Python bbfreeze从已编译库导入函数

在Raspbian和bbfreeze上使用Python2.7，我是否可以编译单个模块，然后将其导入另一个未被混淆的脚本中

大概是这样的：

bbfreeze mylibrary.py (that contain function())

然后：

---

如何做到这一点？

此解决方案适用于Windows。我希望它，或者说是一个小的变体，可以在linux上运行，但我没有访问linux设备来尝试它

使用以下目录结构创建要冻结的应用程序：

-- sample
    -- app
        - __init__.py
        - fns.py
    - main.py
    - setup.py

每个python文件的代码如下：

fns.py

"""Place functions you want to access from the frozen app here"""
def accessible():
    print "This function is accessible"

import app.fns

def inaccessible():
    print "This function is inaccessible; AFAIK"

if __name__ == '__main__':
    inaccessible()
    app.fns.accessible()

from bbfreeze import Freezer

f = Freezer(distdir="frozen")
f.addScript("main.py")
f()

import sys

# Add the path to the frozen applications app module to the python path
sys.path.append("C:\\no_backup\\personal\\sample\\frozen\\library.zip")

import app.fns

if __name__ == '__main__':
    print "Calling a routine from a frozen application"
    app.fns.accessible()

main.py

"""Place functions you want to access from the frozen app here"""
def accessible():
    print "This function is accessible"

import app.fns

def inaccessible():
    print "This function is inaccessible; AFAIK"

if __name__ == '__main__':
    inaccessible()
    app.fns.accessible()

from bbfreeze import Freezer

f = Freezer(distdir="frozen")
f.addScript("main.py")
f()

import sys

# Add the path to the frozen applications app module to the python path
sys.path.append("C:\\no_backup\\personal\\sample\\frozen\\library.zip")

import app.fns

if __name__ == '__main__':
    print "Calling a routine from a frozen application"
    app.fns.accessible()

bb_setup.py

"""Place functions you want to access from the frozen app here"""
def accessible():
    print "This function is accessible"

import app.fns

def inaccessible():
    print "This function is inaccessible; AFAIK"

if __name__ == '__main__':
    inaccessible()
    app.fns.accessible()

from bbfreeze import Freezer

f = Freezer(distdir="frozen")
f.addScript("main.py")
f()

import sys

# Add the path to the frozen applications app module to the python path
sys.path.append("C:\\no_backup\\personal\\sample\\frozen\\library.zip")

import app.fns

if __name__ == '__main__':
    print "Calling a routine from a frozen application"
    app.fns.accessible()

冻结应用程序将导致目录冻结。现在，您可以使用如下代码从“冻结”应用程序访问fns.py中的函数：

useFrozenCode.py

"""Place functions you want to access from the frozen app here"""
def accessible():
    print "This function is accessible"

import app.fns

def inaccessible():
    print "This function is inaccessible; AFAIK"

if __name__ == '__main__':
    inaccessible()
    app.fns.accessible()

from bbfreeze import Freezer

f = Freezer(distdir="frozen")
f.addScript("main.py")
f()

import sys

# Add the path to the frozen applications app module to the python path
sys.path.append("C:\\no_backup\\personal\\sample\\frozen\\library.zip")

import app.fns

if __name__ == '__main__':
    print "Calling a routine from a frozen application"
    app.fns.accessible()

我不知道如何在main.py中调用函数不可访问