使用 Python 自动化无聊的事情 - 第 5 章，幻想游戏清单

Question

下面是我正在测试的代码：

stuff = {'arrow':12, 'gold coin':42, 'rope':1, 'torch':6, 'dagger':1}

def displayInventory(inventory):
    print('Inventory:')
    item_total = 0
    for k, v in inventory.items():
        print(str(v) + ' ' + str(k))
        item_total += v
    print('Total number of items: ' + str(item_total))

displayInventory(stuff)

##
inv = {'gold coin':42, 'rope':1}
dragonLoot = {'gold coin', 'dagger', 'gold coin', 'gold coin', 'ruby'}
def addToInventory(inventory, addedItems):
    for i in addedItems:
        inventory.setdefault(i, 0)
        inventory[i] += 1
    return inventory
dragonLoot = {'gold coin', 'dagger', 'gold coin', 'gold coin', 'ruby'}
inv = addToInventory(inv, dragonLoot)
displayInventory(inv)

我期待它回来

Inventory:
45 gold coin
1 rope
1 dagger
1 ruby
Total number of items: 46

但我只得到 43 的“金币”钥匙。为什么是这样？

Answer 1

问题是，您使用的是set：

dragonLoot = {'gold coin', 'dagger', 'gold coin', 'gold coin', 'ruby'}
print(dragonLoot)

输出：

{'dagger', 'gold coin', 'ruby'}

一套将确保每件物品都是独一无二的，因此重复的“金币”将被丢弃。

解决方案：改用a list：

dragonLoot = ['gold coin', 'dagger', 'gold coin', 'gold coin', 'ruby']

输出：

Inventory:
45 gold coin
1 rope
1 dagger
1 ruby
Total number of items: 48