181

我需要将例程标记为已弃用,但显然没有用于弃用的标准库装饰器。我知道它的配方和警告模块,但我的问题是:为什么这个(常见)任务没有标准库装饰器?

附加问题:标准库中是否有标准装饰器?

4

7 回答 7

80

这是一些片段,根据 Leandro 引用的片段进行了修改:

import warnings
import functools

def deprecated(func):
    """This is a decorator which can be used to mark functions
    as deprecated. It will result in a warning being emitted
    when the function is used."""
    @functools.wraps(func)
    def new_func(*args, **kwargs):
        warnings.simplefilter('always', DeprecationWarning)  # turn off filter
        warnings.warn("Call to deprecated function {}.".format(func.__name__),
                      category=DeprecationWarning,
                      stacklevel=2)
        warnings.simplefilter('default', DeprecationWarning)  # reset filter
        return func(*args, **kwargs)
    return new_func

# Examples

@deprecated
def some_old_function(x, y):
    return x + y

class SomeClass:
    @deprecated
    def some_old_method(self, x, y):
        return x + y

因为在某些解释器中,第一个暴露的解决方案(没有过滤器处理)可能会导致警告抑制。

于 2015-05-15T07:24:28.343 回答
63

这是另一个解决方案:

这个装饰器(实际上是一个装饰器工厂)允许你给出一个原因信息。通过提供源文件名行号来帮助开发人员诊断问题也更有用。

编辑:此代码使用 Zero 的建议:它将warnings.warn_explicitline by替换warnings.warn(msg, category=DeprecationWarning, stacklevel=2)为打印函数调用站点而不是函数定义站点。它使调试更容易。

EDIT2:此版本允许开发人员指定可选的“原因”消息。

import functools
import inspect
import warnings

string_types = (type(b''), type(u''))


def deprecated(reason):
    """
    This is a decorator which can be used to mark functions
    as deprecated. It will result in a warning being emitted
    when the function is used.
    """

    if isinstance(reason, string_types):

        # The @deprecated is used with a 'reason'.
        #
        # .. code-block:: python
        #
        #    @deprecated("please, use another function")
        #    def old_function(x, y):
        #      pass

        def decorator(func1):

            if inspect.isclass(func1):
                fmt1 = "Call to deprecated class {name} ({reason})."
            else:
                fmt1 = "Call to deprecated function {name} ({reason})."

            @functools.wraps(func1)
            def new_func1(*args, **kwargs):
                warnings.simplefilter('always', DeprecationWarning)
                warnings.warn(
                    fmt1.format(name=func1.__name__, reason=reason),
                    category=DeprecationWarning,
                    stacklevel=2
                )
                warnings.simplefilter('default', DeprecationWarning)
                return func1(*args, **kwargs)

            return new_func1

        return decorator

    elif inspect.isclass(reason) or inspect.isfunction(reason):

        # The @deprecated is used without any 'reason'.
        #
        # .. code-block:: python
        #
        #    @deprecated
        #    def old_function(x, y):
        #      pass

        func2 = reason

        if inspect.isclass(func2):
            fmt2 = "Call to deprecated class {name}."
        else:
            fmt2 = "Call to deprecated function {name}."

        @functools.wraps(func2)
        def new_func2(*args, **kwargs):
            warnings.simplefilter('always', DeprecationWarning)
            warnings.warn(
                fmt2.format(name=func2.__name__),
                category=DeprecationWarning,
                stacklevel=2
            )
            warnings.simplefilter('default', DeprecationWarning)
            return func2(*args, **kwargs)

        return new_func2

    else:
        raise TypeError(repr(type(reason)))

您可以将此装饰器用于函数方法

这是一个简单的例子:

@deprecated("use another function")
def some_old_function(x, y):
    return x + y


class SomeClass(object):
    @deprecated("use another method")
    def some_old_method(self, x, y):
        return x + y


@deprecated("use another class")
class SomeOldClass(object):
    pass


some_old_function(5, 3)
SomeClass().some_old_method(8, 9)
SomeOldClass()

你会得到:

deprecated_example.py:59: DeprecationWarning: Call to deprecated function or method some_old_function (use another function).
  some_old_function(5, 3)
deprecated_example.py:60: DeprecationWarning: Call to deprecated function or method some_old_method (use another method).
  SomeClass().some_old_method(8, 9)
deprecated_example.py:61: DeprecationWarning: Call to deprecated class SomeOldClass (use another class).
  SomeOldClass()

EDIT3:这个装饰器现在是弃用库的一部分:

新的稳定版本 v1.2.13

于 2016-10-28T08:55:49.650 回答
20

正如 muon 建议的那样,您可以deprecation为此安装软件包。

deprecation库为您的测试提供了一个deprecated装饰器和一个装饰器。fail_if_not_removed

安装

pip install deprecation

示例用法

import deprecation

@deprecation.deprecated(deprecated_in="1.0", removed_in="2.0",
                        current_version=__version__,
                        details="Use the bar function instead")
def foo():
    """Do some stuff"""
    return 1

