Python基础篇之元组tuple与字典dict用法详解

t1 = (1, 2, 3)
t2 = 1, 2, 3          # 括号可以省略，但不建议
t3 = ()               # 空元组
t4 = (1,)             # 单元素元组，逗号必须有！
t5 = (1)              # ❌ 这不是元组，就是数字 1

t = (1, 2, 3)
t[0] = 100      # ❌ TypeError

t = ([1, 2], "hello")
t[0].append(3)
print(t)   # ([1, 2, 3], 'hello')   ← 能改，因为改的是列表

t = (1, 2, 3, 2)

len(t)          # 4
t[0]            # 1
t[::-1]         # (2, 3, 2, 1)  切片返回元组
t.count(2)      # 2
t.index(3)      # 2
3 in t          # True
t + (4, 5)      # (1, 2, 3, 2, 4, 5)  拼接生成新元组

# 坐标
x, y = (10, 20)

# 交换变量
a, b = b, a

# 函数返回多个值
def get_user():
    return "小庄", 18, "北京"

name, age, city = get_user()

# 用 * 收集剩余
first, *rest = [1, 2, 3, 4]
print(first)   # 1
print(rest)    # [2, 3, 4]

# 只要首尾
head, *_, tail = [1, 2, 3, 4, 5]
print(head, tail)   # 1 5

user = {"name": "小庄", "age": 18, "city": "北京"}
empty = {}
empty2 = dict()

# 从键值对序列创建
dict([("a", 1), ("b", 2)])     # {'a': 1, 'b': 2}
dict(name="小庄", age=18)       # {'name': '小庄', 'age': 18}

user = {"name": "小庄", "age": 18}

user["name"]            # 小庄
user["gender"]          # ❌ KeyError！

user.get("gender")      # None      安全，不会报错
user.get("gender", "未知")  # 未知    带默认值

"name" in user          # True     查 key 是否存在

user["gender"] = "男"          # key 不存在 = 新增
user["age"] = 20               # key 存在 = 修改
user.update({"age": 21, "job": "程序员"})   # 批量更新

user.pop("city")        # 删除并返回该值
user.pop("xxx", None)   # 不存在时返回 None，不报错
del user["age"]         # 直接删
user.clear()            # 清空

user = {"name": "小庄", "age": 18, "city": "北京"}

# 遍历 key（默认）
for k in user:
    print(k)

# 遍历 value
for v in user.values():
    print(v)

# 遍历键值对（最常用）
for k, v in user.items():
    print(f"{k}: {v}")

user.keys()          # 所有 key
user.values()        # 所有 value
user.items()         # 所有键值对
len(user)            # 3
user.setdefault("level", 1)   # 有就返回原值，没有就设置

# 生成平方映射
{i: i**2 for i in range(5)}   # {0:0, 1:1, 2:4, 3:9, 4:16}

# 过滤
{k: v for k, v in user.items() if v}

# 键值互换
{v: k for k, v in d.items()}

# 两个列表合成字典
keys = ["a", "b", "c"]
vals = [1, 2, 3]
dict(zip(keys, vals))          # {'a': 1, 'b': 2, 'c': 3}

text = "apple banana apple orange banana apple"
words = text.split()

count = {}
for word in words:
    count[word] = count.get(word, 0) + 1

print(count)   # {'apple': 3, 'banana': 2, 'orange': 1}

from collections import Counter
count = Counter(words)
print(count.most_common(2))   # [('apple', 3), ('banana', 2)]

# 遍历时删除的正确写法
d = {"a": 1, "b": 2, "c": 3}
for k in list(d.keys()):
    if d[k] < 3:
        d.pop(k)