Question

通常，如果要获取对象的来源，可以通过inspect模块获取它：

import inspect
inspect.getsource(MyObject)

但是，在Jupyter笔记本中，这不起作用：

import inspect

class Foo:
    def __init__(self, info):
        self.info = info

a = Foo("hi")

inspect.getsource(a)

引发错误：

---------------------------------------------------------------------------
TypeError                                 Traceback (most recent call last)
<ipython-input-14-048b6f0c2e9b> in <module>()
      7 a = Foo("hi")
      8 
----> 9 inspect.getsource(a)

/usr/lib/python3.6/inspect.py in getsource(object)
    963     or code object.  The source code is returned as a single string.  An
    964     OSError is raised if the source code cannot be retrieved."""
--> 965     lines, lnum = getsourcelines(object)
    966     return ''.join(lines)
    967 

/usr/lib/python3.6/inspect.py in getsourcelines(object)
    950     raised if the source code cannot be retrieved."""
    951     object = unwrap(object)
--> 952     lines, lnum = findsource(object)
    953 
    954     if ismodule(object):

/usr/lib/python3.6/inspect.py in findsource(object)
    763     is raised if the source code cannot be retrieved."""
    764 
--> 765     file = getsourcefile(object)
    766     if file:
    767         # Invalidate cache if needed.

/usr/lib/python3.6/inspect.py in getsourcefile(object)
    679     Return None if no way can be identified to get the source.
    680     """
--> 681     filename = getfile(object)
    682     all_bytecode_suffixes = importlib.machinery.DEBUG_BYTECODE_SUFFIXES[:]
    683     all_bytecode_suffixes += importlib.machinery.OPTIMIZED_BYTECODE_SUFFIXES[:]

/usr/lib/python3.6/inspect.py in getfile(object)
    661         return object.co_filename
    662     raise TypeError('{!r} is not a module, class, method, '
--> 663                     'function, traceback, frame, or code object'.format(object))
    664 
    665 def getmodulename(path):

TypeError: <__main__.Foo object at 0x7fb9130ee518> is not a module, class, method, function, traceback, frame, or code object

如果我尝试使用Foo查找inspect.getsource(Foo)的来源，则会得到：

TypeError: <module '__main__'> is a built-in class

如何获取在Jupyter笔记本中定义的类的来源？

Answer 1

使用inspect.getsource(inspect.getfile)，我们可以获得处理此问题的一段代码：

...
if isclass(object):
    if hasattr(object, '__module__'):
        object = sys.modules.get(object.__module__)
        if hasattr(object, '__file__'):
            return object.__file__
    raise TypeError('{!r} is a built-in class'.format(object))
...

在ipython或Jupyter笔记本中，似乎定义的类/函数或__main__模块没有与之关联的__file__属性，因此inspect无法检索源文件。在这种情况下，您可以在单独的.py文件中定义类，以便inspect能够检索与其关联的文件。

Answer 2

我找到了一种在 Jupyter Notebook 中获取类的源代码的“hacky 方法”。

假设你有一个单元格：

class MyClass:
    test = 2
    
    def __init__(self):
        self.L = 5
    
    def test(self, x):
        return True
    
    @classmethod
    def forward(cls, x):
        return x

然后您可以使用以下方法提取代码：

import inspect
from IPython.core.magics.code import extract_symbols

obj = MyClass
cell_code = "".join(inspect.linecache.getlines(new_getfile(obj)))
class_code = extract_symbols(cell_code, obj.__name__)[0][0]
print(class_code)

从 here 定义的 new_getfile：

import inspect, sys

def new_getfile(object, _old_getfile=inspect.getfile):
    if not inspect.isclass(object):
        return _old_getfile(object)
    
    # Lookup by parent module (as in current inspect)
    if hasattr(object, '__module__'):
        object_ = sys.modules.get(object.__module__)
        if hasattr(object_, '__file__'):
            return object_.__file__
    
    # If parent module is __main__, lookup by methods (NEW)
    for name, member in inspect.getmembers(object):
        if inspect.isfunction(member) and object.__qualname__ + '.' + member.__name__ == member.__qualname__:
            return inspect.getfile(member)
    else:
        raise TypeError('Source for {!r} not found'.format(object))
inspect.getfile = new_getfile

获取在Jupyter Notebook中定义的对象的来源

2 个答案: