我有一个非常简单的组合问题.我有两个数组(a和b).数组a表示数组b中三个插槽之一可以采用的所有值.数组b中的每个槽可以具有1到5之间的值.其示例是[1,4,5].我想生成一个包含所有可能组合的数组(c).我想扩展大型数组的基本示例.

输入：

a = [1, 2, 3, 4, 5]
b = [1, 2, 3]

输出：

c = [[1, 1, 1], [1, 1, 2],[1, 1, 3], [1, 1, 4], [1, 1, 5],
     [1, 2, 1], [1, 2, 2],[1, 2, 3], [1, 2, 4], [1, 2, 5],
     [1, 3, 1], [1, 3, 2],[1, 3, 3], [1, 3, 4], [1, 3, 5],
     [1, 4, 1], [1, 4, 2],[1, 4, 3], [1, 4, 4], [1, 4, 5],
     [1, 5, 1], [1, 5, 2],[1, 5, 3], [1, 5, 4], [1, 5, 5],
     [2, 1, 1], [2, 1, 2],[2, 1, 3], [2, 1, 4], [2, 1, 5],
     [2, 2, 1], [2, 2, 2],[2, 2, 3], [2, 2, 4], [2, 2, 5],
     [2, 3, 1], [2, 3, 2],[2, 3, 3], [2, 3, 4], [2, 3, 5],
     [2, 4, 1], [2, 4, 2],[2, 4, 3], [2, 4, 4], [2, 4, 5],
     [2, 5, 1], [2, 5, 2],[2, 5, 3], [2, 5, 4], [2, 5, 5],
     [3, 1, 1], [3, 1, 2],[3, 1, 3], [3, 1, 4], [3, 1, 5],
     [3, 2, 1], [3, 2, 2],[3, 2, 3], [3, 2, 4], [3, 2, 5],
     [3, 3, 1], [3, 3, 2],[3, 3, 3], [3, 3, 4], [3, 3, 5],
     [3, 4, 1], [3, 4, 2],[3, 4, 3], [3, 4, 4], [3, 4, 5],
     [3, 5, 1], [3, 5, 2],[3, 5, 3], [3, 5, 4], [3, 5, 5],
     [4, 1, 1], [4, 1, 2],[4, 1, 3], [4, 1, 4], [4, 1, 5],
     [4, 2, 1], [4, 2, 2],[4, 2, 3], [4, 2, 4], [4, 2, 5],
     [4, 3, 1], [4, 3, 2],[4, 3, 3], [4, 3, 4], [4, 3, 5],
     [4, 4, 1], [4, 4, 2],[4, 4, 3], [4, 4, 4], [4, 4, 5],
     [5, 5, 1], [5, 5, 2],[5, 5, 3], [5, 5, 4], [5, 5, 5],
     [5, 1, 1], [5, 1, 2],[5, 1, 3], [5, 1, 4], [5, 1, 5],
     [5, 2, 1], [5, 2, 2],[5, 2, 3], [5, 2, 4], [5, 2, 5],
     [5, 3, 1], [5, 3, 2],[5, 3, 3], [5, 3, 4], [5, 3, 5],
     [5, 4, 1], [5, 4, 2],[5, 4, 3], [5, 4, 4], [5, 4, 5],
     [5, 5, 1], [5, 5, 2],[5, 5, 3], [5, 5, 4], [5, 5, 5]]

解决上述问题：

d = []
for i in range(len(a)):
    for j in range(len(a)):
        for k in range(len(a)):
            e = []
            e.append(i+1)
            e.append(j+1)
            e.append(k+1)
            d.append(e)

我正在寻找一种更通用的方式.一个可以容纳更大的数组(见下文),而不需要使用嵌套的for循环结构.我搜索了一个类似的例子,但无法在*上找到一个.

输入：

a = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20]
b = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20]

解决方法:

您正在寻找itertools.product().

a = [1, 2, 3, 4, 5]
b = 3  # Actually, you just need the length of the array, values do not matter

c = itertools.product(a, repeat=b)

请注意,这将返回一个迭代器,您可能需要使用list()强制转换它,但请注意,如果大小增加,这可能会永远消耗内存.