（我知道该问题的标题可能会误导您，但我找不到其他表达方式-随时进行编辑）
我有两个列表，两个列表的长度相同：
a = [1,2,3] b = [4,5,6]
我想用第二个列表替换所有可能的第一个列表。
output[0] = [1,2,3] # no replacements output[1] = [4,2,3] # first item was replaced output[2] = [1,5,3] # second item was replaced output[3] = [1,2,6] # third item was replaced output[4] = [4,5,3] # first and second items were replaced output[5] = [4,2,6] # first and third items were replaced output[6] = [1,5,6] # second and third items were replaced output[7] = [4,5,6] # all items were replaced
涉及先前链接的答案的可能解决方案是创建多个列表，然后对它们使用 itertools.product 方法。例如，我可以创建2个元素的3个列表，而不是2个元素的3个列表。但是，这会使代码过于复杂，如果可以的话，我宁愿避免这种情况。
有一种简单快捷的方法吗？
创建两个元素的3个列表根本不会使代码复杂化。zip可以轻松地“翻转”多个列表的轴（将Y元素的X序列转换为X元素的Y序列），使其易于使用itertools.product：
zip
itertools.product
import itertools a = [1,2,3] b = [4,5,6] # Unpacking result of zip(a, b) means you automatically pass # (1, 4), (2, 5), (3, 6) # as the arguments to itertools.product output = list(itertools.product(*zip(a, b))) print(*output, sep="\n")
哪个输出：
(1, 2, 3) (1, 2, 6) (1, 5, 3) (1, 5, 6) (4, 2, 3) (4, 2, 6) (4, 5, 3) (4, 5, 6)
与示例输出的排序不同，但是可能的替换集相同。
