列表理解中的类没有属性

Question

当我在列表理解中打印该类时，它说它没有属性“名称”，但它显然有。另外，如果我为一个设置名称，它会为所有设置名称，就好像它们都是同一个类一样？

class A(object):
    def __init__(self, name):
        if name:
            self.name = name
        else:
            self.name = "no name"

    def __repr__(self):
        return self.name
                        
all_a = [A for i in range(5)]
other_a = []
    
for i in range(5):
    name = " name " + str(i)
    a = A(name)
    other_a.append(a)

print(all_a)

for a in all_a:
    print(a)
    print(a.name)

all_a[0].name = "all a zero"
all_a[1].name = "all a one"

for a in all_a:
    print(a)
    print(a.name)
        
for a in other_a:
    print(a)

如果我删除 name 参数，它仍然会崩溃：

class A(object):
    def __init__(self):
        self.name = "no name"

    def __repr__(self):
        return self.name
                        
all_a = [A for i in range(5)]
other_a = []
    
for i in range(5):
    name = " name " + str(i)
    a = A()
    other_a.append(a)

print(all_a)

for a in all_a:
    print(a)
    print(a.name)

all_a[0].name = "all a zero"
all_a[1].name = "all a one"

for a in all_a:
    print(a)
    print(a.name)
        
for a in other_a:
    print(a)

Answer 1

问题在于这一行没有初始化对象：

all_a = [A for i in range(5)]

需要改成这样的：

all_a = [A(str(i)) for i in range(5)]

另一种选择是将name 声明为类属性：

class A(object):

    name = "Undefined"

    def __init__(self, name):
        if name:
            self.name = name
        else:
            self.name = "no name"

    def __repr__(self):
        return self.name

all_a = [A for i in range(5)]

for a in all_a:
    print(a)
    print(a.name)

输出：

<class '__main__.A'>
Undefined
<class '__main__.A'>
Undefined
<class '__main__.A'>
Undefined
<class '__main__.A'>
Undefined
<class '__main__.A'>
Undefined