字典dict

最近学习了字典，笔记整理在代码段中。

# _author:VULCAN
# date:2018/8/14
# 字典
# 增加：
dic = {"8": "3306", "name": "p", "kl": "xzc"}
# print(dic)
#
# dic["8"] = "4212"  # 若键存在,将替换键对应的原来的key，若键不存在,将新增一对键值对
# print(dic)
#
# dic["sex"] = "male"
# print(dic)
# # 键存在,不改动,返回字典中相应的建对应的相应的值
# return1 = dic.setdefault("name", "9910")
# print(return1)
# # 键不存在,在字典中增加新的键值对，并返回相应的值
# return2 = dic.setdefault("height", 186)
# print(return2)


# 删除：
# dic.clear()  # 清除字典
# print(dic)
# dic.pop("8")  # 删除字典中指定键值对,并返回一个该键值对的值
# ret = dic.pop("8")  # 跟列表pop删除操作一样，有返回值
# print(ret)
# ret1 = dic.popitem()  # 随机删除某组键值对,并以元组的方式返回
# print(ret1)
# print(dic)
# print(dic)
# a = dic.copy()  # 拷贝一个跟dic数据一致的字典
# print(a)
# del dic["8"]  # 删除字典中指定键值对
# print(dic)
# del dic  # 删除整个字典
# print(dic)

# 修改：
# dic["name"] = "fok"
# print(dic)
# dic1 = {1: 3, 2: 5, 5: 6}
# dic1.update(dic)
# print(dic1)
# print(dic)
# dic2 = {1: "opg", 9: '"lm'}  # 覆盖时没有相同的键时，直接添加到dic1中，有相同的键是，将values值替换
# dic1.update(dic2)
# print(dic1)
# print(dic2)

# # 查找：通过键来查找值
# print(dic["name"])
# print(dic.keys())
# print(dic.values())
# print(dic.items())
# print(type(dic.values()))
# # 将查到的值转换为列表
# print(list(dic.keys()))
# print(list(dic.values()))
# print(list(dic.items()))

# 这样也可以生成字典，通过dict.fromkeys()方法，将["test1", "test2", "test3"]中的值当做每一对键值对的键,["valuestest"]当做所有键值对的值
# dic_test = dict.fromkeys(["test1", "test2", "test3"], ["valuestest"])
# print(dic_test)  # {'test1': ['valuestest'], 'test2': ['valuestest'], 'test3': ['valuestest']}

# 字典嵌套使用：
# dic_1 = {"1001": {"101": ["101_01", "101_02"], "201": ["201_01", "201_02"]},
#          "1002": {"102": ["102_02", "102_02"], "202": ["202_01", "202_02"]},
#          "1003": {"103": ["103_03", "103_03"], "203": ["203_01", "203_02"]}}
# print(dic_1)
# dic_1["1002"]["202"][1] = "3306"
# print(dic_1)

# 字典的排序
dic_2 = {333: "777", 666: "000", 111: "999"}
# print(sorted(dic_2))  # 默认是对键进行排序
# print(sorted(dic_2.keys()))  # 效果同上
# print(sorted(dic_2.values()))  # 对值进行排序

# 遍历字典：
# for i in dic_2:  # 效率高
#     print(i, dic_2[i])
#
# for i, v in dic_2.items():  # 效率较上低
#     print(i, v)


# 字符串String操作：
# "Hello"跟'Hello'没有任何区别
# a = "les's go!"
# print(a)
# a = "'他''她''它'"
# print(a)

# print("hello" * 100)  # print("hello"*n)    n为几就打印几遍"Hello"