python的list，dict，set

list

# 1.list() 把可迭代对象转换成list，即for循环遍历的可迭代对象
my_str = "abcdef"
new_list = list(my_str)
print(new_list) # ['a', 'b', 'c', 'd', 'e', 'f']
# 此处可以理解为是通过for循环遍历以后，然后存储到列表中,转元组同理
​
new_list = list(range(10)) # [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
​
new_list = list(range(1,11,2)) # [1, 3, 5, 7, 9]
# 因为字典遍历出来得到是key，所以字典转列表得到就是
my_dict = {"key1":"value1","key2":"value2","key3":"value3"}
new_list = list(my_dict)
print(new_list) # ['key1', 'key2', 'key3']
​
new_list = ["小肥牛","哈哈","小肥牛"]
count_num = new_list.count("小肥牛")
print(count_num) # 返回列表中的元素"小肥牛"出现的次数，若元素不存在，返回0
​
# 字符串，元组也有count的功能，用法一致
new_str = "hahaha"
print(new_str.count("h"))

dict

# 创建字典的四种格式
my_dict = {"key1":"value1"}
my_dict = dict()
mydict = dict(key1=value1,key2=value2)
​
keys = ["key1","key2","key3"]
new_dict = {}.fromkeys(keys,"要给key指定的初始值")
print(new_dict) # {'key1': '要给key指定的初始值', 'key2': '要给key指定的初始值', 'key3': '要给key指定的初始值'}
​
# 可迭代类型转换为字典，容器内必须是以嵌套容器且由两个值才可以转换成字典，多余2个或少于两个直接报ValueError
my_list = [("key1", "value1"), ("key2", "value2")]
print(dict(my_list)) # my_list = [("key1", "value1"), ("key2", "value2")]
print(dict(my_list))
​
# 字典的成员运算，in ，not in 是判断key是否存在字典当中的
​
my_dict = {"key1":"value1","key2":"value2"}
​
# setdefault()方法
old_result = my_dict.setdefault("key1","若有key1，就返回对应的值，若没有就把key1添加到字典中，且返回定义的默认值")
print(old_result) # 因为key1存在，所有直接返回对应的value值--value1
result = my_dict.setdefault("key3","默认值")
print(result) # 默认值
print(my_dict) # {'key1': 'value1', 'key2': 'value2', 'key3': '默认值'}

集合set

# 只能存不可变类型，应用场景：去重，和关系运算(共同爱好，共同好友，共同粉丝，共同关注)
# 无序，且不可重复，用{1,2,3}表示，定义空的集合，必须用set()
# 也可用len()计算长度，in not in 判断是否存在，也可用于for循环
hobbies1 = {"吃饭", "睡觉", "打游戏", "学习"}
hobbies2 = {"吃饭", "睡觉", "看书"}
​
result = hobbies1 & hobbies2  # 取两个集合的交集
# hobbies1.intersection(hobbies2)
print(result) # {'睡觉', '吃饭'}
​
result = hobbies1 | hobbies2 # 取并集
# hobbies1.union(hobbies2)
print(result) # {'睡觉', '吃饭', '看书', '学习', '打游戏'}
​
result = hobbies1 - hobbies2 # 取差集，哪个集合在前，就取哪个集合独有的
# hobbies1.difference(hobbies2)
print(result) # {'打游戏', '学习'}
​
result = hobbies1 ^ hobbies2 # 取对称差集，也就是取出每个人分别独有的爱好
# hobbies1.symmetric_difference(hobbies2)
print(result) # {'学习', '打游戏', '看书'}
​
​
# 枚举出列表中的索引位置,enumerate("可迭代对象，如list,tuple")
li = ["a", "b", "c"]

for i in enumerate(li):
    # i的值为
    # 0, 'a')
    # (1, 'b')
    # (2, 'c')
    print(i)

for index, value in enumerate(li):
    # 此处解包
    print(f"列表中索引位置{index}对应的值为{value}")
 """
 列表中索引位置0对应的值为a
列表中索引位置1对应的值为b
列表中索引位置2对应的值为c
 """