问题:检查类是否已定义函数的最快方法是什么?

我正在编写AI状态空间搜索算法,并且有一个通用类可以用于快速实现搜索算法。子类将定义必要的操作,然后算法执行其余操作。

这是我遇到的问题:我想避免一遍又一遍地重新生成父状态,所以我有以下函数,该函数返回可以合法地应用于任何状态的操作:

def get_operations(self, include_parent=True):
    ops = self._get_operations()
    if not include_parent and self.path.parent_op:
        try:
            parent_inverse = self.invert_op(self.path.parent_op)
            ops.remove(parent_inverse)
        except NotImplementedError:
            pass
    return ops

并且invert_op函数默认情况下抛出。

有没有比捕获异常更快的方法来检查函数是否未定义?

我在检查dir中是否存在内容时正在思考,但这似乎不正确。hasattr是通过调用getattr并检查它是否引发来实现的,这不是我想要的。

I’m writing an AI state space search algorithm, and I have a generic class which can be used to quickly implement a search algorithm. A subclass would define the necessary operations, and the algorithm does the rest.

Here is where I get stuck: I want to avoid regenerating the parent state over and over again, so I have the following function, which returns the operations that can be legally applied to any state:

def get_operations(self, include_parent=True):
    ops = self._get_operations()
    if not include_parent and self.path.parent_op:
        try:
            parent_inverse = self.invert_op(self.path.parent_op)
            ops.remove(parent_inverse)
        except NotImplementedError:
            pass
    return ops

And the invert_op function throws by default.

Is there a faster way to check to see if the function is not defined than catching an exception?

I was thinking something on the lines of checking for present in dir, but that doesn’t seem right. hasattr is implemented by calling getattr and checking if it raises, which is not what I want.


回答 0

是的,用于getattr()获取属性并callable()验证它是否为方法:

invert_op = getattr(self, "invert_op", None)
if callable(invert_op):
    invert_op(self.path.parent_op)

请注意,getattr()当属性不存在时,通常会引发异常。但是,如果您指定默认值(None在本例中为),它将返回该值。

Yes, use getattr() to get the attribute, and callable() to verify it is a method:

invert_op = getattr(self, "invert_op", None)
if callable(invert_op):
    invert_op(self.path.parent_op)

Note that getattr() normally throws exception when the attribute doesn’t exist. However, if you specify a default value (None, in this case), it will return that instead.


回答 1

它同时适用于Python 2和Python 3

hasattr(connection, 'invert_opt')

hasattrTrue如果连接对象已invert_opt定义函数,则返回。这是供您放牧的文档

https://docs.python.org/2/library/functions.html#hasattr https://docs.python.org/3/library/functions.html#hasattr

It works in both Python 2 and Python 3

hasattr(connection, 'invert_opt')

hasattr returns True if connection object has a function invert_opt defined. Here is the documentation for you to graze

https://docs.python.org/2/library/functions.html#hasattr https://docs.python.org/3/library/functions.html#hasattr


回答 2

有没有比捕获异常更快的方法来检查函数是否未定义?

你为什么反对那个?在大多数Pythonic情况下,最好是请求宽恕而不是允许。;-)

hasattr是通过调用getattr并检查它是否引发来实现的,这不是我想要的。

同样,为什么呢?以下是相当Pythonic的内容:

    try:
        invert_op = self.invert_op
    except AttributeError:
        pass
    else:
        parent_inverse = invert_op(self.path.parent_op)
        ops.remove(parent_inverse)

要么,

    # if you supply the optional `default` parameter, no exception is thrown
    invert_op = getattr(self, 'invert_op', None)  
    if invert_op is not None:
        parent_inverse = invert_op(self.path.parent_op)
        ops.remove(parent_inverse)

但是请注意,这getattr(obj, attr, default)基本上也是通过捕获异常来实现的。Python领域没有错!

Is there a faster way to check to see if the function is not defined than catching an exception?

Why are you against that? In most Pythonic cases, it’s better to ask forgiveness than permission. ;-)

hasattr is implemented by calling getattr and checking if it raises, which is not what I want.

