我有一个变量x,我想知道它是否指向一个函数。

我希望我能做一些像这样的事情:

>>> isinstance(x, function)

但这给了我:

Traceback (most recent call last):
  File "<stdin>", line 1, in ?
NameError: name 'function' is not defined

我选这个是因为

>>> type(x)
<type 'function'>

当前回答

结果

callable(x) hasattr(x, '__call__') inspect.isfunction(x) inspect.ismethod(x) inspect.isgeneratorfunction(x) inspect.iscoroutinefunction(x) inspect.isasyncgenfunction(x) isinstance(x, typing.Callable) isinstance(x, types.BuiltinFunctionType) isinstance(x, types.BuiltinMethodType) isinstance(x, types.FunctionType) isinstance(x, types.MethodType) isinstance(x, types.LambdaType) isinstance(x, functools.partial)
print × × × × × × × × ×
func × × × × × × × ×
functools.partial × × × × × × × × × ×
<lambda> × × × × × × × ×
generator × × × × × × ×
async_func × × × × × × ×
async_generator × × × × × × ×
A × × × × × × × × × × ×
meth × × × × × × × ×
classmeth × × × × × × × × ×
staticmeth × × × × × × × ×
import types
import inspect
import functools
import typing


def judge(x):
    name = x.__name__ if hasattr(x, '__name__') else 'functools.partial'
    print(name)
    print('\ttype({})={}'.format(name, type(x)))
    print('\tcallable({})={}'.format(name, callable(x)))
    print('\thasattr({}, \'__call__\')={}'.format(name, hasattr(x, '__call__')))
    print()
    print('\tinspect.isfunction({})={}'.format(name, inspect.isfunction(x)))
    print('\tinspect.ismethod({})={}'.format(name, inspect.ismethod(x)))
    print('\tinspect.isgeneratorfunction({})={}'.format(name, inspect.isgeneratorfunction(x)))
    print('\tinspect.iscoroutinefunction({})={}'.format(name, inspect.iscoroutinefunction(x)))
    print('\tinspect.isasyncgenfunction({})={}'.format(name, inspect.isasyncgenfunction(x)))
    print()
    print('\tisinstance({}, typing.Callable)={}'.format(name, isinstance(x, typing.Callable)))
    print('\tisinstance({}, types.BuiltinFunctionType)={}'.format(name, isinstance(x, types.BuiltinFunctionType)))
    print('\tisinstance({}, types.BuiltinMethodType)={}'.format(name, isinstance(x, types.BuiltinMethodType)))
    print('\tisinstance({}, types.FunctionType)={}'.format(name, isinstance(x, types.FunctionType)))
    print('\tisinstance({}, types.MethodType)={}'.format(name, isinstance(x, types.MethodType)))
    print('\tisinstance({}, types.LambdaType)={}'.format(name, isinstance(x, types.LambdaType)))
    print('\tisinstance({}, functools.partial)={}'.format(name, isinstance(x, functools.partial)))


def func(a, b):
    pass


partial = functools.partial(func, a=1)

_lambda = lambda _: _


def generator():
    yield 1
    yield 2


async def async_func():
    pass


async def async_generator():
    yield 1


class A:
    def __call__(self, a, b):
        pass

    def meth(self, a, b):
        pass

    @classmethod
    def classmeth(cls, a, b):
        pass

    @staticmethod
    def staticmeth(a, b):
        pass


for func in [print,
             func,
             partial,
             _lambda,
             generator,
             async_func,
             async_generator,
             A,
             A.meth,
             A.classmeth,
             A.staticmeth]:
    judge(func)

Time

选择三种最常见的方法:

可调用的(x) hasattr (x, __call__) isinstance (x, typing.Callable)

time/s
callable(x) 0.86
hasattr(x, '__call__') 1.36
isinstance(x, typing.Callable) 12.19
import typing
from timeit import timeit


def x():
    pass


def f1():
    return callable(x)


def f2():
    return hasattr(x, '__call__')


def f3():
    return isinstance(x, typing.Callable)


print(timeit(f1, number=10000000))
print(timeit(f2, number=10000000))
print(timeit(f3, number=10000000))
# 0.8643081
# 1.3563508
# 12.193492500000001

其他回答

结合@Sumukh Barve, @Katsu和@tinnick的回答,如果你的动机只是想在控制台中获取可用的内置函数列表,这两个选项是可行的:

(我对j __builtin__.__dict__.items如果j.__class__()。__name__ in ['function', 'builtin_function_or_method']] (我对j __builtin__.__dict__.items()如果str (j)(18): = = ' <内置函数)

在内置命名空间中没有构造函数的内置类型(例如函数、生成器、方法)在types模块中。你可以使用类型。isinstance调用中的函数类型:

>>> import types
>>> types.FunctionType
<class 'function'>

>>> def f(): pass

>>> isinstance(f, types.FunctionType)
True
>>> isinstance(lambda x : None, types.FunctionType)
True

注意,这里使用了一个非常具体的“函数”概念,这通常不是您所需要的。例如,它拒绝zip(严格来说是一个类):

>>> type(zip), isinstance(zip, types.FunctionType)
(<class 'type'>, False)

Open(内置函数有不同类型):

>>> type(open), isinstance(open, types.FunctionType)
(<class 'builtin_function_or_method'>, False)

和随机的。Shuffle(技术上是一种隐藏随机的方法。随机实例):

>>> type(random.shuffle), isinstance(random.shuffle, types.FunctionType)
(<class 'method'>, False)

如果你在做一些特定类型的事情。FunctionType实例,如反编译字节码或检查闭包变量,使用类型。FunctionType,但如果你只是需要一个对象像函数一样可调用,请使用callable。

函数只是一个带有__call__方法的类,所以你可以这样做

hasattr(obj, '__call__')

例如:

>>> hasattr(x, '__call__')
True

>>> x = 2
>>> hasattr(x, '__call__')
False

这是“最好”的方法,但这取决于你为什么需要知道它是可调用的还是注释的,你可以把它放在try/execpt块中:

try:
    x()
except TypeError:
    print "was not callable"

如果try/except比doing if hasattr(x, '__call__'): x()更适合Python,这是有争议的。我会说hasattr更准确,因为你不会意外地捕捉到错误的TypeError,例如:

>>> def x():
...     raise TypeError
... 
>>> hasattr(x, '__call__')
True # Correct
>>> try:
...     x()
... except TypeError:
...     print "x was not callable"
... 
x was not callable # Wrong!

在一些答案中提到的使用hasattr(obj, '__call__')和callable(.)的解决方案有一个主要缺点:对于具有__call__()方法的类和类的实例,它们都返回True。如。

>>> import collections
>>> Test = collections.namedtuple('Test', [])
>>> callable(Test)
True
>>> hasattr(Test, '__call__')
True

检查一个对象是否是用户定义的函数(只有a that)的一个正确方法是使用isfunction(.):

>>> import inspect
>>> inspect.isfunction(Test)
False
>>> def t(): pass
>>> inspect.isfunction(t)
True

如果您需要检查其他类型,请查看inspect -检查活动对象。

任何函数都是一个类,所以你可以取实例x的类名并比较:


if(x.__class__.__name__ == 'function'):
     print "it's a function"