基于文本的游戏项目删除[重复]

Question

在我的大学课程中，我们必须创建一个基于文本的游戏，在这个游戏中，你可以进出房间收集物品。我仍在尝试掌握python和一般编码的诀窍，因此我一直在努力完成这个项目。在大多数情况下，我的代码可以正常工作，但我可以多次添加一个项目。如何将其添加到库存中，然后从文件室中删除，以防止用户多次添加？

import time
#Shows rooms available to move and items in each room
rooms = {
        'Foyer': {'North' : 'Dining Hall', 'East' : 'Admin Office' },
        'Admin Office': {'West': 'Foyer', 'item': 'Key', },
        'Dining Hall': {'North' : 'Exam Room 1', 'South' : 'Foyer', 'East' : 'Bathroom', 'West' : 'Kitchen', 'item' : 'Flashlight', },
        'Bathroom': {'West': 'Dining Hall', 'item': 'Proton Pack' },
        'Kitchen': {'East': 'Dining Hall', 'item': 'Satchel' },
        'Exam Room 1': {'South' : 'Dining Hall', 'East' : 'Isolation Room', 'West': 'Patient Room 1', 'item' : 'Journal' },
        'Patient Room 1': {'East': 'Exam Room 1', 'item': 'Ghost Capsule' },
        'Isolation Room': {'West': 'Exam Room 1', 'item': 'Dr. Finkelstein' },
    }
#Game Instructions
def game_instructions():
   print('Collect 6 items to win the game, or be defeated by the ghost of Dr. Finkelstein.')
   print('Move Commands: North , South , East , or West.')
   print("Add to Inventory: get 'item name'")

print('You’ve been tasked with getting rid of the ghost of Dr. Finkelstein in the old insane asylum up on Academy Hill.')
time.sleep(1)
print('The ghost has taken over the asylum for three centuries and has begun opening a portal for more ghost to pass through.')
time.sleep(1)
print('During your last visit to the asylum, you dropped multiple items during a quick escape.')
time.sleep(1)
print('You need to go back into the asylum, collect all the items, and capture his ghost to close the portal!')
time.sleep(1)
print('The fate of the town is in your hands')
time.sleep(1)
game_instructions()
time.sleep(1)

current_room = 'Foyer' # starts player in the Foyer
inventory = [] # Adds an inventory, starts empty

def get_new_room(current_room, direction):
    new_room = current_room
    for i in rooms:
        if i == current_room:
            if direction in rooms[i]:
                new_room = rooms[i][direction]
    return new_room

def get_item(current_room):
    if 'item' in rooms[current_room]:
        return rooms[current_room]['item']
    else:
        return 'This room has no item!'

while (current_room): # gameplay loop
    print('\n-------------------------')
    print('You are in', current_room)  # tells player what room they are in.
    print('Inventory:', inventory)  # shows player their inventory
    item = get_item(current_room)  # defines item as get item
    print('Item:', item)  # tells the player what item they have found
    print('-------------------------')
    if 'item' == 'Dr. Finkelstein':
        if len(inventory) == 6:
            print('Congratulations!! You have collected all the necessary items and have successfully captured the ghost of Dr. Finkelstein and saved the town!')
        else:
            print('You did not have all the necessary items to defeat the ghost of Dr. Finkelstein, you are defeated.') # notifies player game has ended.
        break  # ends game
    direction = input('Enter direction you would like to move. >>')
    direction = direction.capitalize() # Capitalizes the players input to match what is in the dictionary.

    if (direction == 'North' or direction == 'South' or direction == 'East' or direction == 'West'):
        new_room = get_new_room(current_room, direction)
        if new_room == current_room:
            print('That is a wall not an exit. Try Again!')
        else:
            current_room = new_room
    elif direction == str('get ' + item ).capitalize():
        if item in inventory:
            print('You have already collected this item. Move to another room!')
        else:
            inventory.append(item)
    else:
        print('Not a valid direction!')

Answer 1

使用pop从current_room中删除项目。

def get_item(current_room):
    return rooms[current_room].pop('item', 'This room has no item!')

pop的第二个参数是回退值，在我看来，在这种情况下，最好返回None，而不是字符串。

Answer 2

可以使用del语句从字典中删除键/值对：

del rooms[current_room]['item']

但是有一个快捷方式，允许同时删除和返回项目。这是一种叫做dict.pop的字典方法。使用它，您可以按如下方式重写get_item（）：

def get_item(current_room):
    return rooms[current_room].pop('item', 'This room has no item!')