Python嵌套列表 - 更改和替換單個項目

Question

我正在完成一個簡單的編程練習（我還是新的），其中我通過爲4個不同的字符屬性分配30個點來創建角色配置文件。計劃功能包括：顯示當前配置文件，創建新配置文件或更改現有配置文件。第一個和第二個功能工作正常，但最後一個問題：該程序是爲了解開嵌套列表項目（屬性+分配的分數），要求新的分數，採取舊的和新的差異，並改變數字相應的池中可用點數。最後，在位置0處向列表添加一個新條目（屬性+新分配的分數），然後刪除位置1處的條目，該條目應該是該屬性的舊條目。遍歷列表並完成。但是，一旦執行代碼，您將看到它不起作用。請參閱下面的完整代碼：

options = ["Strength", "Health", "Wisdom", "Dexterity"] 
profile = [] 
points = 30 

choice = None 
while choice != "0": 

    print(
     """ 
    CHARACTER CREATOR PROGRAM 

    0 - Exit 
    1 - See current profile 
    2 - Build new profile 
    3 - Amend existing profile 

    """ 
     ) 

    choice = input("Please choose an option: ") 
    print() 

    if choice == "0": 
     print("Good bye.") 
    elif choice == "1": 
     for item in profile: 
      print(item) 
     input("\nPress the enter key to continue.") 
    elif choice == "2": 
     print("You can now equip your character with attributes for your adventures.") 
     print("You have",points,"points to spent.") 
     print("Now configure your character: \n") 
     #Run the point allocation loop 
     for item in options: 
      point_aloc = int(input("Please enter points for " + str(item) + ":")) 
      if point_aloc <= points: 
       entry = item, point_aloc 
       profile.append(entry) 
       points = points - point_aloc 
       print("\nYour current choice looks like this: ") 
       print(profile) 
       input("\nPress the enter key to continue.") 
      else: 
       print("Sorry, you can only allocate", points," more points!") 
       print("\nYour current choice looks like this: ") 
       print(profile) 
       input("\nPress the enter key to continue.") 
     print("\nWell done, you have configured your character as follows: ") 
     for item in profile: 
      print(item) 
     input("Press the enter key to continue.") 
    elif choice == "3": 
     print("This is your current character profile:\n") 
     for item in profile: 
      print(item) 
     print("\nYou can change the point allocation for each attribute.") 
     for item in profile: 
      point_new = int(input("Please enter new points for " + str(item) + ":")) 
      attribute, points_aloc = item 
      diff = points_aloc - point_new 
      if diff >0: 
       points += diff 
       print("Your point allocation has changed by", -diff,"points.") 
       print(diff,"points have just been added to the pool.") 
       print("The pool now contains", points,"points.") 
       entry = item, point_new 
       profile.insert(0, entry) 
       del profile[1] 
       input("Press the enter key to continue.\n") 
      elif diff <0 and points - diff >=0: 
       points += diff 
       print("Your point allocation has changed by", -diff,"points.") 
       print(-diff,"points have just been taken from the pool.") 
       print("The pool now contains", points,"points.") 
       entry = item, point_new 
       profile.insert(0, entry) 
       del profile[1] 
       input("Press the enter key to continue.\n") 
      elif diff <0 and points - diff <=0: 
       print("Sorry, but you don't have enough points in the pool!") 
       input("Press the enter key to continue.\n") 
    else: 
     print("Sorry, but this is not a valid choice!") 
     input("Press the enter key to continue.\n") 

input("\n\nPress the enter key to exit.") 

注意：您需要先創建配置文件以運行更改。

在此先感謝您的幫助！

Answer 1

正如您對問題的評論所指出的，您並未以最佳方式提出您的問題。但是，我發現它有什麼問題。我可以告訴你如何解決你當前的代碼，但事實是，解決這個問題的方法是完全重寫它。當您這樣做時，您應該採用以下策略：

1. 將代碼分解爲多個部分。在這種情況下，我建議你創建幾個不同的功能。一個可以被稱爲main_loop，並將包含循環菜單的邏輯。它不包含更新或顯示配置文件的任何代碼。相反，它會調用其他函數，display_profile,build_profile和amend_profile。這些函數將接受變量，如options,profile和points，並返回值如options和points。這將極大地簡化您的代碼，並使測試和調試變得更加容易。這是一個什麼樣main_loop可能看起來像一個例子：

   def main_loop(): 
       options = ["Strength", "Health", "Wisdom", "Dexterity"] 
       profile = [] 
       points = 30 
   
       choice = None 
       while choice != "0": 
        print(menu_string) #define menu_string elsewhere 
        choice = input("Please choose an option: ") 
        print() 
   
        if choice == "0": 
         print("Good bye.") 
        elif choice == "1": 
         display_profile(profile) 
        elif choice == "2": 
         profile, points = build_profile(options, profile, points) 
        elif choice == "3": 
         profile, points = amend_profile(profile, points) 
        else: 
         print("Sorry, but this is not a valid choice!") 
         input("Press the enter key to continue.\n") 
   
       input("\n\nPress the enter key to exit.") 
   

   看到多少更好這是什麼？現在您只需定義其他功能即可。像...

   def build_profile(options, profile, points): 
        # insert logic here 
        return profile, points 
   

   這種方法的另一個好處是，現在你可以單獨測試這些功能，而無需運行整個程序。

2. 使用正確的成語進行列表修改。在迭代時修改列表需要格外小心，並且在某些情況下（例如，通過刪除或添加已經迭代的項目來更改列表的長度），它根本無法工作。有些方法可以對profile做些什麼，但對於初學程序員，我會推薦一些更簡單的方法：只需創建一個新列表！然後返回該列表。因此，在您amend_profile功能，做這樣的事情：

   def amend_profile(profile, points): 
        # other code ... 
        new_profile = [] 
        for item in profile: 
         attribute, points_aloc = item 
         # other code ... 
         new_proflie.append(entry) 
        # other code ... 
   
        return new_profile, points 
   

   還請注意，這是你的主要缺陷之一;您將創建一個entry，其中包含(item, point_new)而不是(attribute, point_new)，因此您的新元組中有一個item元組，而不是如預期的那樣只有一個attribute字符串。