Lambda函数传递不需要的自我 [英] Lambda function passing not desired self

问题描述

查看以下代码:

class MyClass_1():
    @staticmethod
    def method_1(func):
        return func(1, 2, 3)

class MyClass_2():
    my_func = lambda a,b,c : a*b*c # I need to call this method

    def method_2(self):
        result = MyClass_1.method_1(self.my_func)
        print(result)

我的错误:

TypeError:()接受3个位置参数，但给出了4个

TypeError: () takes 3 positional arguments but 4 were given

我需要以与上面的代码相同的方式调用lambda函数my_func，但是在我不知道的地方出现self并导致此错误.

I need to call the lambda function my_func in the same way as the code above, but a self is appearing from somewhere I don't know and causing this error.

我想念什么?

推荐答案

由于my_func是MyClass_2的类属性，因此您不应该通过self(该类的实例)访问它.相反，您应该直接通过类访问它:

Since my_func is a class attribute of MyClass_2, you should not be accessing it through self (an instance of the class). Instead, you should be accessing it through the class directly:

result = MyClass_1.method_1(MyClass_2.my_func)
                            ^^^^^^^^^

演示:

>>> class MyClass_1():
...     @staticmethod
...     def method_1(func):
...         return func(1, 2, 3)
...
>>> class MyClass_2():
...     my_func = lambda a,b,c : a*b*c # I need to call this method
...     def method_2(self):
...         result = MyClass_1.method_1(MyClass_2.my_func)
...         print(result)
...
>>> MyClass_2().method_2()
6
>>>

有关更多信息，您可以查看以下来源:

For more information, you can check out these sources:

• https://docs.python.org/3/reference /compound_stmts.html#class-definitions
• Python:类和实例属性之间的区别

• https://docs.python.org/3/reference/compound_stmts.html#class-definitions
• Python: Difference between class and instance attributes

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