Question

我正在尝试使用itertools产品解决问题。这是一个排列问题，但是值不是唯一的，例如：

list = [1,1,2,2,3,3]
#result should be
[1,2,1,2,3,3], [2,2,1,1,3,3], .....

我尝试使用set(itertools.permutations(list))，这是直接的答案，但是对于不同的值和较长的列表，它的处理时间太长了。我也尝试过x = itertools.product(set(list),repeat=len(list))，然后从不满足原始列表值计数的项中清除x（即，生成的列表必须具有两个1，两个2和两个3s），此解决方案是速度更快，但是此答案会引发大量的MemoryError，因为它尝试将输出存储在内存中，然后对其进行处理。

我也尝试遍历乘积结果（即for i in itertools.product(set(list),repeat=len(list))并选择要存储在哪些迭代中以及丢弃哪些迭代），该解决方案解决了内存错误问题，但几乎和第一个一样慢，其中代码可以运行几个小时。

有人对如何解决这个问题有什么建议吗？

Answer 1

这是解决问题的方法。取每个唯一的起始值，然后取值中的所有唯一组合，并删除起始值。这样可以避免生成任何重复项，而只存储每个位置使用的唯一值，因此在最坏的情况下，n个元素列表在内存中是n平方的。

导入itertools

def unique_permutations(values):
    if not values:
        yield []
        return
    seen = set()
    for idx, first in enumerate(values):
        if first in seen:
            continue
        seen.add(first)

        rest = values[:idx] + values[idx+1:]
        for perm in unique_permutations(rest):
            yield [first] + perm

values = [1, 1, 2, 2, 3, 3]
for perm in unique_permutations(values):
    print(perm)

输出：

[1, 1, 2, 2, 3, 3]
[1, 1, 2, 3, 2, 3]
[1, 1, 2, 3, 3, 2]
[1, 1, 3, 2, 2, 3]
[1, 1, 3, 2, 3, 2]
[1, 1, 3, 3, 2, 2]
[1, 2, 1, 2, 3, 3]
[1, 2, 1, 3, 2, 3]
[1, 2, 1, 3, 3, 2]
[1, 2, 2, 1, 3, 3]
[1, 2, 2, 3, 1, 3]
[1, 2, 2, 3, 3, 1]
[1, 2, 3, 1, 2, 3]
[1, 2, 3, 1, 3, 2]
[1, 2, 3, 2, 1, 3]
[1, 2, 3, 2, 3, 1]
[1, 2, 3, 3, 1, 2]
[1, 2, 3, 3, 2, 1]
[1, 3, 1, 2, 2, 3]
[1, 3, 1, 2, 3, 2]
[1, 3, 1, 3, 2, 2]
[1, 3, 2, 1, 2, 3]
[1, 3, 2, 1, 3, 2]
[1, 3, 2, 2, 1, 3]
[1, 3, 2, 2, 3, 1]
[1, 3, 2, 3, 1, 2]
[1, 3, 2, 3, 2, 1]
[1, 3, 3, 1, 2, 2]
[1, 3, 3, 2, 1, 2]
[1, 3, 3, 2, 2, 1]
[2, 1, 1, 2, 3, 3]
[2, 1, 1, 3, 2, 3]
[2, 1, 1, 3, 3, 2]
[2, 1, 2, 1, 3, 3]
[2, 1, 2, 3, 1, 3]
[2, 1, 2, 3, 3, 1]
[2, 1, 3, 1, 2, 3]
[2, 1, 3, 1, 3, 2]
[2, 1, 3, 2, 1, 3]
[2, 1, 3, 2, 3, 1]
[2, 1, 3, 3, 1, 2]
[2, 1, 3, 3, 2, 1]
[2, 2, 1, 1, 3, 3]
[2, 2, 1, 3, 1, 3]
[2, 2, 1, 3, 3, 1]
[2, 2, 3, 1, 1, 3]
[2, 2, 3, 1, 3, 1]
[2, 2, 3, 3, 1, 1]
[2, 3, 1, 1, 2, 3]
[2, 3, 1, 1, 3, 2]
[2, 3, 1, 2, 1, 3]
[2, 3, 1, 2, 3, 1]
[2, 3, 1, 3, 1, 2]
[2, 3, 1, 3, 2, 1]
[2, 3, 2, 1, 1, 3]
[2, 3, 2, 1, 3, 1]
[2, 3, 2, 3, 1, 1]
[2, 3, 3, 1, 1, 2]
[2, 3, 3, 1, 2, 1]
[2, 3, 3, 2, 1, 1]
[3, 1, 1, 2, 2, 3]
[3, 1, 1, 2, 3, 2]
[3, 1, 1, 3, 2, 2]
[3, 1, 2, 1, 2, 3]
[3, 1, 2, 1, 3, 2]
[3, 1, 2, 2, 1, 3]
[3, 1, 2, 2, 3, 1]
[3, 1, 2, 3, 1, 2]
[3, 1, 2, 3, 2, 1]
[3, 1, 3, 1, 2, 2]
[3, 1, 3, 2, 1, 2]
[3, 1, 3, 2, 2, 1]
[3, 2, 1, 1, 2, 3]
[3, 2, 1, 1, 3, 2]
[3, 2, 1, 2, 1, 3]
[3, 2, 1, 2, 3, 1]
[3, 2, 1, 3, 1, 2]
[3, 2, 1, 3, 2, 1]
[3, 2, 2, 1, 1, 3]
[3, 2, 2, 1, 3, 1]
[3, 2, 2, 3, 1, 1]
[3, 2, 3, 1, 1, 2]
[3, 2, 3, 1, 2, 1]
[3, 2, 3, 2, 1, 1]
[3, 3, 1, 1, 2, 2]
[3, 3, 1, 2, 1, 2]
[3, 3, 1, 2, 2, 1]
[3, 3, 2, 1, 1, 2]
[3, 3, 2, 1, 2, 1]
[3, 3, 2, 2, 1, 1]

Python非唯一置换处理

1 个答案: