知识问答

在Python中获取调用函数模块的__name__

内容 隐藏
问题：在Python中获取调用函数模块的__name__
回答 0
回答 1
回答 2

问题：在Python中获取调用函数模块的__name__

假设myapp/foo.py包含：

def info(msg):
    caller_name = ????
    print '[%s] %s' % (caller_name, msg)

并myapp/bar.py包含：

import foo
foo.info('Hello') # => [myapp.bar] Hello

在这种情况下，我想caller_name设置为“ __name__调用函数”模块的属性（即“ myapp.foo”）。如何才能做到这一点？

点击查看英文原文

Suppose myapp/foo.py contains:

def info(msg):
    caller_name = ????
    print '[%s] %s' % (caller_name, msg)

And myapp/bar.py contains:

import foo
foo.info('Hello') # => [myapp.bar] Hello

I want caller_name to be set to the __name__ attribute of the calling functions’ module (which is ‘myapp.foo’) in this case. How can this be done?

回答 0

检出检查模块：

inspect.stack() 将返回堆栈信息。

在函数内部，inspect.stack()[1]将返回调用者的堆栈。从那里，您可以获得有关调用者的函数名称，模块等的更多信息。

有关详细信息，请参阅文档：

http://docs.python.org/library/inspect.html

此外，Doug Hellmann在他的PyMOTW系列中对检查模块做了很好的撰写：

http://pymotw.com/2/inspect/index.html#module-inspect

编辑：这是一些您想要的代码，我认为：

def info(msg):
    frm = inspect.stack()[1]
    mod = inspect.getmodule(frm[0])
    print '[%s] %s' % (mod.__name__, msg)
点击查看英文原文

Check out the inspect module:

inspect.stack() will return the stack information.

Inside a function, inspect.stack()[1] will return your caller’s stack. From there, you can get more information about the caller’s function name, module, etc.

See the docs for details:

http://docs.python.org/library/inspect.html

Also, Doug Hellmann has a nice writeup of the inspect module in his PyMOTW series:

http://pymotw.com/2/inspect/index.html#module-inspect

EDIT: Here’s some code which does what you want, I think:

import inspect 

def info(msg):
    frm = inspect.stack()[1]
    mod = inspect.getmodule(frm[0])
    print '[%s] %s' % (mod.__name__, msg)

回答 1

面对类似的问题，我发现sys模块中的sys._current_frames（）包含有趣的信息，这些信息至少在特定的用例中可以帮助您，而无需导入检查。

>>> sys._current_frames()
{4052: <frame object at 0x03200C98>}

然后，您可以使用f_back“上移”：

>>> f = sys._current_frames().values()[0]
>>> # for python3: f = list(sys._current_frames().values())[0]

>>> print f.f_back.f_globals['__file__']
'/base/data/home/apps/apricot/1.6456165165151/caller.py'

>>> print f.f_back.f_globals['__name__']
'__main__'

对于文件名，您还可以使用f.f_back.f_code.co_filename，如上文Mark Roddy所建议。我不确定此方法的局限性和注意事项（很可能会出现多个线程），但是我打算在自己的情况下使用它。

点击查看英文原文

Confronted with a similar problem, I have found that sys._current_frames() from the sys module contains interesting information that can help you, without the need to import inspect, at least in specific use cases.

>>> sys._current_frames()
{4052: <frame object at 0x03200C98>}

You can then “move up” using f_back :

>>> f = sys._current_frames().values()[0]
>>> # for python3: f = list(sys._current_frames().values())[0]

>>> print f.f_back.f_globals['__file__']
'/base/data/home/apps/apricot/1.6456165165151/caller.py'

>>> print f.f_back.f_globals['__name__']
'__main__'

For the filename you can also use f.f_back.f_code.co_filename, as suggested by Mark Roddy above. I am not sure of the limits and caveats of this method (multiple threads will most likely be a problem) but I intend to use it in my case.

回答 2

我不建议您这样做，但是您可以使用以下方法来实现您的目标：

def caller_name():
    frame=inspect.currentframe()
    frame=frame.f_back.f_back
    code=frame.f_code
    return code.co_filename

然后，如下更新您的现有方法：

def info(msg):
    caller = caller_name()
    print '[%s] %s' % (caller, msg)
点击查看英文原文

I don’t recommend do this, but you can accomplish your goal with the following method:

def caller_name():
    frame=inspect.currentframe()
    frame=frame.f_back.f_back
    code=frame.f_code
    return code.co_filename

Then update your existing method as follows:

def info(msg):
    caller = caller_name()
    print '[%s] %s' % (caller, msg)