我需要将例程标记为已弃用,但显然没有标准的库装饰器可供弃用。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)任务没有标准的库装饰器?
其他问题:标准库中是否有标准装饰器?
答案 0 :(得分:44)
以下是Leandro引用的一些片段:
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。
答案 1 :(得分:28)
这是另一种解决方案:
此装饰器(事实上decorator factory)允许您提供原因消息。通过提供源文件名和行号来帮助开发人员诊断问题也更有用。
编辑:此代码使用了Zero的建议:它将warnings.warn_explicit
行替换为warnings.warn(msg, category=DeprecationWarning, stacklevel=2)
,
它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 :此版本允许开发者指定可选的“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功能,方法和类。
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:此装饰器现在是Deprecated库的一部分:
新稳定版v1.2.5
答案 2 :(得分:8)
我想原因是Python代码无法静态处理(就像它对C ++编译器所做的那样),在实际使用之前你不能得到关于使用某些东西的警告。我不认为用一堆消息向您的脚本用户发送垃圾邮件是一个好主意“警告:此脚本的开发人员正在使用已弃用的API”。
更新:但您可以创建装饰器,将原始功能转换为另一个。新功能将标记/检查开关,告知该功能已被调用,并且仅在转换为开启状态时显示消息。和/或在退出时,它可以打印程序中使用的所有已弃用函数的列表。
答案 3 :(得分:5)
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
然后按如下方式导入弃用装饰器:
from .utils import deprecated
@deprecated("Use method yyy instead")
def some_method()"
pass
答案 4 :(得分:4)
As muon suggested,您可以为此安装deprecation
包。
deprecation
库为您的测试提供deprecated
装饰器和fail_if_not_removed
装饰器。
pip install deprecation
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/。
答案 5 :(得分:0)
增强this answer by Steven Vascellaro:
如果您使用Anaconda,请先安装deprecation
软件包:
conda install -c conda-forge deprecation
然后将以下内容粘贴到文件顶部
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,请参见http://deprecation.readthedocs.io/。