python - 优化python中的递归

Question

我在优化 python 递归时遇到了问题。所以我想生成由 10 个元素组成的数组/列表中的所有可能性，每个元素都可以填充 0-9 数字。 所以，我决定在这种情况下使用递归，这里是:

routes = []
route = []

def generateRoutes(route, floor):
    if floor >= 10:
        routes.append(route)
    else:
        for channel in range(0, 10):
            new_route=route.copy()
            new_route.append(channel)
            generateRoutes(new_route, floor + 1)

generateRoutes(route, 0)

我的代码需要永恒才能完成任务(更不用说占用大量内存)。我的问题是，有没有办法解决/优化我的代码？ (除了递归，我也打开了其他方法)

编辑: 添加了有关函数如何调用的详细信息

Answer 1

itertools 模块已经提供了一个非递归的惰性解决方案:

>>> import itertools
>>> routes = itertools.product(range(10), repeat=10)

每个值都是在您遍历 routes 时按需生成的，而不是一次将所有 100 亿个值都存储在内存中。

>>> print(list(itertools.islice(routes, 20)))
[(0, 0, 0, 0, 0, 0, 0, 0, 0, 0), (0, 0, 0, 0, 0, 0, 0, 0, 0, 1), (0, 0, 0, 0, 0, 0, 0, 0, 0, 2), (0, 0, 0, 0, 0, 0, 0, 0, 0, 3), (0, 0, 0, 0, 0, 0, 0, 0, 0, 4), (0, 0, 0, 0, 0, 0, 0, 0, 0, 5), (0, 0, 0, 0, 0, 0, 0, 0, 0, 6), (0, 0, 0, 0, 0, 0, 0, 0, 0, 7), (0, 0, 0, 0, 0, 0, 0, 0, 0, 8), (0, 0, 0, 0, 0, 0, 0, 0, 0, 9), (0, 0, 0, 0, 0, 0, 0, 0, 1, 0), (0, 0, 0, 0, 0, 0, 0, 0, 1, 1), (0, 0, 0, 0, 0, 0, 0, 0, 1, 2), (0, 0, 0, 0, 0, 0, 0, 0, 1, 3), (0, 0, 0, 0, 0, 0, 0, 0, 1, 4), (0, 0, 0, 0, 0, 0, 0, 0, 1, 5), (0, 0, 0, 0, 0, 0, 0, 0, 1, 6), (0, 0, 0, 0, 0, 0, 0, 0, 1, 7), (0, 0, 0, 0, 0, 0, 0, 0, 1, 8), (0, 0, 0, 0, 0, 0, 0, 0, 1, 9)]
>>> print(list(itertools.islice(routes, 20)))
[(0, 0, 0, 0, 0, 0, 0, 0, 2, 0), (0, 0, 0, 0, 0, 0, 0, 0, 2, 1), (0, 0, 0, 0, 0, 0, 0, 0, 2, 2), (0, 0, 0, 0, 0, 0, 0, 0, 2, 3), (0, 0, 0, 0, 0, 0, 0, 0, 2, 4), (0, 0, 0, 0, 0, 0, 0, 0, 2, 5), (0, 0, 0, 0, 0, 0, 0, 0, 2, 6), (0, 0, 0, 0, 0, 0, 0, 0, 2, 7), (0, 0, 0, 0, 0, 0, 0, 0, 2, 8), (0, 0, 0, 0, 0, 0, 0, 0, 2, 9), (0, 0, 0, 0, 0, 0, 0, 0, 3, 0), (0, 0, 0, 0, 0, 0, 0, 0, 3, 1), (0, 0, 0, 0, 0, 0, 0, 0, 3, 2), (0, 0, 0, 0, 0, 0, 0, 0, 3, 3), (0, 0, 0, 0, 0, 0, 0, 0, 3, 4), (0, 0, 0, 0, 0, 0, 0, 0, 3, 5), (0, 0, 0, 0, 0, 0, 0, 0, 3, 6), (0, 0, 0, 0, 0, 0, 0, 0, 3, 7), (0, 0, 0, 0, 0, 0, 0, 0, 3, 8), (0, 0, 0, 0, 0, 0, 0, 0, 3, 9)]