python - 在 ipython 控制台中处理用户定义的类时出现错误

Question

首先我应该说我正在使用Python 3.5.2。当我定义自己的类时，我认为 ipython 有问题。因此，当我在标准 python 控制台中运行以下代码时，我没有收到错误:

class A(dict):
    def __getattribute__(self, item):
        return self[item]
a = A({'x' : 1})
a['x']

但是当我在 ipython 控制台中运行上述代码时，出现以下错误:

KeyError                                  Traceback (most recent call last)
/usr/local/lib/python3.5/dist-packages/IPython/core/prefilter.py in prefilter_lines(self, lines, continue_prompt)
    333                              for lnum, line in enumerate(llines) ])
    334         else:
--> 335             out = self.prefilter_line(llines[0], continue_prompt)
    336 
    337         return out

/usr/local/lib/python3.5/dist-packages/IPython/core/prefilter.py in prefilter_line(self, line, continue_prompt)
    308             return normal_handler.handle(line_info)
    309 
--> 310         prefiltered = self.prefilter_line_info(line_info)
    311         # print "prefiltered line: %r" % prefiltered
    312         return prefiltered

/usr/local/lib/python3.5/dist-packages/IPython/core/prefilter.py in prefilter_line_info(self, line_info)
    250         """
    251         # print "prefilter_line_info: ", line_info
--> 252         handler = self.find_handler(line_info)
    253         return handler.handle(line_info)
    254 

/usr/local/lib/python3.5/dist-packages/IPython/core/prefilter.py in find_handler(self, line_info)
    257         for checker in self.checkers:
    258             if checker.enabled:
--> 259                 handler = checker.check(line_info)
    260                 if handler:
    261                     return handler

/usr/local/lib/python3.5/dist-packages/IPython/core/prefilter.py in check(self, line_info)
    414     def check(self, line_info):
    415         obj = self.shell.user_ns.get(line_info.ifun)
--> 416         if isinstance(obj, Macro):
    417             return self.prefilter_manager.get_handler_by_name('macro')
    418         else:

<ipython-input-1-cb6806333b6b> in __getattribute__(self, item)
      1 class A(dict):
      2     def __getattribute__(self, item):
----> 3         return self[item]
      4 

KeyError: '__class__'

如果您对此问题有任何想法，请提供帮助。

Answer 1

看起来 IPython 正在尝试访问 a.__class__，但失败了。它可能是为了诊断目的而将其作为某种标准的对象扫描来执行的。如果您尝试 print(a.__class__)，您会在常规 Python 控制台中收到相同的错误。

您的__getattribute__应该有一个后备，以防字典中没有与该键匹配的条目。 the documentation如果您需要访问实际属性，建议调用基类的 __getattribute__ 方法。像这样的东西:

class A(dict):
    def __getattribute__(self, item):
        if item in self:
            return self[item]
        else:
            return object.__getattribute__(self, item)
            #or possibly:
            #return super().__getattribute__(item)
            #... Depending on one's interpretation of "base class"

a = A({'x' : 1})
a['x']
print(a.__class__)

覆盖__getattr__也可能有意义。而不是__getattribute__。这样，仍然可以访问对象的所有属性，而无需在函数中对它们进行任何显式检查。

class A(dict):
    def __getattr__(self, item):
        if item in self:
            return self[item]
        else:
            raise AttributeError

a = A({'x' : 1})
a['x']
print(a.__class__)