python - x 不在集合中与 x != 集合中的每个元素？

Question

假设我们有 x = ['a', 'b']。

声明的幕后情况:

x not in {None, False}

这引发了 unhashable type: 'list' 错误？

我找到的解决方法是改写这个:

x != None and x!= False

但我很困惑，因为在数学上，两个 bool 表达式是等价的。

Answer 1

理由

官方文档是这样写的:

1. [Python 3]: class set([iterable]) :

   Return a new set or frozenset object whose elements are taken from iterable. The elements of a set must be hashable.

2. [Python 3]: hashable :

   An object is hashable if it has a hash value which never changes during its lifetime (it needs a __hash__() method), and can be compared to other objects (it needs an __eq__() method). Hashable objects which compare equal must have the same hash value.
   ...
   All of Python’s immutable built-in objects are hashable; mutable containers (such as lists or dictionaries) are not.

3. [Python 3]: object.__contains__(self, item) (就在 anchor 上方):

   The membership test operators (in and not in) are normally implemented as an iteration through a sequence. However, container objects can supply the following special method with a more efficient implementation, which also does not require the object be a sequence.

进入[GitHub]: python/cpython - (v3.5.4) cpython/Objects/setobject.c :

• 第 #1991 行:

  static PyMethodDef set_methods[] = {
      {"add",             (PyCFunction)set_add,           METH_O,
       add_doc},
      {"clear",           (PyCFunction)set_clear,         METH_NOARGS,
       clear_doc},
      {"__contains__",(PyCFunction)set_direct_contains,           METH_O | METH_COEXIST,  // @TODO - cfati: MARK THIS LINE
  

• 第 #1843 行:

  static PyObject *
  set_direct_contains(PySetObject *so, PyObject *key)
  {
      long result;
  
      result = set_contains(so, key);  // @TODO - cfati: MARK THIS LINE
      if (result == -1)
          return NULL;
      return PyBool_FromLong(result);
  }
  

• 第 #1823 行:

  static int
  set_contains(PySetObject *so, PyObject *key)
  {
      PyObject *tmpkey;
      int rv;
  
      rv = set_contains_key(so, key);    // @TODO - cfati: MARK THIS LINE
      if (rv == -1) {
          if (!PySet_Check(key) || !PyErr_ExceptionMatches(PyExc_TypeError))
              return -1;
          PyErr_Clear();
          tmpkey = make_new_set(&PyFrozenSet_Type, key);
          if (tmpkey == NULL)
              return -1;
          rv = set_contains_key(so, tmpkey);  // @TODO - cfati: MARK THIS LINE
          Py_DECREF(tmpkey);
      }
      return rv;
  }
  

• 第 #627 行:

  static int
  set_contains_key(PySetObject *so, PyObject *key)
  {
      setentry entry;
      Py_hash_t hash;
  
      if (!PyUnicode_CheckExact(key) ||
          (hash = ((PyASCIIObject *) key)->hash) == -1) {  // @TODO - cfati: MARK THIS LINE
          hash = PyObject_Hash(key);
          if (hash == -1)
              return -1;
      }
      entry.key = key;
      entry.hash = hash;
      return set_contains_entry(so, &entry);  // @TODO - cfati: MARK THIS LINE
  }
  

• 第 #614 行:

  static int
  set_contains_entry(PySetObject *so, setentry *entry)
  {
      PyObject *key;
      setentry *lu_entry;
  
      lu_entry = set_lookkey(so, entry->key, entry->hash);  // @TODO - cfati: MARK THIS LINE
      if (lu_entry == NULL)
          return -1;
      key = lu_entry->key;
      return key != NULL && key != dummy;
  }
  

从“调用堆栈”(以相反顺序呈现)可以看出，为了测试成员资格(in/not in)，hash 正在对候选成员(“includee”)执行(在所有代码路径上)，并且由于 list 实例没有 hash 功能，解释器吐出 TypeError。

决议

有很多方法可以解决这个问题(正如许多其他人已经指出的那样):

• 使用不要求其元素可哈希的容器(列表、元组)
• 测试__hash__成员
• 将成员资格测试包装在try/except block 中
• 为元素使用可散列的容器(元组):x = ('a', 'b')

但是(通常)这些只是解决问题的方法(这是我个人的意见)，因为如果您最终将列表与None 和< em>错误，代码(生成该列表)可以使用一些重构。