Python 检查OrderedICT中是否存在密钥

我正在处理从python2到python3的代码迁移。我对python中的OOP和

OrderedDict

没有太多经验。这是我无法解决的问题。有人问过类似的问题，但对这个案子似乎没有任何效果。我有以下代码：

for row in rows:
    new_ = dummy(row.attr, row.error_type, row.array_agg_1,
                   row.with_error, row.total, row.level, row.order_)

    if new_ in result[n.id]: # <- never becomes true
         print('I am in the if')
         if new_.with_error is not None:
             result[n.id][new_][1] += new_.with_error
             result['total'][new_][1] += new_.with_error
    else:
         print('I am in the else')
         if new_.with_error is not None:
             result[n.id][new_] = [new_.attr, new_.with_error]

新建

如下所示：

#an example of result[n.id]
>>> ([(Eclipse Required Items'nce'[1540972], ['Eclipse Required Items', 1, 1, [1540972], 'nce', 1, 1681]), 
(Other Story Tab Info learned/discovered documentation and accuracy'nce'[1540973], ['Other Story Tab Info learned/discovered documentation and accuracy', 1, 1, [1540973], 'nce', 1, 1684]),
(Other (please provide detail in Comments section)'bce'[1541001], ['Other (please provide detail in Comments section', 1, 1, [1540973], 'bce', 1, 1684]
(Other Static bar information was documented'nce'[1540974], ['Other Static bar information was documented', 1, 1, [1541001], 'nce', 1, 1707])])

#type of result[n.id]
>>> <class 'collections.OrderedDict'>

#print(new_)
Other (please provide detail in Comments section)'bce'[1541001]

#print(type(new_))
<class 'smd.lib.asynch.calibration.dummy'>

class dummy(object):

    def __init__(self, attr, error_type, path, with_error=None, total=None,
                 level=None, order=None):
        self.attr = attr
        self.error_type = error_type
        self.path = path
        self.with_error = with_error
        self.total = total
        self.level = level
        self.order = order

    def __cmp__(self, other):
        return not (
            self.attr == other.attr and self.error_type == other.error_type
            and self.path == other.path
        )

    def __base_repr__(self):
        return "{0}{1}{2}".format(self.attr, self.error_type, self.path)

    def __hash__(self):
        return hash(self.__base_repr__())

    def __repr__(self):
        return self.__base_repr__()

但是，它在使用

KeyError

时失败，我不确定这是否是最好的方法。此外，它还可能破坏软件的其他功能

if new_ in result[n.id].keys():

不起作用。我们非常感谢您为解决此问题提供的任何帮助和指导

dummy

类如下所示：

#an example of result[n.id]
>>> ([(Eclipse Required Items'nce'[1540972], ['Eclipse Required Items', 1, 1, [1540972], 'nce', 1, 1681]), 
(Other Story Tab Info learned/discovered documentation and accuracy'nce'[1540973], ['Other Story Tab Info learned/discovered documentation and accuracy', 1, 1, [1540973], 'nce', 1, 1684]),
(Other (please provide detail in Comments section)'bce'[1541001], ['Other (please provide detail in Comments section', 1, 1, [1540973], 'bce', 1, 1684]
(Other Static bar information was documented'nce'[1540974], ['Other Static bar information was documented', 1, 1, [1541001], 'nce', 1, 1707])])

#type of result[n.id]
>>> <class 'collections.OrderedDict'>

#print(new_)
Other (please provide detail in Comments section)'bce'[1541001]

#print(type(new_))
<class 'smd.lib.asynch.calibration.dummy'>

class dummy(object):

    def __init__(self, attr, error_type, path, with_error=None, total=None,
                 level=None, order=None):
        self.attr = attr
        self.error_type = error_type
        self.path = path
        self.with_error = with_error
        self.total = total
        self.level = level
        self.order = order

    def __cmp__(self, other):
        return not (
            self.attr == other.attr and self.error_type == other.error_type
            and self.path == other.path
        )

    def __base_repr__(self):
        return "{0}{1}{2}".format(self.attr, self.error_type, self.path)

    def __hash__(self):
        return hash(self.__base_repr__())

    def __repr__(self):
        return self.__base_repr__()

---

在Python3中，cmp不再使用。

---

你需要在你的类中实现

\uuuuuuuuuuuueq\uuuuu

，然后删除

\uuuucmp\uuuuuu
删除
\uuuuuucmp\uuuuuuuuuuuu
并改为实现
\uuuuuuuuuuuueq\uuu
。@balderman感谢你的洞察力。我会调查一下的。你能推荐一个好的学习资源吗？是一个很好的资源吗？请看@balderman谢谢你的帮助。现在可以了。你介意写个答案吗？我很乐意接受并结束这个问题。很乐意帮忙。写了一份答复。