有关完整文档,请参阅http://deprecation.readthedocs.io/

于 2018-05-16T17:49:37.050 回答
18

我猜原因是 Python 代码不能被静态处理(就像它对 C++ 编译器所做的那样),在实际使用它之前你不会收到关于使用某些东西的警告。我认为向您的脚本用户发送大量消息“警告:此脚本的此开发人员正在使用已弃用的 API”向用户发送垃圾邮件并不是一个好主意。

更新:但是您可以创建将原始功能转换为另一个功能的装饰器。新功能将标记/检查开关,告知该功能已被调用,并且仅在将开关变为开启状态时才会显示消息。和/或在退出时,它可能会打印程序中使用的所有已弃用函数的列表。

于 2010-03-29T07:36:28.793 回答
12

您可以创建一个 utils 文件

import warnings

def deprecated(message):
  def deprecated_decorator(func):
      def deprecated_func(*args, **kwargs):
          warnings.warn("{} is a deprecated function. {}".format(func.__name__, message),
                        category=DeprecationWarning,
                        stacklevel=2)
          warnings.simplefilter('default', DeprecationWarning)
          return func(*args, **kwargs)
      return deprecated_func
  return deprecated_decorator

然后导入 deprecation 装饰器,如下所示:

from .utils import deprecated

@deprecated("Use method yyy instead")
def some_method()"
 pass
于 2018-02-05T21:50:52.457 回答
0

更新:我认为更好的是,当我们只为每个代码行第一次显示 DeprecationWarning并且我们可以发送一些消息时:

import inspect
import traceback
import warnings
import functools

import time


def deprecated(message: str = ''):
    """
    This is a decorator which can be used to mark functions
    as deprecated. It will result in a warning being emitted
    when the function is used first time and filter is set for show DeprecationWarning.
    """
    def decorator_wrapper(func):
        @functools.wraps(func)
        def function_wrapper(*args, **kwargs):
            current_call_source = '|'.join(traceback.format_stack(inspect.currentframe()))
            if current_call_source not in function_wrapper.last_call_source:
                warnings.warn("Function {} is now deprecated! {}".format(func.__name__, message),
                              category=DeprecationWarning, stacklevel=2)
                function_wrapper.last_call_source.add(current_call_source)

            return func(*args, **kwargs)

        function_wrapper.last_call_source = set()

        return function_wrapper
    return decorator_wrapper


@deprecated('You must use my_func2!')
def my_func():
    time.sleep(.1)
    print('aaa')
    time.sleep(.1)


def my_func2():
    print('bbb')


warnings.simplefilter('always', DeprecationWarning)  # turn off filter
print('before cycle')
for i in range(5):
    my_func()
print('after cycle')
my_func()
my_func()
my_func()

结果:

before cycle
C:/Users/adr-0/OneDrive/Projects/Python/test/unit1.py:45: DeprecationWarning: Function my_func is now deprecated! You must use my_func2!
aaa
aaa
aaa
aaa
aaa
after cycle
C:/Users/adr-0/OneDrive/Projects/Python/test/unit1.py:47: DeprecationWarning: Function my_func is now deprecated! You must use my_func2!
aaa
C:/Users/adr-0/OneDrive/Projects/Python/test/unit1.py:48: DeprecationWarning: Function my_func is now deprecated! You must use my_func2!
aaa
C:/Users/adr-0/OneDrive/Projects/Python/test/unit1.py:49: DeprecationWarning: Function my_func is now deprecated! You must use my_func2!
aaa

Process finished with exit code 0

我们只需点击警告路径并转到 PyCharm 中的行。

于 2016-11-30T22:34:08.753 回答
0

Python是一种动态类型语言。不必将类型静态声明为函数的变量或参数类型。

由于它是动态的,如果在运行时处理每件事。即使一个方法已被弃用,它也会在运行时或仅在解释期间知道。

使用弃用模块来弃用方法。

deprecation 是一个支持自动弃用的库。它提供deprecated()装饰器来包装函数,在文档中和通过 Python 的警告系统提供适当的警告,以及用于测试方法的deprecation.fail_if_not_removed()装饰器以确保最终删除不推荐使用的代码。

安装:

python3.10 -m pip install deprecation

小示范:

import deprecation

@deprecation.deprecated(details="Use bar instead")
def foo():
    print("Foo")


def bar():
    print("Bar")


foo()

bar()

输出:

test.py: DeprecatedWarning: foo is deprecated. Use bar instead
  foo()

Foo

Bar
于 2022-02-21T04:20:08.313 回答