介绍Python中内置的itertools模块

> import itertools
> natuals = itertools.count(1)
> for n in natuals:
...   print n
...
1
2
3
...

> 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'
...

> ns = itertools.repeat('A', 10)
> for n in ns:
...   print n
...

> natuals = itertools.count(1)
> ns = itertools.takewhile(lambda x: x <= 10, natuals)
> for n in ns:
...   print n
...

for c in chain('ABC', 'XYZ'):
  print c
# 迭代效果：'A' 'B' 'C' 'X' 'Y' 'Z'

> 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']

> 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']

> for x in itertools.imap(lambda x, y: x * y, [10, 20, 30], itertools.count(1)):
...   print x
...
10
40
90

> r = map(lambda x: x*x, [1, 2, 3])
> r # r已经计算出来了
[1, 4, 9]

> 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

> for x in r:
...   print x
...
1
4
9

> r = itertools.imap(lambda x: x*x, itertools.count(1))
> for n in itertools.takewhile(lambda x: x<100, r):
...   print n
...

> r = map(lambda x: x*x, itertools.count(1))