Python itertools.product 重新排序生成

我有这个：

shape = (2, 4) # arbitrary, could be 3 dimensions such as (3, 5, 7), etc...

for i in itertools.product(*(range(x) for x in shape)):
    print(i)

# output: (0, 0) (0, 1) (0, 2) (0, 3) (1, 0) (1, 1) (1, 2) (1, 3)

到目前为止，一切都很好，itertools.product在每次迭代中前进最右边的元素。但现在我希望能够根据以下内容指定迭代顺序：

axes = (0, 1) # normal order
# output: (0, 0) (0, 1) (0, 2) (0, 3) (1, 0) (1, 1) (1, 2) (1, 3)

axes = (1, 0) # reversed order
# output: (0, 0) (1, 0) (2, 0) (3, 0) (0, 1) (1, 1) (2, 1) (3, 1)

If shapes具有三个维度，axes例如可能是(0, 1, 2) or (2, 0, 1)等等，所以这不是简单使用的问题reversed()。所以我写了一些代码来做到这一点，但似乎效率很低：

axes = (1, 0)

# transposed axes
tpaxes = [0]*len(axes)
for i in range(len(axes)):
    tpaxes[axes[i]] = i

for i in itertools.product(*(range(x) for x in shape)):
    # reorder the output of itertools.product
    x = (i[y] for y in tpaxes)
    print(tuple(x))

关于如何正确执行此操作有什么想法吗？

嗯，这其实是一本专门的手册product。它应该更快，因为轴仅重新排序一次：

def gen_chain(dest, size, idx, parent):
    # iterate over the axis once
    # then trigger the previous dimension to update
    # until everything is exhausted
    while True:
        if parent: next(parent) # StopIterator is propagated upwards

        for i in xrange(size):
            dest[idx] = i
            yield 

        if not parent: break

def prod(shape, axes):
    buf = [0] * len(shape)
    gen = None

    # EDIT: fixed the axes order to be compliant with the example in OP 
    for s, a in zip(shape, axes):
        # iterate over the axis and put to transposed
        gen = gen_chain(buf, s, a, gen)

    for _ in gen:
        yield tuple(buf)


print list(prod((2,4), (0,1)))
# [(0, 0), (0, 1), (0, 2), (0, 3), (1, 0), (1, 1), (1, 2), (1, 3)]
print list(prod((2,4), (1,0)))
# [(0, 0), (1, 0), (2, 0), (3, 0), (0, 1), (1, 1), (2, 1), (3, 1)]
print list(prod((4,3,2),(1,2,0)))
# [(0, 0, 0), (1, 0, 0), (0, 0, 1), (1, 0, 1), (0, 0, 2), (1, 0, 2), ...