Again, why is that? The following is quite Pythonic:

    try:
        invert_op = self.invert_op
    except AttributeError:
        pass
    else:
        parent_inverse = invert_op(self.path.parent_op)
        ops.remove(parent_inverse)

Or,

    # if you supply the optional `default` parameter, no exception is thrown
    invert_op = getattr(self, 'invert_op', None)  
    if invert_op is not None:
        parent_inverse = invert_op(self.path.parent_op)
        ops.remove(parent_inverse)

Note, however, that getattr(obj, attr, default) is basically implemented by catching an exception, too. There is nothing wrong with that in Python land!


回答 3

这里的响应检查字符串是否是对象的属性的名称。需要一个额外的步骤(使用callable)来检查属性是否为方法。

因此,可以归结为:检查对象obj是否具有属性attrib的最快方法是什么。答案是

'attrib' in obj.__dict__

之所以如此,是因为dict对其键进行了哈希处理,因此可以快速检查键的存在。

请参见下面的时序比较。

>>> class SomeClass():
...         pass
...
>>> obj = SomeClass()
>>>
>>> getattr(obj, "invert_op", None)
>>>
>>> %timeit getattr(obj, "invert_op", None)
1000000 loops, best of 3: 723 ns per loop
>>> %timeit hasattr(obj, "invert_op")
The slowest run took 4.60 times longer than the fastest. This could mean that an intermediate result is being cached.
1000000 loops, best of 3: 674 ns per loop
>>> %timeit "invert_op" in obj.__dict__
The slowest run took 12.19 times longer than the fastest. This could mean that an intermediate result is being cached.
10000000 loops, best of 3: 176 ns per loop

The responses herein check if a string is the name of an attribute of the object. An extra step (using callable) is needed to check if the attribute is a method.

So it boils down to: what is the fastest way to check if an object obj has an attribute attrib. The answer is

'attrib' in obj.__dict__

This is so because a dict hashes its keys so checking for the key’s existence is fast.

See timing comparisons below.

>>> class SomeClass():
...         pass
...
>>> obj = SomeClass()
>>>
>>> getattr(obj, "invert_op", None)
>>>
>>> %timeit getattr(obj, "invert_op", None)
1000000 loops, best of 3: 723 ns per loop
>>> %timeit hasattr(obj, "invert_op")
The slowest run took 4.60 times longer than the fastest. This could mean that an intermediate result is being cached.
1000000 loops, best of 3: 674 ns per loop
>>> %timeit "invert_op" in obj.__dict__
The slowest run took 12.19 times longer than the fastest. This could mean that an intermediate result is being cached.
10000000 loops, best of 3: 176 ns per loop

回答 4

我喜欢内森·奥斯特加德的回答,并对此进行了投票。但是解决问题的另一种方法是使用记忆修饰符,该修饰符将缓存函数调用的结果。因此,您可以继续使用具有昂贵功能的功能来解决某些问题,但是当您一遍又一遍地调用它时,后续调用很快。函数的记忆版本会在字典中查找参数,然后从实际函数计算结果时开始在字典中查找结果,然后立即返回结果。

这是雷蒙德·海廷格(Raymond Hettinger)称为“ lru_cache”的记忆修饰器的食谱。现在,此版本是Python 3.2中functools模块的标准版本。

http://code.activestate.com/recipes/498245-lru-and-lfu-cache-decorators/

http://docs.python.org/release/3.2/library/functools.html

I like Nathan Ostgard’s answer and I up-voted it. But another way you could solve your problem would be to use a memoizing decorator, which would cache the result of the function call. So you can go ahead and have an expensive function that figures something out, but then when you call it over and over the subsequent calls are fast; the memoized version of the function looks up the arguments in a dict, finds the result in the dict from when the actual function computed the result, and returns the result right away.

