野餐用品计数
之前写过一段，字符在字符串中出现次数的代码
import pprint
message = 'It was a bright cold day in April, ' \
'and the clocks were striking thirteen.'
count = {}
for character in message:
count.setdefault(character, 0)
count[character] += 1
pprint.pprint(count)
在写野餐用品计数代码时，发现多了个return
由于一直对于返回值的概念不甚了解，又把两段代码拿出来做了对比
并且把野餐用品代码拆分后，每个判断后加上print来查看流程及参数值
发现如果不用return给totalBrought（）设一个返回值，
那当apple = totalBrought（allGuests，‘apples’）时，就没有一个值可以返回给apple
由此对return又多了点了解
原代码是没有apple=这一句的，直接用
print("Apples: " + str(totalBrought(allGuests, 'apples')))
来替代了Apples打印行
allGuests = {'Alice': {'apples': 5, 'pretzels': 12},
'Bob': {"ham sandwiches": 3, 'apples': 2},
'Carol': {'cups': 3, 'apple pies': 1}}
def totalBrought(guests, item):
numBrought = 0
for k, v in guests.items():
numBrought = numBrought + v.get(item, 0)
return numBrought
apple = totalBrought(allGuests, 'apples')
print("Number of things being brought:")
print("Apples: " + str(apple))
print("Cups: " + str(totalBrought(allGuests, 'cups')))
...
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物品清单统计：
1. 一开始写，很呆滞，每个物品单独一个打印语句
2. 还把return写倒了total行同缩进，导致返回的数值一直是1，修改后：
inventory = {'rope': 1, 'torch': 6, 'gold coin': 42, 'dagger': 1, 'arrow': 12}
def totalitem(allItems):
total = 0
for k, v in allItems.items():
total = allItems.get(k, 0) + total
return total
items = totalitem(inventory)
print('Inventory:' + "\n"
+ str(inventory['arrow']) + ' arrow' + "\n"
+ str(inventory['gold coin']) + ' gold coin' + "\n"
+ str(inventory['rope']) + ' rope' + "\n"
+ str(inventory['torch']) + ' torch' + "\n"
+ str(inventory['dagger']) + ' dagger' + "\n"
+ 'Total number of items: ' + str(items))
3. 参考了提示并整合后，打印加入到循环中
inventory = {'rope': 1, 'torch': 6, 'gold coin': 42, 'dagger': 1, 'arrow': 12}
def displayInventory(inventory):
print('Inventory')
total = 0
for k, v in inventory.items():
print(str(v) + ' ' + k)
total += v
print('Total number of items:' + str(total))
displayInventory(inventory)
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假设征服一条龙的战利品表示为这样的字符串列表：
dragonLoot = ['gold coin', 'dagger', 'gold coin', 'gold coin', 'ruby']
写一个名为 addToInventory(inventory, addedItems)的函数，
其中 inventory 参数是一个字典，表示玩家的物品清单（像前面项目一样），
addedItems 参数是一个列表，就像 dragonLoot。
addToInventory()函数应该返回一个字典，表示更新过的物品清单。
def displayInventory(inventory):
print('Inventory')
total = 0
for k, v in inventory.items():
print(str(v) + ' ' + k)
total += v
print('Total number of items:' + str(total))
def addToInventory(inventory, addedItems):
for k in addedItems:
if k in inventory.keys():
inventory[k] += 1
else:
inventory[k] = 1
return inventory
inv = {'gold coin': 42, 'rope': 1}
dragonLoot = ['gold coin', 'dagger', 'gold coin', 'gold coin', 'ruby']
inv = addToInventory(inv, dragonLoot)
displayInventory(inv)
1. 开始想到将列表转换成字典，再将两个字典比较后拼接
2. 若字典中不存在key，则可用dic[key]=value来增加键值对
3. 两个字典相加：相同key，值相加；不同key，值保留
dica = {'a': 1, 'b': 2, 'c': 3, 'f': "hello"}
dicb = {'b': 3, 'k': 'world'}
dic = {}
for key in dica:
if dicb.get(key):
dic[key] = dica[key] + dicb[key]
else:
dic[key] = dica[key]
for key in dicb:
if dica.get(key):
pass
else:
dic[key] = dicb[key]
print(dic)
4. 起初代码一直如下，试了很久v+=1，但v的值始终无变化（其实是混淆了v和dic[k]的意义，v并不是dic[k]）
重看定义：key（）、values（）、items（）方法返回的值是元组，即不可变（设定如此，其实是搅屎棍，变不变的都无所谓）
因为k，v只是字典在使用方法时返回的值（返回的键值值赋值给了变量k，v），所以即使修改变量v的值，对字典本身的值是不起作用的
5. 所以需要循环累加字典内键的值，必须用dict[k]+=1
dictA = {'a': 1, 'b': 42}
listB = ['b', 'c', 'b']
for i in listB:
for k, v in dictA.items():
print(i, k, v, dictA[k])
v += 1 # ×
# dictA[k] += 1 # √
