组合
import itertools
nums = [1, 2, 3, 4, 5, 6, 7, 8]
for i in itertools.combinations(nums, 4):
print(i)
(1, 2, 3, 4)
(1, 2, 3, 5)
(1, 2, 3, 6)
(1, 2, 3, 7)
(1, 2, 3, 8)
。。。
(4, 5, 6, 8)
(4, 5, 7, 8)
(4, 6, 7, 8)
(5, 6, 7, 8)排列
for i in itertools.permutations(nums, 4):
print(i)
(1, 2, 3, 4)
(1, 2, 3, 5)
(1, 2, 3, 6)
(1, 2, 3, 7)
。。。
(8, 7, 6, 2)
(8, 7, 6, 3)
(8, 7, 6, 4)
(8, 7, 6, 5)product
从给定的序列中各取一个生成序列，可重复的排列数
product('ab', range(3)) --> ('a',0) ('a',1) ('a',2) ('b',0) ('b',1) ('b',2)
product((0,1), (0,1), (0,1)) --> (0,0,0) (0,0,1) (0,1,0) (0,1,1) (1,0,0) ...
for i in itertools.product([0, 1], repeat=3):
print(i)
(0, 0, 0)
(0, 0, 1)
(0, 1, 0)
(0, 1, 1)
(1, 0, 0)
(1, 0, 1)
(1, 1, 0)
(1, 1, 1)combinations_with_replacement
combinations_with_replacement(iterable, r) --> combinations_with_replacement object
Return successive r-length combinations of elements in the iterable
allowing individual elements to have successive repeats.
combinations_with_replacement('ABC', 2) --> AA AB AC BB BC CC
可重复的组合数
for i in itertools.combinations_with_replacement([1, 2, 3, 4], 3):
print(i)
(1, 1, 1)
(1, 1, 2)
(1, 1, 3)
(1, 1, 4)
(1, 2, 2)
(1, 2, 3)
(1, 2, 4)
(1, 3, 3)
(1, 3, 4)
(1, 4, 4)
(2, 2, 2)
(2, 2, 3)
(2, 2, 4)
(2, 3, 3)
(2, 3, 4)
(2, 4, 4)
(3, 3, 3)
(3, 3, 4)
(3, 4, 4)
(4, 4, 4)对比product
for i in itertools.product([1, 2, 3, 4], repeat=3):
print(i)
(1, 1, 1)
(1, 1, 2)
(1, 1, 3)
(1, 1, 4)
(1, 2, 1)
(1, 2, 2)
(1, 2, 3)
(1, 2, 4)
(1, 3, 1)
(1, 3, 2)
(1, 3, 3)
(1, 3, 4)
(1, 4, 1)
(1, 4, 2)
(1, 4, 3)
(1, 4, 4)
(2, 1, 1)
(2, 1, 2)
(2, 1, 3)
(2, 1, 4)
(2, 2, 1)
(2, 2, 2)
(2, 2, 3)
(2, 2, 4)
(2, 3, 1)
(2, 3, 2)
(2, 3, 3)
(2, 3, 4)
(2, 4, 1)
(2, 4, 2)
(2, 4, 3)
(2, 4, 4)
(3, 1, 1)
(3, 1, 2)
(3, 1, 3)
(3, 1, 4)
(3, 2, 1)
(3, 2, 2)
(3, 2, 3)
(3, 2, 4)
(3, 3, 1)
(3, 3, 2)
(3, 3, 3)
(3, 3, 4)
(3, 4, 1)
(3, 4, 2)
(3, 4, 3)
(3, 4, 4)
(4, 1, 1)
(4, 1, 2)
(4, 1, 3)
(4, 1, 4)
(4, 2, 1)
(4, 2, 2)
(4, 2, 3)
(4, 2, 4)
(4, 3, 1)
(4, 3, 2)
(4, 3, 3)
(4, 3, 4)
(4, 4, 1)
(4, 4, 2)
(4, 4, 3)
(4, 4, 4)
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