 grades 
       print(sum(grade_list[-3:])/3) 
       break 

輸出

what class would you like? 1 
name? Abby 
score? 11 
do you want to enter another student? y 
name? Bobby 
score? 22 
do you want to enter another student? y 
name? Charles 
score? 33 
do you want to enter another student? y 
name? Damon 
score? 44 
do you want to enter another student? n 
[44, 33, 22, 11] 
[22, 33, 44] 
33.0