4楼: Originally posted by 额纪 at 2013-04-03 23:00:13
再问个问题，怎么知道有这些现成的函数呀？...

>>> import itertools
>>> dir(itertools)
['__doc__', '__name__', '__package__', 'chain', 'combinations', 'combinations_with_replacement', 'compress', 'count', 'cycle', 'dropwhile', 'groupby', 'ifilter', 'ifilterfalse', 'imap', 'islice', 'izip', 'izip_longest', 'permutations', 'product', 'repeat', 'starmap', 'takewhile', 'tee']
>>> help(itertools.permutations)
Help on class permutations in module itertools:

class permutations(__builtin__.object)
|  permutations(iterable[, r]) --> permutations object
|  
|  Return successive r-length permutations of elements in the iterable.
|  
|  permutations(range(3), 2) --> (0,1), (0,2), (1,0), (1,2), (2,0), (2,1)
|  
|  Methods defined here:
|  
|  __getattribute__(...)
|      x.__getattribute__('name') <==> x.name
|  
|  __iter__(...)
|      x.__iter__() <==> iter(x)
|  
|  next(...)
|      x.next() -> the next value, or raise StopIteration
|  
|  ----------------------------------------------------------------------
|  Data and other attributes defined here:
|  
|  __new__ =
|      T.__new__(S, ...) -> a new object with type S, a subtype of T

>>>  

>>> from itertools import permutations
>>> a = [1,2,3,4,5,6]
>>> b = [x for x in permutations(a)]
>>> len(b)
720
>>> b[0]
(1, 2, 3, 4, 5, 6)
>>> b[1]
(1, 2, 3, 4, 6, 5)
>>>

2楼: Originally posted by libralibra at 2013-04-03 00:07:23
>>> from itertools import permutations
>>> a =
>>> b =
>>> len(b)
720
>>> b
(1, 2, 3, 4, 5, 6)
>>> b
(1, 2, 3, 4, 6, 5)
>>>