day5

#dict字典
#dict字典

#数据类型划分；可变数据和不可变数据
#不可变数据：元组、bool、 int、str 可哈西
#可变数据：list dict set   不可哈西

# k--value
# key 不可变数据类型
# value任意数据类型
#dict优点：二分查找速度快，储存量关系

# dic1={
#     "name":["hsaj","hjsahsa","sjkjasll"],
#     "age":[{"num":12 ,"avg-age":13}],
#     True:1,
#     (1,2,3):"hsjhjs",
# }
# print(dic1)

#增1
# dic1["high"]=28918#无则添加
# dic1["name"]=21898#有则覆盖
# print(dic1)

#增2 setdefault(增加key，value)
# dic1.setdefault("weight")#没有添加
# print(dic1)
# dic1.setdefault("weight",123)#没有添加，有不做改变
# print(dic1)
# dic1.setdefault("name",23)#此处有name，不做任何改变
# print(dic1)

#删除按索引
# dic1={'name':"hsaj","age":12,"askha":"snakns","weight":"hska"}
# print(dic1.pop('name'))#有就删，没有就报错
# print(dic1.pop("age",None))#有就删，没有不报错

#删除popitem
# dic1.popitem()#从后方删除，有返回值，元组没有
# print(dic1)

#删除del没有返回值
# del dic1['name']
# print(dic1)

# #清空字典
# dic1.clear()

# #改1
# dic1['age']=13

#改2
# dic1={'name':"hsaj","age":12,"askha":"snakns","weight":"hska"}
# dic2={'name':"hs","age":12,"askha":"snakns"}
# dic2.update(dic1)#有则覆盖，无则添加
# print(dic1)
# print(dic2)

#查
#查
# print(dic1.values(),type(dic1.values()))
# print(dic1.items())
# print(dic1.keys(),type(dic1.keys()))

# print(dic1["name"])#输出对应value#此处没有报错
# print(dic1.get("name1","没有这个key"))#get(查询值，查不到时的返回值)

#for i
# for i in dic1:#默认打印key
#     print(i)
# for i in dic1.values():
#     print(i)
# for k,v in dic1.items():#items()
#     print(k,v)

# #面试题 转换ab值
# a=1
# b=2
# a,b=b,a
# print(a,b)
# a,b=[1,2]
# a,b=[1,2],[2,3]
# a,b=(1,2)
# a,b=[1,2,3]#错误写法

#嵌套
#嵌套

# dic1={
#     "name":["hsaj","hjsahsa","sjkjasll"],
#     "age":{"num":12 ,
#            "avg-age":13,
#            },
#     True:1,
#     (1,2,3):"hsjhjs",
# }

# #改
# dic1["age"]=12
# print(dic1)
# # 改大写
# dic1["name"][1]=dic1["name"][1].upper()#只能用于列表内全是字符串的才能大写
# print(dic1)
# #字典内列表添加
# dic1["name"].append("zhangsan")
# print(dic1)
# #字典内字典添加
# dic1["age"]["ages"]=3#无则添加
# print(dic1)
# dic1["age"]["num"]=3#有则覆盖
# print(dic1)

##题目
info=input('>>>>')
for i in info:
    if i.isalpha():
        info=info.replace(i,' ')
l=info.split()
print(len(l))