Python迭代器有has_next方法吗?


当前回答

不,没有这样的方法。迭代的结束由异常表示。请参见文档。

其他回答

也许只有我这么想,但虽然我喜欢https://stackoverflow.com/users/95810/alex-martelli的答案,但我发现这个更容易读:

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

class MyIterator(Iterator):  # need to subclass Iterator rather than object
  def __init__(self, it):
    self._iter = iter(it)
    self._sentinel = object()
    self._next = next(self._iter, self._sentinel)
    
  def __iter__(self): 
    return self
  
  def __next__(self):        # __next__ vs next in python 2
    if not self.has_next():
      next(self._iter)  # raises StopIteration

    val = self._next
    self._next = next(self._iter, self._sentinel)
    return val
  
  def has_next(self):
    return self._next is not self._sentinel

建议的方法是StopIteration。 请参阅tutorialspoint中的斐波那契示例

#!usr/bin/python3

import sys
def fibonacci(n): #generator function
   a, b, counter = 0, 1, 0
   while True:
      if (counter > n): 
         return
      yield a
      a, b = b, a + b
      counter += 1
f = fibonacci(5) #f is iterator object

while True:
   try:
      print (next(f), end=" ")
   except StopIteration:
      sys.exit()

不。最类似的概念很可能是StopIteration异常。

除了所有提到的StopIteration, Python的“for”循环只是做你想要的:

>>> it = iter("hello")
>>> for i in it:
...     print i
...
h
e
l
l
o

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