如何根据项目条件删除字典条目

Question

我希望从游戏开始时从我的字典中删除一个房间，而雪地靴 = False。当雪地靴 = True 时，我希望房间可以到达，我想拿起雪地靴使它们成为 True。

如果这是有道理的。

roomDirections = {
    "hallEnt":{"e":"hallMid"},
    "hallMid":{"s":"snowRoom", "e":"giantNature", "w":"hallEnt"},
    "snowRoom":{"n":"hallMid"},
    "giantNature":{"s":"strangeWall", "e":"riverBank", "w":"hallMid"},
    "strangeWall":{"s":"hallOuter", "e":"riverBank", "n":"giantNature"},
    "riverBank":{"e":"lilyOne", "w":"giantNature"},
    "lilyOne":{"e":"lilyTwo", "w":"riverBank", "n":"riverBank", "s":"riverBank"},
    "lilyTwo":{"e":"riverBank", "w":"lilyThree", "n":"riverBank", "s":"riverBank"},
    "lilyThree":{"e":"riverBank", "w":"lilyFour", "n":"riverBank", "s":"riverBank"},
    "lilyFour":{"e":"riverBank", "w":"treasureRoom", "n":"riverBank", "s":"riverBank"},
    "treasureRoom":{"w":"hallEnt"},
}

roomItems = {
    "hallEnt":["snowboots"],
    "snowRoom":["lamp"],
    "treasureRoom":["treasure"],
    }
snowboots = lamp = treasure = False

这些是我的字典和我所谓的变量。

if "snowboots" == False:
            del roomDirections["hallMid"]
        else:
            print ("you cannot go that way")

这是为了将大厅从房间中移除方向，因此不可能从中移动，直到......

elif playerInput in roomItems[currentRoom]:
        print("picked up", playerInput)
        invItems.append(playerInput)
        playerInput == True
        for i in range(0, len(roomItems[currentRoom])):
            if playerInput == roomItems[currentRoom][i]:
                del roomItems[currentRoom][i]
                break

雪地靴=是的，这是这个块应该做的，但它似乎不起作用，我是接近还是完全偏离轨道？

编辑 -- 我的主要游戏循环 --

while True:
    playerInput = input("What do you want to do? ")
    playerInput = playerInput.lower()
    if playerInput == "quit":
        break
    elif playerInput == "look":
        print(roomDescriptions[currentRoom])


    elif playerInput in dirs:
        playerInput = playerInput[0]
        if playerInput in roomDirections[currentRoom]:

            currentRoom = roomDirections[currentRoom][playerInput]
            print(roomEntrance [currentRoom])
        else:
            print("You can't go that way")
    elif playerInput == "lookdown":
        if currentRoom in roomItems.keys():
            print ("You see", roomItems[currentRoom])
        else:
            print ("You see nothing on the ground")
    elif playerInput == "inventory" or playerInput == "inv":
        print (invItems)


    elif playerInput in roomItems[currentRoom]:
        print("picked up", playerInput)
        invItems.append(playerInput)       
        for i in range(0, len(roomItems[currentRoom])):
            if playerInput == roomItems[currentRoom][i]:
                del roomItems[currentRoom][i]
                break
    elif playerInput in invItems:
        print("dropped", playerInput)
        roomItems[currentRoom].append (playerInput)
        for i in range (0, len(invItems)):
            if playerInput == invItems[i]:
                del invItems[i]
                break
    else:
        print ("I don't understand")

Answer 1

据我了解，您想添加一些条件来决定是否允许通过特定出口。您目前有映射每个房间方向的字典，用于保存玩家是否拥有每个物品的变量，以及每个房间中的物品列表和玩家物品栏。请注意，项目变量是多余的;您可以简单地检查库存。

您建议的方法是在获得或丢失所需物品时添加和删除房间的出口。这是可以做到的，但是找到要删除的出口的复杂性是您首先忽略它们所需要的;如果它们被删除，则恢复它们比根据需要过滤它们更难。这里有一种方法：

requirements = {'snowRoom': 'snowboots', 'darkCave': 'lamp'}
reasons = {'snowboots': "You'd sink into the snow.",
           'lamp': "It would be too dark to see."}

然后，如果不满足条件，您可以使用这些来忽略方向：

elif playerInput in dirs:
    playerInput = playerInput[0]
    if playerInput in roomDirections[currentRoom]:
        newRoom = roomDirections[currentRoom][playerInput]
        required = requirements.get(newRoom)
        if required and required not in invItems:
            print("You can't go that way. " + reasons[required])
        else:
            currentRoom = newRoom
            print(roomEntrance [currentRoom])
    else:
        print("You can't go that way")

你也可以这样做，让玩家无法移除房间中的所需物品：

elif playerInput in invItems:
    if playerInput != requirements[currentRoom]:
        print("dropped", playerInput)
        roomItems[currentRoom].append (playerInput)
        invItems.remove(playerInput)
    else:
        print("You still need " + playerInput + ". " + reasons[required])

采用更面向对象的方法可能是有意义的，其中房间实际上包含其项目列表、指向其他房间的链接和要求。您还可以执行诸如向项目添加同义词之类的操作。

Answer 2

让我在这里指出你的错误。您有snowboots（这是一个对象），但您正在比较"snowboots"（这是一个字符串）。因此，当您在if条件下比较"snowboots"时，它将返回 false，因此它将始终转到else部分。尝试以下代码，让我知道它是否对您有帮助。

snowboots = False
if snowboots == False:              ## notice the absence of double quotes here 
    del roomDirections["hallMid"]
else:
    print ("you cannot go that way")

编辑：为了更清楚起见，我编辑了您的代码以表明del有效。我已删除

roomDirections = {
    "hallMid":{"s":"snowRoom", "e":"giantNature", "w":"hallEnt"},
}
print roomDirections  ## print dictionary before your deleting.
snowboots = lamp = treasure = False
if snowboots == False:              ## notice the absence of double quotes here 
    del roomDirections["hallMid"]
else:
    print ("you cannot go that way")
print roomDirections                ## print dictionary after deleting

输出：

    {'hallMid': {'s': 'snowRoom', 'e': 'giantNature', 'w': 'hallEnt'}}   ## before deleting
     {}     ## after deleting.

Answer 3

playerInput是一个字符串。 看起来您需要将字符串映射到布尔值。

items = {'snowboots' : False,
         'lamp' : False,
         'treasure' : False}

然后更改

if "snowboots" == False到if not items['snowboots']

playerInput == True到items[playerInput] = True

snowboots = lamp = treasure = False至

for item in items:
    items[item] = False

看起来您在字符串和名称方面存在概念问题，并且似乎将它们混合在代码中。 "foo"与foo不同.

>>> 
>>> foo = 2
>>> "foo" = 2
SyntaxError: can't assign to literal
>>> foo = "foo"
>>> foo
'foo'
>>> "foo" = foo
SyntaxError: can't assign to literal
>>> foo = False
>>> bool("foo")
True
>>> bool(foo)
False
>>>

您将需要遍历所有代码并解析所有类似的代码。