如何解决列表推导中的类没有属性
当我在列表推导式中打印该类时,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,它会为所有名称设置名称,好像它们都是同一个类?
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)
解决方法
问题是这一行没有初始化对象:
all_a = [A for i in range(5)]
需要改成这样:
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
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