如何导入所有子模块？ [英] How to import all submodules?

问题描述

我的目录结构如下：

| main.py
| scripts
|--| __init__.py
   | script1.py
   | script2.py
   | script3.py

从 main.py ，导入模块脚本。我尝试将 pkgutils.walk_packages 与 __ all __ 结合使用，但是使用它，我只能直接导入所有子模块从脚本导入* 使用 main 。我想把它们都放在脚本下。导入脚本的所有子模块的最简洁方法是什么，以便我可以从<$访问 scripts.script1 c $ c> main ？

From main.py, the module scripts is imported. I tried using pkgutils.walk_packages in combination with __all__, but using that, I can only import all the submodules directly under main using from scripts import *. I would like to get them all under scripts. What would be the cleanest way to import all the submodules of scripts so that I could access scripts.script1 from main?

编辑：对不起，我有点模糊。我想在运行时导入子模块，而不在 __ init __。py 中明确指定它们。我可以使用 pkgutils.walk_packages 来获取子模块名称（除非有人知道更好的方法），但我不确定使用这些名称的最简洁方法（或者可能） walk_packages 的I​​mpImporters返回？）导入它们。

I am sorry that I was a bit vague. I would like to import the submodules on run-time without specifying them explicitly in __init__.py. I can use pkgutils.walk_packages to get the submodule names (unless someone knows of a better way), but I am not sure of the cleanest way to use these names (or maybe the ImpImporters that walk_packages returns?) to import them.

推荐答案

编辑：这是在运行时以递归方式导入所有内容的一种方法...

Here's one way to recursively import everything at runtime...

它使用exec，因此几乎可以肯定有更好的方法，但它确实如此工作（即使对于任意嵌套的子包，我认为）。

It uses exec, so there's almost certainly a better way, but it does work (even for arbitrarily nested sub-packages, I think).

（顶部的 __ init __。py 的内容包目录）

(Contents of __init__.py in top package directory)

import pkgutil

__all__ = []
for loader, module_name, is_pkg in  pkgutil.walk_packages(__path__):
    __all__.append(module_name)
    module = loader.find_module(module_name).load_module(module_name)
    exec('%s = module' % module_name)

我没有使用 __ import __（__ path __ +如果您没有嵌套的子包，并且想要避免 exec / eval ，那么它就是一个这样做的方法。

I'm not using __import__(__path__+'.'+module_name) here, as it's difficult to properly recursively import packages using it. If you don't have nested sub-packages, and wanted to avoid the exec/eval, though, it's one way to do it.

可能有更好的方法，但无论如何这是我能做的最好的。

There's probably a better way, but this is the best I can do, anyway.

原始答案（对于上下文，请忽略。我最初误解了这个问题）：

Original Answer (For context, ignore othwerwise. I misunderstood the question initially):

你的脚本是什么/ __ init __。py 看起来像？它应该类似于：

What does your scripts/__init__.py look like? It should be something like:

import script1
import script2
import script3
__all__ = ['script1', 'script2', 'script3']

你甚至可以不定义 __ all __ ，但是如果你定义的东西（pydoc，如果没有别的话）会更干净地工作，即使它只是你导入的列表。

You could even do without defining __all__, but things (pydoc, if nothing else) will work more cleanly if you define it, even if it's just a list of what you imported.

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