Python список объектов, разделяющих ссылки на другие объекты?

Question

У меня есть следующий код, который на первый взгляд должен создать 10 заданий с 3 заданиями в каждом.

class Job:
    id = None
    tasks = {}

class Task:
    id = None

cnt = 0
jobs = []
for i in range(0, 10):
    job = Job()
    job.id = i

    for ii in range(0, 3):
        task = Task()    
        task.id = cnt
        job.tasks[task.id] = task
        cnt += 1

    jobs.append(job)

for job in jobs:
    print("job {}, tasks: {}".format(job.id, job.tasks.keys()))

Результат несколько удивителен - у нас есть 30 заданий, поделенных каждым заданием:

job 0, tasks: dict_keys([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29])
job 1, tasks: dict_keys([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29])
job 2, tasks: dict_keys([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29])
job 3, tasks: dict_keys([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29])
job 4, tasks: dict_keys([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29])
job 5, tasks: dict_keys([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29])
job 6, tasks: dict_keys([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29])
job 7, tasks: dict_keys([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29])
job 8, tasks: dict_keys([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29])
job 9, tasks: dict_keys([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29])

Может кто-нибудь объяснить, что здесь происходит?

Задачи UPDATE - это переменная класса, общая для всех экземпляров.

Answer 1

В вашем классе Job вы должны сделать это

class Job:
    id = None
    def __init__(self):
        self.tasks = {}

tasks в вашем классе, и каждый раз, когда вы добавляете класс tasks, который является общим для всех экземпляров.