Python—itertools库使用

itertools.product()

重复使用words 3次进行元素间的组合，总共是3×3×3=27中情况。

words = ["word", "good", "best"]
print(list(itertools.product(words, repeat=3)))

结果：
[('word', 'word', 'word'), ('word', 'word', 'good'), ('word', 'word', 'best'), ('word', 'good', 'word'), ('word', 'good', 'good'),
('word', 'good', 'best'), ('word', 'best', 'word'), ('word', 'best', 'good'), ('word', 'best', 'best'), ('good', 'word', 'word'),
('good', 'word', 'good'), ('good', 'word', 'best'), ('good', 'good', 'word'), ('good', 'good', 'good'), ('good', 'good', 'best'),
('good', 'best', 'word'), ('good', 'best', 'good'), ('good', 'best', 'best'), ('best', 'word', 'word'), ('best', 'word', 'good'),
('best', 'word', 'best'), ('best', 'good', 'word'), ('best', 'good', 'good'), ('best', 'good', 'best'), ('best', 'best', 'word'),
('best', 'best', 'good'), ('best', 'best', 'best')]

itertools.combinations()

内部有序自己组合，要求长度小于列表自身长度。

words = ["word", "good", "best"]
print(list(itertools.combinations(words, 3)))

结果：
[('word', 'good', 'best')]

words = ["word", "good", "best"]
print(list(itertools.combinations(words, 2)))

结果：
[('word', 'good'), ('word', 'best'), ('good', 'best')]

itertools.permutations()

内部无序自组合

words = ["word", "good", "best"]
print(list(itertools.permutations(words, 3))) # 3为列表长度

结果：
[('word', 'good', 'best'), ('word', 'best', 'good'), ('good', 'word', 'best'), ('good', 'best', 'word'), ('best', 'word', 'good'), ('best', 'good', 'word')]