如何导入 .pyc 编译的 python 文件并使用它

How can I import a .pyc compiled python file and use it(如何导入 .pyc 编译的 python 文件并使用它)

问题描述

我试图弄清楚如何在 python 脚本中包含 .pyc 文件.

Im trying to figure out how to include a .pyc file in a python script.

例如我的脚本被称为:

myscript.py

我想包含的脚本被称为:

and the script I would like to include is called:

included_script.pyc

那么，我只是使用:

import included_script

那会自动执行 included_script.pyc 吗?或者我还需要做些什么来让我的 included_script.pyc 在 myscript.py 中运行?

And will that automatically execute the included_script.pyc ? Or is there something further I need to do, to get my included_script.pyc to run inside the myscript.py?

我还需要传递 included_script.pyc 中使用的变量吗?如果可以，如何实现?

Do I need to pass the variables used in included_script.pyc also? If so, how might this be achieved?

推荐答案

很遗憾，不，这不能自动完成.当然，您也可以以一种非常丑陋的方式手动完成.

Unfortunately, no, this cannot be done automatically. You can, of course, do it manually in a gritty ugly way.

出于演示目的，我将首先生成一个 .pyc 文件.为此，我们首先需要一个 .py 文件.我们的示例 test.py 文件如下所示:

For demonstration purposes, I'll first generate a .pyc file. In order to do that, we first need a .py file for it. Our sample test.py file will look like:

def foo():
    print("In foo")

if __name__ == "__main__":
    print("Hello World")

超级简单.生成 .pyc 文件可以使用 py_compile 模块在标准库中找到.我们只需通过以下方式传入 .py 文件的名称和 .pyc 文件的名称:

Super simple. Generating the .pyc file can done with the py_compile module found in the standard library. We simply pass in the name of the .py file and the name for our .pyc file in the following way:

 py_compile.compile('test.py', 'mypyc.pyc')

这会将 mypyc.pyc 放在我们当前的工作目录中.

This will place mypyc.pyc in our current working directory.

现在，.pyc 文件包含以下结构的字节:

Now, .pyc files contain bytes that are structured in the following way:

• 前 4 个字节表示幻数"
• 接下来的 4 个字节包含修改时间戳
• 其余内容是编组的 code 对象.

我们需要的是编组的 code 对象，所以我们需要 import marshal 解组并执行它.此外，我们真的不关心/不需要前 8 个字节，并且不允许使用它们解组 .pyc 文件，因此我们将忽略它们(seek过去):

What we're after is that marshalled code object, so we need to import marshal to un-marshall it and execute it. Additionally, we really don't care/need the 8 first bytes, and un-marshalling the .pyc file with them is disallowed, so we'll ignore them (seek past them):

import marshal

s = open('mypyc.pyc', 'rb')
s.seek(8)  # go past first eight bytes
code_obj = marshal.load(s)

所以，现在我们有了用于 test.py 的花哨的 code 对象，它是有效的，可以按照我们的意愿执行.这里有两种选择:

So, now we have our fancy code object for test.py which is valid and ready to be executed as we wish. We have two options here:

1. 在当前的 global 命名空间中执行它.这将绑定当前命名空间中我们的 .pyc 文件中的所有定义，并将充当一种:from file import * 语句.

1. Execute it in the current global namespace. This will bind all definitions inside our .pyc file in the current namespace and will act as a sort of: from file import * statement.

创建一个新的模块对象并执行模块内的代码.这类似于 import file 语句.

Create a new module object and execute the code inside the module. This will be like the import file statement.

<小时>

模拟 from file import * 类似的行为:

执行此操作非常简单，只需执行以下操作:

---

Emulating from file import * like behaviour:

Performing this is pretty simple, just do:

exec(code_obj)

这将在当前命名空间中执行 code_obj 中包含的代码并绑定那里的所有内容.调用之后，我们可以像调用任何其他函数一样调用 foo:

This will execute the code contained inside code_obj in the current namespace and bind everything there. After the call we can call foo like any other funtion:

foo()
# prints: In foo!

注意:exec()是内置的.

这包括另一个要求，types 模块.这包含 ModuleType<的类型/a> 我们可以使用它来创建一个新的模块对象.它有两个参数，模块的名称(强制)和它的文档(可选):

This includes another requirement, the types module. This contains the type for ModuleType which we can use to create a new module object. It takes two arguments, the name for the module (mandatory) and the documentation for it (optional):

m = types.ModuleType("Fancy Name", "Fancy Documentation")

print(m)
<module 'Fancy Name' (built-in)>

现在我们有了模块对象，我们可以再次使用 exec 来执行模块命名空间内的 code_obj 中包含的代码(即 m.__dict__):

Now that we have our module object, we can again use exec to execute the code contained in code_obj inside the module namespace (namely, m.__dict__):

exec(code_obj, m.__dict__)

现在，我们的模块 m 已经在 code_obj 中定义了所有内容，您可以通过运行来验证这一点:

Now, our module m has everything defined in code_obj, you can verify this by running:

m.foo() 
# prints: In foo

<小时>

这些是您可以在模块中包含".pyc 文件的方法.至少，我能想到的方法.我真的看不出这有什么实用性，但是嘿，我不是来评判的.

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These are the ways you can 'include' a .pyc file in your module. At least, the ways I can think of. I don't really see the practicality in this but hey, I'm not here to judge.

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