不。最类似的概念很可能是StopIteration异常。

我相信python只有next()，根据文档，如果没有更多的元素，它就会抛出异常。

http://docs.python.org/library/stdtypes.html#iterator-types

在任意迭代器对象中尝试__length_hint__()方法:

iter(...).__length_hint__() > 0

如果您确实需要一个has-next功能，那么使用一个小包装器类很容易获得它。例如:

class hn_wrapper(object):
  def __init__(self, it):
    self.it = iter(it)
    self._hasnext = None
  def __iter__(self): return self
  def next(self):
    if self._hasnext:
      result = self._thenext
    else:
      result = next(self.it)
    self._hasnext = None
    return result
  def hasnext(self):
    if self._hasnext is None:
      try: self._thenext = next(self.it)
      except StopIteration: self._hasnext = False
      else: self._hasnext = True
    return self._hasnext

现在就像

x = hn_wrapper('ciao')
while x.hasnext(): print next(x)

发出

c
i
a
o

是必需的。

Note that the use of next(sel.it) as a built-in requires Python 2.6 or better; if you're using an older version of Python, use self.it.next() instead (and similarly for next(x) in the example usage). [[You might reasonably think this note is redundant, since Python 2.6 has been around for over a year now -- but more often than not when I use Python 2.6 features in a response, some commenter or other feels duty-bound to point out that they are 2.6 features, thus I'm trying to forestall such comments for once;-)]]

===

对于Python3，您将进行以下更改:

from collections.abc import Iterator  # since python 3.3 Iterator is here

class hn_wrapper(Iterator):  # need to subclass Iterator rather than object
  def __init__(self, it):
    self.it = iter(it)
    self._hasnext = None
    
  def __iter__(self): 
    return self
  
  def __next__(self):        # __next__ vs next in python 2
    if self._hasnext:
      result = self._thenext
    else:
      result = next(self.it)
    self._hasnext = None
    return result
  
  def hasnext(self):
    if self._hasnext is None:
      try: 
        self._thenext = next(self.it)
      except StopIteration: 
        self._hasnext = False
      else: self._hasnext = True
    return self._hasnext

为了读取所有迭代，基于处理“StopIteration”执行的解决方法非常简单:

    end_cursor = False
    while not end_cursor:
        try:
            print(cursor.next())
        except StopIteration:
            print('end loop')
            end_cursor = True
        except:
            print('other exceptions to manage')
            end_cursor = True

非常有趣的问题，但是这个“hasnext”的设计已经放进了leetcode: https://leetcode.com/problems/iterator-for-combination/

这是我的实现:

class CombinationIterator:

def __init__(self, characters: str, combinationLength: int):
    from itertools import combinations
    from collections import deque
    self.iter = combinations(characters, combinationLength)
    self.res = deque()


def next(self) -> str:
    if len(self.res) == 0:
        return ''.join(next(self.iter))
    else:
        return ''.join(self.res.pop())


def hasNext(self) -> bool:
    try:
        self.res.insert(0, next(self.iter))
        return True
    except:
        return len(self.res) > 0