python itertool

itertools.product()

この関数はデカルト積を出力してくれる

>>> from itertools import product
>>>
>>> print list(product([1,2,3],repeat = 2))
[(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)]
>>>
>>> print list(product([1,2,3],[3,4]))
[(1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)]
>>>
>>> A = [[1,2,3],[3,4,5]]
>>> print list(product(*A))
[(1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5), (3, 3), (3, 4), (3, 5)]

itertools.permutations（iterable [、r]）

イテラブル内の要素で、長さrの重複を許さない順列を出力する

>>> from itertools import permutations
>>> print permutations(['1','2','3'])
<itertools.permutations object at 0x02A45210>
>>> 
>>> print list(permutations(['1','2','3']))
[('1', '2', '3'), ('1', '3', '2'), ('2', '1', '3'), ('2', '3', '1'), ('3', '1', '2'), ('3', '2', '1')]
>>> 
>>> print list(permutations(['1','2','3'],2))
[('1', '2'), ('1', '3'), ('2', '1'), ('2', '3'), ('3', '1'), ('3', '2')]
>>>
>>> print list(permutations('abc',3))
[('a', 'b', 'c'), ('a', 'c', 'b'), ('b', 'a', 'c'), ('b', 'c', 'a'), ('c', 'a', 'b'), ('c', 'b', 'a')]

itertools.combinations(iterable, r)

長さrのタプル列をソートされた順で重複なしで出力する

>>> from itertools import combinations
>>> 
>>> print list(combinations('12345',2))
[('1', '2'), ('1', '3'), ('1', '4'), ('1', '5'), ('2', '3'), ('2', '4'), ('2', '5'), ('3', '4'), ('3', '5'), ('4', '5')]
>>> 
>>> A = [1,1,3,3,3]
>>> print list(combinations(A,4))
[(1, 1, 3, 3), (1, 1, 3, 3), (1, 1, 3, 3), (1, 3, 3, 3), (1, 3, 3, 3)]

itertools.combinations_with_replacement()

長さrのタプル列をソートされた順で重複ありで出力する

>>> from itertools import combinations_with_replacement
>>> 
>>> print list(combinations_with_replacement('12345',2))
[('1', '1'), ('1', '2'), ('1', '3'), ('1', '4'), ('1', '5'), ('2', '2'), ('2', '3'), ('2', '4'), ('2', '5'), ('3', '3'), ('3', '4'), ('3', '5'), ('4', '4'), ('4', '5'), ('5', '5')]
>>> 
>>> A = [1,1,3,3,3]
>>> print list(combinations(A,2))
[(1, 1), (1, 3), (1, 3), (1, 3), (1, 3), (1, 3), (1, 3), (3, 3), (3, 3), (3, 3)]