Python：itertools模块 combinations和product的使用

1.combinations(iterable, r)  创建一个迭代器，返回iterable中所有长度为r的子序列，返回的子序列中的项按输入iterable中的顺序排序:

官方文档

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def combinations(iterable, r):     # combinations('ABCD', 2) --> AB AC AD BC BD CD     # combinations(range(4), 3) --> 012 013 023 123     pool = tuple(iterable)     n = len(pool)     if r > n:         return     indices = range(r)     yield tuple(pool[i] for i in indices)     while True:         for i in reversed(range(r)):             if indices[i] != i + n - r:                 break         else:             return         indices[i] += 1         for j in range(i+1, r):             indices[j] = indices[j-1] + 1         yield tuple(pool[i] for i in indices)

　　

def combinations(iterable, r):
    pool = tuple(iterable)
    n = len(pool)
    for indices in permutations(range(n), r):
        if sorted(indices) == list(indices):
            yield tuple(pool[i] for i in indices)

例：

>>> list(combinations(range(3),2))
[(0, 1), (0, 2), (1, 2)]
>>> list(combinations(range(3),3))
[(0, 1, 2)]
>>> list(combinations(range(3),1))
[(0,), (1,), (2,)]

2.product(*iterables[, repeat])  创建一个迭代器，生成表示iterables中的项目的笛卡尔积的元组，repeat表示重复生成序列的次数。

def product(*args, **kwds):
    # product('ABCD', 'xy') --> Ax Ay Bx By Cx Cy Dx Dy
    # product(range(2), repeat=3) --> 000 001 010 011 100 101 110 111
    pools = map(tuple, args) * kwds.get('repeat', 1)
    result = [[]]
    for pool in pools:
        result = [x+[y] for x in result for y in pool]
    for prod in result:
        yield tuple(prod)

 例:

>>> list(product(range(3),repeat=1))
[(0,), (1,), (2,)]
>>> list(product(range(3),repeat=2))
[(0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (1, 2), (2, 0), (2, 1), (2, 2)]
>>> list(product(range(3),repeat=3))
[(0, 0, 0), (0, 0, 1), (0, 0, 2), (0, 1, 0), (0, 1, 1), (0, 1, 2), (0, 2, 0), (0, 2, 1), (0, 2, 2), (1, 0, 0), (1, 0, 1), (1, 0, 2), (1, 1, 0), (1, 1, 1), (1, 1, 2), (1, 2, 0), (1, 2, 1), (1, 2, 2), (2, 0, 0), (2, 0, 1), (2, 0, 2), (2, 1, 0), (2, 1, 1), (2, 1, 2), (2, 2, 0), (2, 2, 1), (2, 2, 2)]