employee.py

class Employee(object):

def __init__(self, id, name, members=None):
    self.id = id
    self.name = name
    self.members = members
    self.is_team = members is not None

通过数据馈送，实例创建一个Employee对象。

feed.py

from employee import Employee

anne = Employee(0, 'Anne')
bob = Employee(1, 'Bob')
carlos = Employee(2, 'Carlos')
carol = Employee(3, 'Carol')
charlie = Employee(4, 'Charlie')
cherry = Employee(5, 'Cherry')
dave = Employee(6, 'Dave')
emma = Employee(7, 'Emma')
mary = Employee(8, 'Mary')
peggy = Employee(9, 'Peggy')
trent = Employee(10, 'Trent')

admin = Employee(90, 'Admin', [anne, bob, carlos])
engineering = Employee(91, 'Engineering', [carlos, trent, bob])
catering = Employee(92, 'Catering', [emma, anne, bob])

people = [anne, bob, carlos, carol, charlie, cherry, dave, emma, mary,
          peggy, trent, admin, engineering, catering]

到目前为止，我的解决方案，

get_names.py

import feed

def get_names(person, e_list):
    try:
        for p in e_list:
            if p == person and p in e_list:
                print(p.name)
    except AttributeError:
        print('Not found')

print([e.name for e in get_names(feed.carlos, feed.people)])

我希望返回的数据是：

['Admin', 'Catering']

因为卡洛斯属于两者。

Answer 1

我可以用OOP解决此问题：

class Employee(object):
    def __init__(self, id, name):
        self.id = id
        self.name = name
        self.teams = []

class Team(Employee):
    teams = []

    def __init__(self, id, name, members):
        super().__init__(id, name)
        self.members = members
        for member in self.members:
            member.teams.append(name)
        self.__class__.teams.append(self)

    @classmethod
    def find_employee_teams(cls, person):
        found_teams = []
        if isinstance(person, str):
            for team in cls.teams:
                for member in team.members:
                    if member.name == person:
                        found_teams.append(team.name)
            return found_teams
        elif isinstance(person, Employee):
            return person.teams
        # TODO error handling in case person is neither

carlos = Employee(2, 'Carlos')
admin = Team(90, 'Admin', [carlos])
engineering = Team(91, 'Engineering', [carlos])

print(Team.find_employee_teams('Carlos'))
print(Team.find_employee_teams(carlos))

输出

['Admin', 'Engineering']
['Admin', 'Engineering']

find_employee_teams应该用Employee而不是Team来实现。

Answer 2

您应该检查该人的成员列表。

import feed

def get_names(person, e_list):
    try:
        for p in e_list:
            all_members = p.members
            if all_members and person in all_members:
                print(p.name)
                yield p
    except AttributeError:
        print('Not found')


print([e.name for e in get_names(feed.carlos, feed.people)])

Answer 3

您应该在自己的班级中代表一个团队。团队不是员工。这样，您可以使代码更易于阅读和理解。

class Employee(object):    
    employees = set()

    def __init__(self, id, name):
        self.id = id
        self.name = name
        Employee.employees.add(self)

    def in_teams(self):
        """ Return the teams an employee is member of. """
        teams = []
        for team, members in Team.teams.items():
            if self in members:
                teams.append(team)
                continue
        return teams

class Team(object):
    teams = dict()

    def __init__(self, name, members=None):
        self.name = name
        if members:
            self.members = members
        else:
            self.members = set()
        Team.teams[self.name] = self.members

    def add_member(self, employee):
        """ Add an employee to the team. """
        if employee not in self.members:
            self.members.add(employee)

    def rm_member(self, employee):
        """ Remove an employee from the team. """
        if employee in self.members:
            self.members.remove(employee)        

if __name__ == "__main__":
# Employees
anne = Employee(0, "Anne")
bob = Employee(1, "Bob")
carlos = Employee(2, "Carlos")
trent = Employee(10, "Trent")
dave = Employee(6, "Dave")

# Teams
admins = Team("admin", {anne, bob, carlos})
engineering = Team("engineering", {carlos, trent, bob})
engineering.rm_member(trent)
engineering.add_member(dave)

for employee in Employee.employees:
    teams = employee.in_teams()

    if teams:
        print(f"{employee.name} is in the teams {teams}.")
    else:
        print(f"{employee.name} is in no team.")

Answer 4

在打印和返回函数值之间，您会感到困惑。您当前的代码会打印出一些内容（无论如何都不是您想要的内容），但返回None。

此外，您的测试是完全错误的：您应扫描<IfModule mod_rewrite.c> RewriteEngine on RewriteCond %{REQUEST_FILENAME} !-d RewriteCond %{REQUEST_FILENAME} !-f RewriteRule ^preview/(.+)$ /preview/index.php?tslug=$1&cat=$2 [QSA,L] </IfModule>以查看团队，并查看e_list是否是团队成员。

代码可能变为：

person

话虽这么说，团队不应该是名为 Employee 的类的成员。您最好构建一个类的层次结构：

def get_names(person, e_list):
    return [t.name for t in e_list if t.is_team and person in t.members]

这样，角色可以是团队或员工。

我有一个Employee类，我想返回“姓名”列表

employee.py

feed.py

get_names.py

4 个答案: