Question

这是我想要做的事情

class BaseClass(object):
    successify = lambda x: "<Success>%s</Success>" % x
    errorify = lambda x: "<Error>%s</Error>" % x
    def try1(self):
        print successify("try1")
    def try2(self):
        print self.successify("try2")

但这两种方法似乎都不起作用..

>>> BaseClass().try1()
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "<stdin>", line 5, in try1
NameError: global name 'successify' is not defined
>>> BaseClass().try2()
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "<stdin>", line 7, in try2
TypeError: <lambda>() takes exactly 1 argument (2 given)

如何在类中使用lambdas作为方法？

Answer 1

使用/访问lambdas的类变量几乎没有可能。其中三个是：

class BaseClass(object):
    successify = lambda x: "<Success>%s</Success>" % x
    errorify = lambda x: "<Error>%s</Error>" % x
    def try1(self):
        print(self.__class__.successify("try1"))
    def try2(self):
        print(self.__class__.successify("try2"))

 # or 

class BaseClass(object):
    successify = lambda x: "<Success>%s</Success>" % x
    errorify = lambda x: "<Error>%s</Error>" % x
    def try1(self):
        print(BaseClass.successify("try1"))
    def try2(self):
        print(BaseClass.successify("try2"))

# or Please not changes to lambda definitions below

class BaseClass(object):
    successify = lambda self,x: "<Success>%s</Success>" % x
    errorify = lambda self,x: "<Error>%s</Error>" % x
    def try1(self):
        print(self.successify("try1"))
    def try2(self):
        print(self.successify("try2"))

Answer 2

使用lambda self, x: "...%s..." % x

Answer 3

如果要访问successify作为对象函数，则第一个参数为self表示对象实例。

如果必须使用self.successify

，则必须更改lambada函数

successify = lambda self, x: "<Success>%s</Success>" % x

Answer 4

不幸的是，Marcin's的第一个答案在Python 2中不起作用。总而言之，他写道

class BaseClass(object):
    successify = lambda x: "<Success>%s</Success>" % x
    errorify = lambda x: "<Error>%s</Error>" % x
    def try1(self):
        print(self.__class__.successify("try1"))
    def try2(self):
        print(self.__class__.successify("try2"))

但是这在Python 2中不起作用。一个有趣的技巧是使类认为它是静态方法而不是静态变量。适当更改代码后，它变为：

class BaseClass(object):
    successify = staticmethod(lambda x: "<Success>%s</Success>" % x)
    errorify = staticmethod(lambda x: "<Error>%s</Error>" % x)
    def try1(self):
        print(self.__class__.successify("try1")) # Works!
    def try2(self):
        print(self.__class__.successify("try2")) # Works!

如何在类中使用lambda作为方法？

4 个答案: