Python3:如何创建同一类的不同对象的dict?

2024-05-16 04:20:30 发布

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我想创建一个dict,包含同一类的几个对象。每个对象必须是独立的。 比如:

#!/usr/bin/python3

class myReserve():

    myList = dict()

    def __init__(self, initName):
        self.myName = initName
        self.setList()

    def setList(self):
        if self.myName == "fruit":
            self.myList[0] = "Orange"
            self.myList[1] = "Lemon"
        elif self.myName == "vegetable":
            self.myList[0] = "Tomato"
            self.myList[1] = "Carrot"
        #If neither fruit nor vegetable
        #myList should be empty.


myStore = dict()
myStore[0] = myReserve("fruit")
myStore[1] = myReserve("vegetable")
myStore[2] = myReserve("spices")

print(myStore[0].myList)

这张照片:

{0: 'Tomato', 1: 'Carrot'}

我以为它会打印:

{0: 'Orange', 1: 'Lemon'}

我知道对象在Python中是通过引用传递的。你知道吗

dict1 = {"var": 128}
dict2 = dict1
dict2["var"] = 0
print(dict1["var"])

将打印:

0

通过创建一个类,我想为不同的对象创建一个结构。我不理解第一个代码示例的行为。有可能用Python的方式做这样的事情吗?你知道吗


Tags: 对象selfvardefdictsetlistfruitorange
1条回答
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1楼 · 发布于 2024-05-16 04:20:30

您的问题是在类级别上定义myList,因此myReserve的每个实例都共享它。请尝试在myReserve.__init__中定义它:

class myReserve():
    def __init__(self, initName):
        self.myList = dict()

完整代码:

#!/usr/bin/python3

class myReserve():
    def __init__(self, initName):
        self.myList = dict()
        self.myName = initName
        self.setList()

    def setList(self):
        if self.myName == "fruit":
            self.myList[0] = "Orange"
            self.myList[1] = "Lemon"
        elif self.myName == "vegetable":
            self.myList[0] = "Tomato"
            self.myList[1] = "Carrot"
        #If neither fruit nor vegetable
        #myList should be empty.


myStore = dict()
myStore[0] = myReserve("fruit")
myStore[1] = myReserve("vegetable")
myStore[2] = myReserve("spices")

print(myStore[0].myList)

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