为什么或者为什么不呢?


当前回答

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…

其他回答

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…

选择范围有以下几个原因:

1) xrange将在新的Python版本中消失。这为您提供了方便的未来兼容性。

2) range将具有与xrange相关的效率。

书中给出了一个很好的例子:Practical Python By Magnus Lie Hetland

>>> zip(range(5), xrange(100000000))
[(0, 0), (1, 1), (2, 2), (3, 3), (4, 4)]

我不建议在前面的例子中使用range而不是xrange——尽管如此 只需要前五个数字,range计算所有数字,这可能需要很多时间 的时间。使用xrange,这不是问题,因为它只计算所需的数字。

是的,我读了@Brian的回答:在python 3中,range()是一个生成器,xrange()不存在。

这里的每个人对于xrange和range的利弊都有不同的看法。它们大多是正确的,xrange是一个迭代器,而range充实并创建了一个实际的列表。在大多数情况下,您不会真正注意到两者之间的区别。(你可以在range中使用map,但不能在xrange中使用,但这会占用更多内存。)

但是,我认为您可能希望听到的是首选的选项是xrange。由于Python 3中的range是一个迭代器,代码转换工具2to3将正确地将xrange的所有使用转换为range,并将抛出一个使用range的错误或警告。如果您希望确保将来可以轻松地转换代码,您将只使用xrange,当您确定需要一个列表时使用list(xrange)。我是在今年(2008年)芝加哥PyCon的CPython冲刺中了解到这一点的。

虽然在大多数情况下xrange比range快,但性能上的差异非常小。下面的小程序比较了range和xrange的迭代:

import timeit
# Try various list sizes.
for list_len in [1, 10, 100, 1000, 10000, 100000, 1000000]:
  # Time doing a range and an xrange.
  rtime = timeit.timeit('a=0;\nfor n in range(%d): a += n'%list_len, number=1000)
  xrtime = timeit.timeit('a=0;\nfor n in xrange(%d): a += n'%list_len, number=1000)
  # Print the result
  print "Loop list of len %d: range=%.4f, xrange=%.4f"%(list_len, rtime, xrtime)

下面的结果显示xrange确实更快,但还不足以让人担心。

Loop list of len 1: range=0.0003, xrange=0.0003
Loop list of len 10: range=0.0013, xrange=0.0011
Loop list of len 100: range=0.0068, xrange=0.0034
Loop list of len 1000: range=0.0609, xrange=0.0438
Loop list of len 10000: range=0.5527, xrange=0.5266
Loop list of len 100000: range=10.1666, xrange=7.8481
Loop list of len 1000000: range=168.3425, xrange=155.8719

所以无论如何都要使用xrange,但除非您在受限的硬件上,否则不要太担心它。