Question

我是Python的新手，我写了一个函数：

def f(x, y, z):
    ret = []
    for i in range(x):
        for j in range(y):
            for k in range(z):
                ret.append((i, j, k))
    return ret


print f(2, 3, 4)

输出：

[(0, 0, 0), (0, 0, 1), (0, 0, 2), (0, 0, 3), (0, 1, 0), (0, 1, 1), (0, 1, 2), (0, 1, 3), (0, 2, 0), (0, 2, 1), (0, 2, 2), (0, 2, 3), (1, 0, 0), (1, 0, 1), (1, 0, 2), (1, 0, 3), (1, 1, 0), (1, 1, 1), (1, 1, 2), (1, 1, 3), (1, 2, 0), (1, 2, 1), (1, 2, 2), (1, 2, 3)]

但我对此并不满意，因为我认为必须缩短实施时间。

那么有人可以给我一些暗示吗？

Answer 1

您可以使用itertools.product，因为这基本上就是您所追求的Cartesian product

>>> from itertools import product
>>> list(product(range(2), range(3), range(4)))
[(0, 0, 0), (0, 0, 1), (0, 0, 2), (0, 0, 3), (0, 1, 0), (0, 1, 1), (0, 1, 2), (0, 1, 3), (0, 2, 0), (0, 2, 1), (0, 2, 2), (0, 2, 3), (1, 0, 0), (1, 0, 1), (1, 0, 2), (1, 0, 3), (1, 1, 0), (1, 1, 1), (1, 1, 2), (1, 1, 3), (1, 2, 0), (1, 2, 1), (1, 2, 2), (1, 2, 3)]

因此，您可以替换现有的功能

def f(x, y, z):
    return list(product(range(x), range(y), range(z)))

要删除必须输入range的次数，您可以接受单个列表参数，然后使用生成器表达式，例如

def f(l):
    return list(product(*(range(i) for i in l)))

那么你可以把它称为

>>> f([2,3,4])
[(0, 0, 0), (0, 0, 1), (0, 0, 2), (0, 0, 3), (0, 1, 0), (0, 1, 1), (0, 1, 2), (0, 1, 3), (0, 2, 0), (0, 2, 1), (0, 2, 2), (0, 2, 3), (1, 0, 0), (1, 0, 1), (1, 0, 2), (1, 0, 3), (1, 1, 0), (1, 1, 1), (1, 1, 2), (1, 1, 3), (1, 2, 0), (1, 2, 1), (1, 2, 2), (1, 2, 3)]

Answer 2

您可以使用list comprehension

 >>> [(i, j, k) for i in range(2) for j in range(3) for k in range(4)]
[(0, 0, 0), (0, 0, 1), (0, 0, 2), (0, 0, 3), (0, 1, 0), (0, 1, 1), (0, 1, 2), (0, 1, 3), (0, 2, 0), (0, 2, 1), (0, 2, 2), (0, 2, 3), (1, 0, 0), (1, 0, 1), (1, 0, 2), (1, 0, 3), (1, 1, 0), (1, 1, 1), (1, 1, 2), (1, 1, 3), (1, 2, 0), (1, 2, 1), (1, 2, 2), (1, 2, 3)]

您可以编写如下函数：

def f(x, y, z):
    return [(i, j, k) for i in range(x) for j in range(y) for k in range(z)]

Python：如何优雅地实现这一目标？

2 个答案: