python程序中的算法错误

Question

当我运行它时，金币的数量只增加了一个，尽管有三个额外的金币。我不知道我的逻辑有什么问题。我创建了两个函数：addToInventory，第一个参数是dictionary，第二个参数是list。如果字典中不存在键，则函数会将键添加到字典中，并将值递增1。displayInventory方法打印字典中的键/值。以下是我的源代码：

#fantasyGameInventory.py - a function that displays the inventory from a
#dictionary data structure

stuff = {'rope': 1, 'torch': 6, 'gold coin': 40, 'dagger': 1, 'arrow': 12}
dragonLoot = {'gold coin', 'dagger', 'gold coin', 'gold coin', 'ruby'}

def addToInventory(inventory, addedItems):
    #Iterate through the list 
    for k in addedItems:
       #if key not in the dictionary, add to the dictionary and set its value to zero
        inventory.setdefault(k,0)
        inventory[k] = inventory[k] + 1
        print(str(inventory[k]))

def displayInventory(inventory):
    print('Inventory:')
    itemTotal = 0
    #iterate through the dictionary and print the key/values
    for k, v in inventory.items():
        print(k + ': ' + str(v))
        itemTotal = itemTotal + v
        print ('Total number of items: ' + str(itemTotal))

addToInventory(stuff, dragonLoot)
displayInventory(stuff) 

它说有41枚金币，尽管显然应该有42枚：原来的40枚来自stuff，加上另外两枚在dragonLoot。你知道吗

Answer 1

首先需要将set更改为tuple，因为它在遍历列表时跳过重复的条目。你知道吗

以下是工作代码：

stuff = {'rope': 1, 'torch': 6, 'gold coin': 40, 'dagger': 1, 'arrow': 12}
dragonLoot = ('gold coin', 'dagger', 'gold coin', 'gold coin', 'ruby')

def addToInventory(inventory, addedItems):
    #Iterate through the list
    for k in addedItems:
        #if key not in the dictionary, add to the dictionary and set its value to zero
        inventory.setdefault(k,0)
        #increment value by one
        inventory[k] = inventory[k] + 1
        print(str(inventory[k]))

def displayInventory(inventory):
    print('Inventory:')
    itemTotal = 0
    #iterate through the dictionary and print the key/values
    for k, v in inventory.items():
        print(k + ': ' + str(v))
        itemTotal = itemTotal + v
    print ('Total number of items: ' + str(itemTotal))

addToInventory(stuff, dragonLoot)
displayInventory(stuff)