python - 使用 itertools 进行格雷码顺序的笛卡尔积？

Question

有没有类似Python的itertools.product()的东西它提供了一组按格雷码顺序的笛卡尔积的迭代？例如，假设存在这样一个假设的生成器，它被称为 gray_code_product() ，然后gray_code_product(['a','b','c'], [0,1], ['x','y'])将按顺序生成:

('a',0,'x')
('a',0,'y')
('a',1,'y')
('a',1,'x')
('b',1,'x')
('b',1,'y')
('b',0,'y')
('b',0,'x')
('c',0,'x')
('c',0,'y')
('c',1,'y')
('c',1,'x')

Answer 1

根据documentation itertools.product 的函数相当于以下 Python 代码:

def product(*args, repeat=1):
    pools = [tuple(pool) for pool in args] * repeat
    result = [[]]
    for pool in pools:
        result = [x+[y] for x in result for y in pool]
    for prod in result:
        yield tuple(prod)

由于格雷码产品是关于反转每个池的前面序列的顺序，因此您可以在迭代它以确定之前的 result 列表时使用 enumerate 来确定如果索引是奇数或偶数，如果是奇数则反转池的顺序:

def gray_code_product(*args, repeat=1):
    pools = [tuple(pool) for pool in args] * repeat
    result = [[]]
    for pool in pools:
        result = [x+[y] for i, x in enumerate(result) for y in (
            reversed(pool) if i % 2 else pool)]
    for prod in result:
        yield tuple(prod)

这样:

for p in gray_code_product(['a','b','c'], [0,1], ['x','y']):
    print(p)

输出:

('a', 0, 'x')
('a', 0, 'y')
('a', 1, 'y')
('a', 1, 'x')
('b', 1, 'x')
('b', 1, 'y')
('b', 0, 'y')
('b', 0, 'x')
('c', 0, 'x')
('c', 0, 'y')
('c', 1, 'y')
('c', 1, 'x')