Python class: 为什么我不能在 eq(self, other) 中使用 len() 方法？

Question

https://gist.github.com/rjurney/1e8454af8e44312d02d7

class FrozenSortedTuple:
  """A frozenset that cares about order of tuples. And is not actually frozen."""
  def __init__(self, vals):
    if type(vals) not in [list, set, tuple]:
      raise Exception('Value not a list or set')
    self.vals = list(vals)

  def __eq__(self, other):
    if len(self.vals) != len(other):
      return False
    if type(self.vals) != type(other):
      return False
    if type(other) not in [list, set, tuple]:
      return False
    other_list = list(other)
    for i,item in enumerate(self.vals):
      if item != other_list[i]:
        return False
    return True

在 iPython 调用：

In [2]: a = ['a','b']

In [3]: b = ['b','a']

In [4]: c = ['a','b']

In [5]: a == b
Out[5]: False

In [6]: FrozenSortedTuple(a)
Out[6]: <__main__.FrozenSortedTuple instance at 0x103c56200>

In [7]: fa = FrozenSortedTuple(a)

In [8]: fb = FrozenSortedTuple(b)

In [9]: fa == fb

错误：

---------------------------------------------------------------------------
AttributeError                            Traceback (most recent call last)
<ipython-input-9-317181571e4d> in <module>()
----> 1 fa == fb

<ipython-input-1-ef99f0af5061> in __eq__(self, other)
     15 
     16   def __eq__(self, other):
---> 17     if len(self.vals) != len(other):
     18       return False
     19     if type(self.vals) != type(other):

AttributeError: FrozenSortedTuple instance has no attribute '__len__'

我很困惑。

Answer 1

如果您尝试直接比较两个 FrozenSortedTuples 值的结构和内容，请将 other 的所有实例更改为 other.vals。

def __eq__(self, other):
    if len(self.vals) != len(other.vals):
      return False
    if type(self.vals) != type(other.vals):
      return False
    if type(other.vals) not in [list, set, tuple]:
      return False
    other_list = list(other.vals)
    for i,item in enumerate(self.vals):
      if item != other_list[i]:
        return False
    return True

当然，如果 other 不是 FrozenSortedTuple，这将无法工作。例如，fa == 23 将不起作用，因为数字 23 没有 "vals" 属性。

Answer 2

您定义 __eq__(self, other) 的方式，只有当 other 是您正在包装的类型（即列表、集合或元组）的实例时才能实现平等。您通过比较 FrozenSortedTuple 的两个实例来触发错误。错误消息告诉您无法在此类实例上调用 len()，那是因为您尚未在 class.

中定义方法 __len__(self)

如果您为 class 定义 __len__()，它将起作用。请参阅 the Python Documentation（这链接到 2.7 文档，但它在 Python 3.x 中应该是一样的）或者，您可以比较 len(self.vals) == len(other.vals) 来比较 FrozenSortedTuple 个实例。

Answer 3

谢谢，根据其他答案，我想要的是：

class FrozenSortedTuple:
  """A frozenset that cares about order of tuples. And is not actually frozen."""
  def __init__(self, vals):
    if type(vals) not in [list, set, tuple]:
      raise Exception('Value not a list or set')
    self.vals = vals

  def __len__(self):
    return len(self.vals)

  def __iter__(self):
    return iter(self.vals)

  def __getitem__(self, key):
    return list(self.vals)[key]

  def __eq__(self, other):
    if len(self) != len(other):
      print "len(self)"
      return False

    for i,item in enumerate(self.vals):
      if item != other[i]:
        return False
    return True

  def __str__(self):
    str_val = str()
    for val in self:
      str_val += str(val)
    return str_val

  def __hash__(self):
    return hash(str(self))

测试：

# FrozenSortedTuple Tests
a = ['a','b']
b = ['b','a']
c = ['a','b']

fa = FrozenSortedTuple(a)
fb = FrozenSortedTuple(b)
fc = FrozenSortedTuple(c)

fa == fb
fa == fc
fa == ['a','b']
fa == ['b','a']
fa == ('a','b')
fa == ('b','a')

a = set([fa, fb, fc])
b = set([fa, fb, fc])
c = set([fa, fc, fb])
a == b
b == c
fa in a
fb in b

d = set([fb])
fa in d

Python class: 为什么我不能在 eq(self, other) 中使用 len() 方法？

Python class: Why can't I use the method len() inside eq(self, other)?

python

class

python-2.7

oop

variable-length

Python class: 为什么我不能在 __eq__(self, other) 中使用 len() 方法？

Python class: Why can't I use the method len() inside __eq__(self, other)?

python

class

python-2.7

oop

variable-length

Python class: 为什么我不能在 eq(self, other) 中使用 len() 方法？

Python class: Why can't I use the method len() inside eq(self, other)?