Python迭代和迭代器详解

In [38]: s = 'ab'

In [39]: it = iter(s)

In [40]: it
Out[40]: <iterator at 0x1068e6d50>

In [41]: print it
<iterator object at 0x1068e6d50>

In [42]: it.next()
Out[42]: 'a'

In [43]: it.next()
Out[43]: 'b'

In [44]: it.next()
---------------------------------------------------------------------------
StopIteration               Traceback (most recent call last)
<ipython-input-44-54f0920595b2> in <module>()
----> 1 it.next()

StopIteration:

class Reverse:
  """Iterator for looping over a sequence backwards."""
  def __init__(self, data):
    self.data = data
    self.index = len(data)

  def __iter__(self):
    return self

  def next(self):
    if self.index == 0:
      raise StopIteration
    self.index = self.index - 1
    return self.data[self.index]

rev = Reverse('spam')
for char in rev:
  print char

[output]
m
a
p
s

def reverse(data):
  for index in range(len(data)-1, -1, -1):
    yield data[index]

>>> for char in reverse('golf'):
...   print char
...
f
l
o
g

Help on built-in function range in module __builtin__:

range(...)
  range(stop) -> list of integers
  range(start, stop[, step]) -> list of integers

  Return a list containing an arithmetic progression of integers.
  range(i, j) returns [i, i+1, i+2, ..., j-1]; start (!) defaults to 0.
  When step is given, it specifies the increment (or decrement).
  For example, range(4) returns [0, 1, 2, 3]. The end point is omitted!
  These are exactly the valid indices for a list of 4 elements.

class xrange(object)
 | xrange(stop) -> xrange object
 | xrange(start, stop[, step]) -> xrange object
 |
 | Like range(), but instead of returning a list, returns an object that
 | generates the numbers in the range on demand. For looping, this is
 | slightly faster than range() and more memory efficient.
iter()

In [113]: s = 'abc'

In [114]: s.next()
---------------------------------------------------------------------------
AttributeError              Traceback (most recent call last)
<ipython-input-114-5e5e6532ea26> in <module>()
----> 1 s.next()

AttributeError: 'str' object has no attribute 'next'

In [115]: it = iter(s)

In [116]: it.next()
Out[116]: 'a'

In [119]: num = (i for i in range(10))

In [120]: sum(num)
Out[120]: 45