Question

我不知道为什么在尝试访问类中的函数时得到NameError。

这是我遇到问题的代码。我想念什么吗？

class ArmstrongNumber:

    def cubesum(num):
        return sum([int(i)**3 for i in list(str(num))])

    def PrintArmstrong(num):
        if cubesum(num) == num:
            return "Armstrong Number"
        return "Not an Armstrong Number"

    def Armstrong(num):
        if cubesum(num) == num:
            return True
        return False

[i for i in range(1000) if ArmstrongNumber.Armstrong(i)] # this return NameError

错误消息：

NameError                                 Traceback (most recent call last)
<ipython-input-32-f3d39f24a48c> in <module>
----> 1 ArmstrongNumber.Armstrong(153)

<ipython-input-31-fd21586166ed> in Armstrong(num)
     10 
     11     def Armstrong(num):
---> 12         if cubesum(num) == num:
     13             return True
     14         return False

NameError: name 'cubesum' is not defined

Answer 1

在方法前使用classname

class ArmstrongNumber:

    def cubesum(num):
        return sum([int(i)**3 for i in list(str(num))])

    def PrintArmstrong(num):
        if ArmstrongNumber.cubesum(num) == num:
            return "Armstrong Number"
        return "Not an Armstrong Number"

    def Armstrong(num):
        if ArmstrongNumber.cubesum(num) == num:
            return True
        return False

print([i for i in range(1000) if ArmstrongNumber.Armstrong(i)])

请避免将self传递给那些不是instance methods的函数。即使您在类中定义了该类，您仍然需要使用classname访问它们。

Answer 2

如果您真的想使用类，这应该是您的实际解决方案

class ArmstrongNumber(object):

    def cubesum(self, num):
        return sum([int(i)**3 for i in list(str(num))])

    def PrintArmstrong(self, num):
        if self.cubesum(num) == num:
            return "Armstrong Number"
        return "Not an Armstrong Number"

    def Armstrong(self, num):
        if self.cubesum(num) == num:
            return True
        return False


a = ArmstrongNumber()

print([i for i in range(1000) if a.Armstrong(i)])

输出

[0, 1, 153, 370, 371, 407]

第二种方法：

如果您不想使用类，请使用这样的静态方法

def cubesum(num):
    return sum([int(i)**3 for i in list(str(num))])


def PrintArmstrong(num):
    if cubesum(num) == num:
        return "Armstrong Number"
    return "Not an Armstrong Number"


def Armstrong(num):
    if cubesum(num) == num:
        return True
    return False


# a = ArmstrongNumber()

print([i for i in range(1000) if Armstrong(i)])

为什么在访问函数时出现NameError

2 个答案: