如何从Python中的代码对象生成模块对象 [英] How to generate a module object from a code object in Python

问题描述

鉴于我有一个模块的代码对象，如何获得相应的模块对象？

Given that I have the code object for a module, how do I get the corresponding module object?

它看起来像 moduleNames = { }; moduleNames 中的exec代码执行的操作与我想要的非常接近。它将模块中声明的全局变量返回到字典中。但是，如果我想要实际的模块对象，该如何获得它？

It looks like moduleNames = {}; exec code in moduleNames does something very close to what I want. It returns the globals declared in the module into a dictionary. But if I want the actual module object, how do I get it?

编辑：
看起来您可以滚动自己的模块对象。模块类型不方便记录，但是您可以执行以下操作：

It looks like you can roll your own module object. The module type isn't conveniently documented, but you can do something like this:

import sys
module = sys.__class__
del sys
foo = module('foo', 'Doc string')
foo.__file__ = 'foo.pyc'
exec code in foo.__dict__

推荐答案

正如评论已经表明的那样，在当今的Python中，实例化没有内置名称的类型是调用通过类型模块：

As a comment already indicates, in today's Python the preferred way to instantiate types that don't have built-in names is to call the type obtained via the types module from the standard library:

>>> import types
>>> m = types.ModuleType('m', 'The m module')

请注意，这确实不是会自动将新模块插入 sys.modules ：

note that this does not automatically insert the new module in sys.modules:

>>> import sys
>>> sys.modules['m']
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
KeyError: 'm'

这是您必须手动执行的任务：

That's a task you must perform by hand:

>>> sys.modules['m'] = m
>>> sys.modules['m']
<module 'm' (built-in)>

这很重要，因为模块的代码对象通常在之后执行模块已添加到 sys.modules 中-例如，此类代码引用 sys.modules [__ name __] ，如果您忘记了此步骤，将失败（ KeyError ）。 此步骤之后，将 m .__ file __ 设置为您已经在编辑中得到的值，

This can be important, since a module's code object normally executes after the module's added to sys.modules -- for example, it's perfectly correct for such code to refer to sys.modules[__name__], and that would fail (KeyError) if you forgot this step. After this step, and setting m.__file__ as you already have in your edit,

>>> code = compile("a=23", "m.py", "exec")
>>> exec code in m.__dict__
>>> m.a
23

（或Python 3等效项，其中 exec 是一个函数，当然，如果您使用的是Python 3，-）是正确的（当然，通常您会通过精巧的方式而不是编译字符串来获取代码对象，但这就是

(or the Python 3 equivalent where exec is a function, if Python 3 is what you're using, of course;-) is correct (of course, you'll normally have obtained the code object by subtler means than compiling a string, but that's not material to your question;-).

在旧版本的Python中，您应该使用 new 模块来代替的 types 模块在开始时就创建了一个新的模块对象，但是 new 从Python 2.6开始就已弃用，并在Python 3。

In older versions of Python you would have used the new module instead of the types module to make a new module object at the start, but new is deprecated since Python 2.6 and removed in Python 3.

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