我只是想说，获得一个具有切片和索引功能的xrange对象真的不是那么困难。我写了一些代码，工作得很好，就像xrange一样快，当它计数(迭代)。

from __future__ import division

def read_xrange(xrange_object):
    # returns the xrange object's start, stop, and step
    start = xrange_object[0]
    if len(xrange_object) > 1:
       step = xrange_object[1] - xrange_object[0]
    else:
        step = 1
    stop = xrange_object[-1] + step
    return start, stop, step

class Xrange(object):
    ''' creates an xrange-like object that supports slicing and indexing.
    ex: a = Xrange(20)
    a.index(10)
    will work

    Also a[:5]
    will return another Xrange object with the specified attributes

    Also allows for the conversion from an existing xrange object
    '''
    def __init__(self, *inputs):
        # allow inputs of xrange objects
        if len(inputs) == 1:
            test, = inputs
            if type(test) == xrange:
                self.xrange = test
                self.start, self.stop, self.step = read_xrange(test)
                return

        # or create one from start, stop, step
        self.start, self.step = 0, None
        if len(inputs) == 1:
            self.stop, = inputs
        elif len(inputs) == 2:
            self.start, self.stop = inputs
        elif len(inputs) == 3:
            self.start, self.stop, self.step = inputs
        else:
            raise ValueError(inputs)

        self.xrange = xrange(self.start, self.stop, self.step)

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

    def __getitem__(self, item):
        if type(item) is int:
            if item < 0:
                item += len(self)

            return self.xrange[item]

        if type(item) is slice:
            # get the indexes, and then convert to the number
            start, stop, step = item.start, item.stop, item.step
            start = start if start != None else 0 # convert start = None to start = 0
            if start < 0:
                start += start
            start = self[start]
            if start < 0: raise IndexError(item)
            step = (self.step if self.step != None else 1) * (step if step != None else 1)
            stop = stop if stop is not None else self.xrange[-1]
            if stop < 0:
                stop += stop

            stop = self[stop]
            stop = stop

            if stop > self.stop:
                raise IndexError
            if start < self.start:
                raise IndexError
            return Xrange(start, stop, step)

    def index(self, value):
        error = ValueError('object.index({0}): {0} not in object'.format(value))
        index = (value - self.start)/self.step
        if index % 1 != 0:
            raise error
        index = int(index)


        try:
            self.xrange[index]
        except (IndexError, TypeError):
            raise error
        return index

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

老实说，我认为整个问题有点傻，xrange应该做所有这些…

不，它们都有自己的用途:

迭代时使用xrange()，因为它节省内存。说:

for x in xrange(1, one_zillion):

而不是:

for x in range(1, one_zillion):

另一方面，如果实际需要一个数字列表，则使用range()。

multiples_of_seven = range(7,100,7)
print "Multiples of seven < 100: ", multiples_of_seven

对于性能而言，尤其是在大范围内迭代时，xrange()通常更好。然而，仍然有一些情况下你可能更喜欢range():

In python 3, range() does what xrange() used to do and xrange() does not exist. If you want to write code that will run on both Python 2 and Python 3, you can't use xrange(). range() can actually be faster in some cases - eg. if iterating over the same sequence multiple times. xrange() has to reconstruct the integer object every time, but range() will have real integer objects. (It will always perform worse in terms of memory however) xrange() isn't usable in all cases where a real list is needed. For instance, it doesn't support slices, or any list methods.

[编辑]有几篇文章提到了如何通过2to3工具升级range()。为了记录，下面是在range()和xrange()的一些示例用法上运行该工具的输出。

RefactoringTool: Skipping implicit fixer: buffer
RefactoringTool: Skipping implicit fixer: idioms
RefactoringTool: Skipping implicit fixer: ws_comma
--- range_test.py (original)
+++ range_test.py (refactored)
@@ -1,7 +1,7 @@

 for x in range(20):
-    a=range(20)
+    a=list(range(20))
     b=list(range(20))
     c=[x for x in range(20)]
     d=(x for x in range(20))
-    e=xrange(20)
+    e=range(20)

如您所见，当在for循环或推导式中使用时，或者已经使用list()包装时，range保持不变。

Xrange()更有效，因为它每次只生成一个对象，而不是生成一个对象列表。而不是100个整数，以及它们所有的开销，以及将它们放入的列表，你每次只需要一个整数。更快的生成，更好的内存使用，更高效的代码。

除非我特别需要一个列表，否则我总是喜欢xrange()

Range (): Range(1,10)返回1到10个数字的列表，并将整个列表保存在内存中。 xrange():类似于range()，但不是返回一个列表，而是返回一个根据需要生成范围内数字的对象。对于循环，这比range()略快，内存效率更高。xrange()对象类似于迭代器，并根据需要生成数字(惰性求值)。

In [1]: range(1,10)
Out[1]: [1, 2, 3, 4, 5, 6, 7, 8, 9]

In [2]: xrange(10)
Out[2]: xrange(10)

In [3]: print xrange.__doc__
Out[3]: xrange([start,] stop[, step]) -> xrange object

range()所做的事情与Python 3中xrange()所做的事情相同，并且Python 3中不存在术语xrange()。 如果多次迭代同一个序列，那么Range()实际上在某些场景中更快。Xrange()每次都必须重构整数对象，但range()将拥有真正的整数对象。

另一个区别是Python 2实现的xrange()不支持大于C int的数字，所以如果你想使用Python内置的大数字支持来获得一个范围，你必须使用range()。

Python 2.7.3 (default, Jul 13 2012, 22:29:01) 
[GCC 4.7.1] on linux2
Type "help", "copyright", "credits" or "license" for more information.
>>> range(123456787676676767676676,123456787676676767676679)
[123456787676676767676676L, 123456787676676767676677L, 123456787676676767676678L]
>>> xrange(123456787676676767676676,123456787676676767676679)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
OverflowError: Python int too large to convert to C long

Python 3没有这个问题:

Python 3.2.3 (default, Jul 14 2012, 01:01:48) 
[GCC 4.7.1] on linux2
Type "help", "copyright", "credits" or "license" for more information.
>>> range(123456787676676767676676,123456787676676767676679)
range(123456787676676767676676, 123456787676676767676679)