使用for循环将@classmethod创建的对象添加到Python列表中

Question

我为变量homeTasks分配了以下值列表：

[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12]

我有一个for循环将类Task的对象列表分配给变量allTasks

# create empty allTasks list
allTasks=[]

# create thisTask for every value of homeTasks and append to allTasks
for y in range(len(homeTasks)):
        thisTask = Task.get_task(homeTasks[y])
        allTasks.append(thisTask)
        print(allTasks[y].taskName)

# print task names from allTasks for testing:
for z in range(len(allTasks)):
        print(z)
        print(allTasks[z].taskName)

这给了我以下信息：

Do Dishes
Make Bed
Laundry
Study
Laundry
Elliptical
Elliptical
Laundry
Elliptical
Elliptical
Elliptical
test
0
test
1
test
2
test
3
test
4
test
5
test
6
test
7
test
8
test
9
test
10
test
11
test

由于某些原因，分配给allTasks的最后一个对象被分配给列表中的每个索引。这适用于thisTask的所有属性

作为参考，以下是我的任务类和get_任务类方法的外观：

class Task(object):
    def __init__(self, taskID, taskName, points, active, complete, approved, assignedUserID, createdByUserID, dateCreated, dateCompleted, frequency):
        self.taskID = taskID
        self.taskName = taskName
        self.points = points
        self.complete = complete
        self.approved = approved
        self.assignedUserID = assignedUserID
        self.createdByUserID = createdByUserID
        self.dateCreated = dateCreated
        self.dateCompleted = dateCompleted
        self.frequency = frequency
        self.homeID = homeID

    @classmethod

    def get_task(self, taskID):
        t = (taskID,)
        cur = mysql.connection.cursor()
        cur.execute("SELECT * FROM tasks WHERE taskID=%s", t)
        columns = [col[0] for col in cur.description]
        task = [dict(zip(columns, row)) for row in cur.fetchall()]
        self.taskID = taskID 
        self.taskName = task[0]['taskName']
        self.points = task[0]['points']
        self.aproved = task[0]['approved']
        self.assignedUserID = task[0]['assignedUserID']
        self.createdByUserID = task[0]['createdByUserID']
        self.dateCreated = task[0]['dateCreated']
        self.dateCompleted = task[0]['dateCompleted']
        self.frequency = task[0]['frequency']
        self.dueDate = task[0]['dueDate']
        self.homeID = task[0]['homeID']
        return self

我知道该方法工作正常，因为每次循环迭代时thisTask都有正确的值。似乎.append不起作用，因为一旦我退出循环，allTasks中的值就错了

编辑：根据@barny，我为以下内容创建了一个最小的可复制示例：

import copy

# initialize Task object

class Task(object):
    def __init__(self, taskID, taskName, points, active, complete, approved, assignedUserID, createdByUserID, dateCreated, dateCompleted, frequency):
        self.taskID = taskID
        self.taskName = taskName

# define method to return values

    @classmethod

    def get_task(self, index):
        taskIDList = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12]
        taskNames = ['Do Dishes', 'Make Bed', 'Laundry', 'Study', 'Laundry', 'Elliptical', 'Elliptical', 'Laundry', 'Elliptical', 'Elliptical', 'Elliptical', 'test']
        self.taskID = taskIDList[index]
        self.taskName = taskNames[index]
        return self

# create empty allTasks list
allTasks=[]

homeTasks = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12]

# create thisTask for every value of homeTasks and append to allTasks
for y in range(len(homeTasks)):
        thisTask = Task.get_task(y)
        allTasks.append(thisTask)
        print(allTasks[y].taskName)

# print task names from allTasks for testing:
for z in range(len(allTasks)):
        print(z)
        print(allTasks[z].taskName)

此外，根据反馈，我现在了解到thisTask是对对象的引用，而不是get_task结果的新副本。为此，我尝试在{}和{}上使用{}和{}都没有效果。似乎连副本都是参考资料。还有别的办法吗

编辑2：以下是我尝试使用copy和deepcopy的方法：

for y in range(len(homeTasks)):
        print(y)
        thisTask = copy.copy(Task.get_task(y))
        allTasks.append(thisTask)
        print(allTasks[y].taskName)

for y in range(len(homeTasks)):
        print(y)
        thisTask = copy.deepcopy(Task.get_task(y))
        allTasks.append(thisTask)
        print(allTasks[y].taskName)

for y in range(len(homeTasks)):
        print(y)
        thisTask = Task.get_task(y)
        allTasks.append(copy.copy(thisTask))
        print(allTasks[y].taskName)

for y in range(len(homeTasks)):
        print(y)
        thisTask = Task.get_task(y)
        allTasks.append(copy.deepcopy(thisTask))
        print(allTasks[y].taskName)

在每次测试后：

for z in range(len(allTasks)):
        print(z)
        print(allTasks[z].taskName)

这四种方法产生了完全相同的结果：

0
Do Dishes
1
Make Bed
2
Laundry
3
Study
4
Laundry
5
Elliptical
6
Elliptical
7
Laundry
8
Elliptical
9
Elliptical
10
Elliptical
11
test
0
test
1
test
2
test
3
test
4
test
5
test
6
test
7
test
8
test
9
test
10
test
11
test

Answer 1

getTask不会在每次调用时创建新的Task；在使用当前元素allTasks中的值修改后，它总是返回相同的值。因此，您已经将相同的Task重复添加到列表中，最后得到了上次调用getTask的值

解决这一问题的一种方法是每次{}时从{}的当前内容生成一个新的{}；例如：

Task(thisTask.taskID, thisTask.taskName, 1,1,1,1,1,1,1,1,1)

或者更好的方法是，您可以修改getTask来执行此操作：

return Task( taskIDList[index], taskNames[index], 1,1,1,1,1,1,1,1,1 )