判断是否为函数

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#1 Created at ... [Delete] [Delete and Lock User]

# -*- coding: utf-8 -*-

import types

def is_func(fn):

return type(fn) == types.FunctionType or type(fn) == types.BuiltinFunctionType or type(fn) == types.LambdaType \

or type(fn) == types.GeneratorType

print(is_func(abs), is_func(is_func), is_func(lambda x: x**2))

#2 Created at ... [Delete] [Delete and Lock User]

>>> callable(abs)
True