# 基础款
user ={"name":"Alice","age":30}
# 嵌套高阶款  
student ={
"id":1001,
"courses":{"Python":95,"SQL":88},
"hobbies":["阅读","摄影"]
}

# 关键字参数  
user =dict(name="Bob", age=25)
# 可迭代对象转字典（列表/元组的键值对）  
items =[("name","Charlie"),("age",35)]
user =dict(items)

empty_dict ={}# 推荐  
empty_dict =dict()# 等价，但前者更简洁  

# 将两个列表转换为字典
keys =['a','b','c']
values =[1,2,3]
new_dict ={k: v for k, v inzip(keys, values)}
# 输出: {'a': 1, 'b': 2, 'c': 3}

# 根据条件过滤创建字典
numbers =[1,2,3,4,5,6]
result ={x: x **2for x in numbers if x %2==0}
# 输出: {2: 4, 4: 16, 6: 36}

# 创建包含三个键，且值都为 None 的字典
default_dict =dict.fromkeys(['x','y','z'])
# 输出: {'x': None, 'y': None, 'z': None}

# 创建包含四个键，且值都为 0 的字典
counter_dict =dict.fromkeys(['apple','banana','cherry','date'],0)
# 输出: {'apple': 0, 'banana': 0, 'cherry': 0, 'date': 0}

# 危险！键不存在时直接报错  
print(user["address"])# KeyError: 'address'  

# 安全模式：`get()`方法+默认值  
address = user.get("address","未填写")# 返回 "未填写"  

# 添加新键值对（键不存在时）  
user["address"]="上海"
# 修改现有值（键存在时）  
user["age"]=26# 原值25被覆盖  

# 删除指定键并获取值（键不存在时可设默认值）  
age = user.pop("age",None)# 返回26，字典中移除"age"  
# 随机删除一个键值对（3.7+按插入顺序删除最后一个）  
last_item = user.popitem()# 返回 ("address", "上海")  
# 清空字典  
user.clear()# 字典变为 {}  

# 遍历所有键值对（最常用）  
for key, value in student.items():
print(f"{key}: {value}")

# 单独遍历键或值  
for key in student.keys():# 等价于直接for key in student  
for value in student.values():

# 场景：生成数字1-5的平方字典  
squares ={x: x**2for x inrange(1,6)}
# 输出：{1: 1, 2: 4, 3: 9, 4: 16, 5: 25}  

# 高阶：过滤+转换（提取成绩≥90的科目）  
grades ={"Python":95,"SQL":88,"Java":92}
high_grades ={k: v for k, v in grades.items()if v >=90}
# 输出：{"Python": 95, "Java": 92}  

# 存储多个用户信息  
users ={
"user1":{"name":"Alice","age":30,"score":85},
"user2":{"name":"Bob","age":25,"score":92},
}
# 访问嵌套值  
print(users["user2"]["score"])# 输出92  

# 方法1：update() 原地修改  
a ={"a":1,"b":2}
b ={"b":3,"c":4}
a.update(b)# a变为 {"a": 1, "b": 3, "c": 4}  

# 方法2：`|` 操作符（Python 3.9+）生成新字典  
c = a | b  # c是新字典，a和b保持不变  

valid_key =("name","Alice")# 元组可作为键  
invalid_key =["name","Alice"]# 列表不能作为键，会报错  

text ="Python is great. Python is simple. Python is powerful."
word_count ={}
for word in text.split():
# 去除标点（简化处理）  
    cleaned_word = word.strip(".")
    word_count[cleaned_word]= word_count.get(cleaned_word,0)+1
print(word_count)
# 输出：{'Python': 3, 'is': 3, 'great': 1, 'simple': 1, 'powerful': 1}  

students ={
"S001":{"name":"张三","math":85,"english":90},
"S002":{"name":"李四","math":92,"english":88},
}
# 需求：计算每个学生的总分  
for student_id, info in students.items():
    total = info["math"]+ info["english"]
    info["total"]= total
print(students["S001"]["total"])# 输出175  

# 模拟配置文件内容  
config_str ="""
host=localhost
port=8080
debug=True
"""
config ={}
for line in config_str.strip().split("\n"):
    key, value = line.split("=")
# 自动转换类型（简化逻辑）  
    config[key]= value if value.isalpha()elseeval(value)
print(config["port"])# 输出8080（数字类型）  

# ❌ 错误：直接访问不存在的键  
print(user["nonexistent_key"])# 运行时报错  

# ✅ 正确：先检查或用get()  
if"email"in user:
print(user["email"])
else:
print("邮箱未注册")

key =("name","Alice")
data ={key:"value"}
# ❌ 错误：修改元组中的元素（会导致键失效）  
key[0]="Bob"# 元组不可变，此操作本身会报错  

a ={"user":"admin","level":1}
b ={"user":"guest","mode":"read"}
a.update(b)# "user"键被覆盖为"guest"  

# ❌ 反模式：值为重复列表，浪费内存  
bad_dict ={f"item_{i}":[1,2,3]for i inrange(1000)}
# ✅ 优化：存储公共列表的引用（需确保不可变）  
common_list =[1,2,3]
good_dict ={f"item_{i}": common_list for i inrange(1000)}

# ❌ 错误：遍历中删除元素导致异常  
data ={"a":1,"b":2,"c":3}
for key in data:
    data.pop(key)# 报错：字典大小改变期间迭代  

# ✅ 正确：遍历副本或使用列表存储键  
for key inlist(data.keys()):
    data.pop(key)