python 字典所有操作

Posted 2022-05-28 ericblog1992

# 字典的创建
# dict1 = 
# print(type(dict1))
#
# dict2 = 
#         ‘name‘:‘汪峰‘,
#         ‘sex‘:‘男‘,
#         ‘hiredate‘:‘1997-10-20‘
#         
# print(dict2)
#
# dict3 = dict(name = ‘eric‘,age = 28)
# print(dict3)

#  取值操作
# employee = 
#         ‘name‘:‘汪峰‘,
#         ‘sex‘:‘男‘,
#         ‘grade‘:‘A‘,
#         ‘hiredate‘:‘1997-10-20‘,
#         ‘salary‘: 1000,
#         ‘welfare‘:100
#         
# name = employee[‘name‘]
# print(name)
#
# sex = employee.get(‘sex‘)
# print(sex)
#
# dept = employee.get(‘dept‘,‘其他部门‘)
# print(dept)

# in 成员运算符  判断 key 是否存在
# print(‘name‘ in employee)  # True
# print(‘dept‘ in employee)  # False
# print(‘dept‘ not in employee)  # True

# 遍历字典
# 汪峰
# 男
# 1997-10-20
# for key in employee:
#     v = employee[key]
#     print(v)

# name 汪峰
# sex 男
# hiredate 1997-10-20
# for k,v in employee.items():
#     print(k,v)


# 单个更新
# employee[‘grade‘] = ‘B‘
# print(employee)
# # 多个更新
# employee.update(salary = 1200, welfare = 150)
# print(employee)
#
# # 新增
# employee[‘dept‘] = ‘研发部‘
# print(employee)
#
# #  删除操作
# employee.pop(‘dept‘)
# print(employee)
# #  删除操作  删除最后一个 kv
# employee.popitem()
# print(employee)
#
# # 清空字典
# employee.clear()
# print(employee)


# emp1 = ‘name‘:‘jacky‘,‘grade‘:‘B‘,‘hiredate‘:‘1989-10-12‘
# emp2 = ‘name‘:‘Lily‘,‘grade‘:‘A‘,‘hiredate‘:‘1989-10-12‘
#
# # setdefault 设置默认值 存在就忽略 不存在就创建
# emp2.setdefault(‘grade‘,‘C‘)
# print(emp2)

#  获取字典的视图
# k = emp1.keys()
# # dict_keys([‘name‘, ‘grade‘])
# print(k)
#
# v = emp1.values()
# # dict_values([‘jacky‘, ‘B‘])
# print(v)
#
# items = emp1.items()
# # dict_items([(‘name‘, ‘jacky‘), (‘grade‘, ‘B‘)])
# print(items)

# 格式化字符串
# emp_str = "姓名:name,评级:grade,入职时间hiredate".format_map(emp1)
# print(emp_str)

# 散列值
# h1 = hash("dabc")
# # print(h1)
# # h2 = hash("1231231")
# # print(h2)

# 例子
source = "12345,jace,ercd,php,5000$2345,clead,ercd,dfd,5000$22345,clead,ercd,dfd,5000$32345,clead,ercd,dfd,5000"
employee_list = source.split("$")
print(employee_list)

all_emp = 
for i in range(0,len(employee_list)):
    # print(i)
    e = employee_list[i].split(",")
    # print(e)
    employee = ‘no‘:e[0],‘name‘:e[1],‘www‘:e[2],‘jod‘:e[3],‘salary‘:e[4]
    print(employee)
    all_emp[employee[‘no‘]] = employee
print(all_emp)

empno = input(‘编号:‘)
if empno in all_emp:
    emp = all_emp.get(empno)
    print(emp);
else:
    print(‘编号不存在‘)