介绍Python中内置的itertools模块
Python的内建模块itertools提供了非常有用的用于操作迭代对象的函数。
首先，我们看看itertools提供的几个“无限”迭代器：
>>> import itertools >>> natuals = itertools.count(1) >>> for n in natuals: ... print n ... 1 2 3 ...
因为count()会创建一个无限的迭代器，所以上述代码会打印出自然数序列，根本停不下来，只能按Ctrl+C退出。
cycle()会把传入的一个序列无限重复下去：
>>> import itertools
>>> cs = itertools.cycle('ABC') # 注意字符串也是序列的一种
>>> for c in cs:
... print c
...
'A'
'B'
'C'
'A'
'B'
'C'
...
同样停不下来。
repeat()负责把一个元素无限重复下去，不过如果提供第二个参数就可以限定重复次数：
>>> ns = itertools.repeat('A', 10)
>>> for n in ns:
... print n
...
打印10次'A'
无限序列只有在for迭代时才会无限地迭代下去，如果只是创建了一个迭代对象，它不会事先把无限个元素生成出来，事实上也不可能在内存中创建无限多个元素。
无限序列虽然可以无限迭代下去，但是通常我们会通过takewhile()等函数根据条件判断来截取出一个有限的序列：
>>> natuals = itertools.count(1) >>> ns = itertools.takewhile(lambda x: x <= 10, natuals) >>> for n in ns: ... print n ...
打印出1到10
itertools提供的几个迭代器操作函数更加有用：
chain()
chain()可以把一组迭代对象串联起来，形成一个更大的迭代器：
for c in chain('ABC', 'XYZ'):
print c
# 迭代效果：'A' 'B' 'C' 'X' 'Y' 'Z'
groupby()
groupby()把迭代器中相邻的重复元素挑出来放在一起：
>>> for key, group in itertools.groupby('AAABBBCCAAA'):
... print key, list(group) # 为什么这里要用list()函数呢？
...
A ['A', 'A', 'A']
B ['B', 'B', 'B']
C ['C', 'C']
A ['A', 'A', 'A']
实际上挑选规则是通过函数完成的，只要作用于函数的两个元素返回的值相等，这两个元素就被认为是在一组的，而函数返回值作为组的key。如果我们要忽略大小写分组，就可以让元素'A'和'a'都返回相同的key：
>>> for key, group in itertools.groupby('AaaBBbcCAAa', lambda c: c.upper()):
... print key, list(group)
...
A ['A', 'a', 'a']
B ['B', 'B', 'b']
C ['c', 'C']
A ['A', 'A', 'a']
imap()
imap()和map()的区别在于，imap()可以作用于无穷序列，并且，如果两个序列的长度不一致，以短的那个为准。
>>> for x in itertools.imap(lambda x, y: x * y, [10, 20, 30], itertools.count(1)): ... print x ... 10 40 90
注意imap()返回一个迭代对象，而map()返回list。当你调用map()时，已经计算完毕：
>>> r = map(lambda x: x*x, [1, 2, 3]) >>> r # r已经计算出来了 [1, 4, 9]
当你调用imap()时，并没有进行任何计算：
>>> r = itertools.imap(lambda x: x*x, [1, 2, 3]) >>> r <itertools.imap object at 0x103d3ff90> # r只是一个迭代对象
必须用for循环对r进行迭代，才会在每次循环过程中计算出下一个元素：
>>> for x in r: ... print x ... 1 4 9
这说明imap()实现了“惰性计算”，也就是在需要获得结果的时候才计算。类似imap()这样能够实现惰性计算的函数就可以处理无限序列：
>>> r = itertools.imap(lambda x: x*x, itertools.count(1)) >>> for n in itertools.takewhile(lambda x: x<100, r): ... print n ...
结果是什么?
如果把imap()换成map()去处理无限序列会有什么结果？
>>> r = map(lambda x: x*x, itertools.count(1))
结果是什么?
ifilter()
不用多说了，ifilter()就是filter()的惰性实现。
小结
itertools模块提供的全部是处理迭代功能的函数，它们的返回值不是list，而是迭代对象，只有用for循环迭代的时候才真正计算。
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