pymotw 英文版模块介绍链接 http://pymotw.com/2/itertools/
基本是基于文档的翻译和补充，相当于翻译了。
itertools用于高效循环的迭代函数集合
无限迭代器
迭代器 参数 结果 例子
count() start, [step] start, start+step, start+2*step, ... count(10) --> 10 11 12 13 14 ...
cycle() p p0, p1, ... plast, p0, p1, ... cycle('ABCD') --> A B C D A B C D ...
repeat() elem [,n] elem, elem, elem, ... endlessly or up to n times repeat(10, 3) --> 10 10 10
处理输入序列迭代器
迭代器 参数 结果 例子
chain() p, q, ... p0, p1, ... plast, q0, q1, ... chain('ABC', 'DEF') --> A B C D E F
compress() data, selectors (d[0] if s[0]), (d[1] if s[1]), ... compress('ABCDEF', [1,0,1,0,1,1]) --> A C E F
dropwhile() pred, seq seq[n], seq[n+1], starting when pred fails dropwhile(lambda x: x<5, [1,4,6,4,1]) --> 6 4 1
groupby() iterable[, keyfunc] sub-iterators grouped by value of keyfunc(v)
ifilter() pred, seq elements of seq where pred(elem) is True ifilter(lambda x: x%2, range(10)) --> 1 3 5 7 9
ifilterfalse() pred, seq elements of seq where pred(elem) is False ifilterfalse(lambda x: x%2, range(10)) --> 0 2 4 6 8
islice() seq, [start,] stop [, step] elements from seq[start:stop:step] islice('ABCDEFG', 2, None) --> C D E F G
imap() func, p, q, ... func(p0, q0), func(p1, q1), ... imap(pow, (2,3,10), (5,2,3)) --> 32 9 1000
starmap() func, seq func(*seq[0]), func(*seq[1]), ... starmap(pow, [(2,5), (3,2), (10,3)]) --> 32 9 1000
tee() it, n it1, it2 , ... itn splits one iterator into n
takewhile() pred, seq seq[0], seq[1], until pred fails takewhile(lambda x: x<5, [1,4,6,4,1]) --> 1 4
izip() p, q, ... (p[0], q[0]), (p[1], q[1]), ... izip('ABCD', 'xy') --> Ax By
izip_longest() p, q, ... (p[0], q[0]), (p[1], q[1]), ... izip_longest('ABCD', 'xy', fillvalue='-') --> Ax By C- D-
组合生成器
迭代器 参数 结果
product() p, q, ... [repeat=1] cartesian product, equivalent to a nested for-loop
permutations() p[, r] r-length tuples, all possible orderings, no repeated elements
combinations() p, r r-length tuples, in sorted order, no repeated elements
combinations_with_replacement() p, r r-length tuples, in sorted order, with repeated elements
product('ABCD', repeat=2) AA AB AC AD BA BB BC BD CA CB CC CD DA DB DC DD
permutations('ABCD', 2) AB AC AD BA BC BD CA CB CD DA DB DC
combinations('ABCD', 2) AB AC AD BC BD CD
combinations_with_replacement('ABCD', 2) AA AB AC AD BB BC BD CC CD DD
第一部分
itertools.count(start=0, step=1)
创建一个迭代器，生成从n开始的连续整数，如果忽略n，则从0开始计算（注意：此迭代器不支持长整数）
如果超出了sys.maxint，计数器将溢出并继续从-sys.maxint-1开始计算。
定义
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def count(start=0, step=1):
# count(10) --> 10 11 12 13 14 ...
# count(2.5, 0.5) -> 2.5 3.0 3.5 ...
n = start
while True:
yield n
n += step
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等同于(start + step * i for i in count())
使用 代码如下:
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from itertools import *
for i in izip(count(1), ['a', 'b', 'c']):
print i
(1, 'a')
(2, 'b')
(3, 'c')
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itertools.repeat(object[, times])
创建一个迭代器，重复生成object，times（如果已提供）指定重复计数，如果未提供times，将无止尽返回该对象。
定义
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def repeat(object, times=None):
# repeat(10, 3) --> 10 10 10
if times is None:
while True:
yield object
else:
for i in xrange(times):
yield object
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使用 代码如下:
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from itertools import *
for i in repeat('over-and-over', 5):
print i
over-and-over
over-and-over
over-and-over
over-and-over
over-and-over
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