Defining a @classmethod lambda in python

Question

class TestClass(object):
    aa = lambda x: 35
    def __init__(self):
        self.k = self.aa()


o = TestClass()
print o.k

This gives me 35, which I understand why.

But this:

class TestClass(object):
    @classmethod
    aa = lambda x: 35
    print type(aa)
    def __init__(self):
        self.k = TestClass.aa()


o = TestClass()
print o.k

This gives me

  File "test1.py", line 3
    aa = lambda x: 35
     ^
SyntaxError: invalid syntax

Why so ?

BrenBarn · Accepted Answer · 2017-07-10 18:25:41Z

6

Decorators are only syntactically valid on def and class statements. But the decorator syntax is just shorthand for calling the decorator with the decorated function (or class) as its argument, so you can achieve the same result with:

class TestClass(object):
    aa = classmethod(lambda x: 35)
    # etc.

answered Jul 10, 2017 at 18:25

BrenBarn

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Comments

LVFToENlCx · Accepted Answer · 2017-07-10 18:27:27Z

1

You can't use a decorator on a lambda. You could replace it with

aa = classmethod(lambda x:35)

answered Jul 10, 2017 at 18:27

LVFToENlCx

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