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
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。
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。
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…
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。
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)
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