Here is a recipe for a memoizing decorator called “lru_cache” by Raymond Hettinger. A version of this is now standard in the functools module in Python 3.2.

http://code.activestate.com/recipes/498245-lru-and-lfu-cache-decorators/

http://docs.python.org/release/3.2/library/functools.html


回答 5

像Python中的任何东西一样,如果您尽力而为,那么您就可以直截了当地去做一些令人讨厌的事情。现在,这是令人讨厌的部分:

def invert_op(self, op):
    raise NotImplementedError

def is_invert_op_implemented(self):
    # Only works in CPython 2.x of course
    return self.invert_op.__code__.co_code == 't\x00\x00\x82\x01\x00d\x00\x00S'

请帮我们一个忙,只要继续解决您的问题,就不要使用它,除非您是PyPy团队的黑客,他们正在侵入Python解释器。您所拥有的是Pythonic,我在这里拥有的是纯EVIL

Like anything in Python, if you try hard enough, you can get at the guts and do something really nasty. Now, here’s the nasty part:

def invert_op(self, op):
    raise NotImplementedError

def is_invert_op_implemented(self):
    # Only works in CPython 2.x of course
    return self.invert_op.__code__.co_code == 't\x00\x00\x82\x01\x00d\x00\x00S'

Please do us a favor, just keep doing what you have in your question and DON’T ever use this unless you are on the PyPy team hacking into the Python interpreter. What you have up there is Pythonic, what I have here is pure EVIL.


回答 6

您也可以遍历类:

import inspect


def get_methods(cls_):
    methods = inspect.getmembers(cls_, inspect.isfunction)
    return dict(methods)

# Example
class A(object):
    pass

class B(object):
    def foo():
        print('B')


# If you only have an object, you can use `cls_ = obj.__class__`
if 'foo' in get_methods(A):
    print('A has foo')

if 'foo' in get_methods(B):
    print('B has foo')

You can also go over the class:

import inspect


def get_methods(cls_):
    methods = inspect.getmembers(cls_, inspect.isfunction)
    return dict(methods)

# Example
class A(object):
    pass

class B(object):
    def foo():
        print('B')


# If you only have an object, you can use `cls_ = obj.__class__`
if 'foo' in get_methods(A):
    print('A has foo')

if 'foo' in get_methods(B):
    print('B has foo')

回答 7

虽然在__dict__属性中检查属性确实非常快,但是您不能将其用于方法,因为它们不会出现在__dict__哈希中。但是,如果性能至关重要,则可以在课堂上采用棘手的解决方法:

class Test():
    def __init__():
        # redefine your method as attribute
        self.custom_method = self.custom_method

    def custom_method(self):
        pass

然后检查方法为:

t = Test()
'custom_method' in t.__dict__

时间比较getattr

>>%timeit 'custom_method' in t.__dict__
55.9 ns ± 0.626 ns per loop (mean ± std. dev. of 7 runs, 10000000 loops each)

>>%timeit getattr(t, 'custom_method', None)
116 ns ± 0.765 ns per loop (mean ± std. dev. of 7 runs, 10000000 loops each)

我并不是在鼓励这种方法,但是它似乎有效。

[EDIT]当方法名称不在给定的类中时,性能提升甚至更高:

>>%timeit 'rubbish' in t.__dict__
65.5 ns ± 11 ns per loop (mean ± std. dev. of 7 runs, 10000000 loops each)

>>%timeit getattr(t, 'rubbish', None)
385 ns ± 12.9 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

While checking for attributes in __dict__ property is really fast, you cannot use this for methods, since they do not appear in __dict__ hash. You could however resort to hackish workaround in your class, if performance is that critical:

class Test():
    def __init__():
        # redefine your method as attribute
        self.custom_method = self.custom_method

    def custom_method(self):
        pass

Then check for method as:

t = Test()
'custom_method' in t.__dict__

Time comparision with getattr:

>>%timeit 'custom_method' in t.__dict__
55.9 ns ± 0.626 ns per loop (mean ± std. dev. of 7 runs, 10000000 loops each)

>>%timeit getattr(t, 'custom_method', None)
116 ns ± 0.765 ns per loop (mean ± std. dev. of 7 runs, 10000000 loops each)

Not that I’m encouraging this approach, but it seems to work.

[EDIT] Performance boost is even higher when method name is not in given class:

>>%timeit 'rubbish' in t.__dict__
65.5 ns ± 11 ns per loop (mean ± std. dev. of 7 runs, 10000000 loops each)

>>%timeit getattr(t, 'rubbish', None)
385 ns ± 12.9 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

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