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谁能向我解释一下如何在列表列表上执行自定义交叉？假设我有这样的候选人：

candidate = [[0,1,2,3][4,5,6,7,8][9,10,11]]

我知道如何为单个列表进行交叉。但是实际上如何为列表列表执行此操作？

以下是单个列表的交叉方法：

@crossover
def partially_matched_crossover(random, mom, dad, args):
    """Return the offspring of partially matched crossover on the candidates.

    This function performs partially matched crossover (PMX). This type of
    crossover assumes that candidates are composed of discrete values that
    are permutations of a given set (typically integers). It produces offspring
    that are themselves permutations of the set.

    .. Arguments:
       random -- the random number generator object
       mom -- the first parent candidate
       dad -- the second parent candidate
       args -- a dictionary of keyword arguments

    Optional keyword arguments in args:

    - *crossover_rate* -- the rate at which crossover is performed 
      (default 1.0)

    """
    crossover_rate = args.setdefault('crossover_rate', 1.0)
    if random.random() < crossover_rate:
        size = len(mom)
        points = random.sample(range(size), 2)
        x, y = min(points), max(points)
        bro = copy.copy(dad)
        bro[x:y+1] = mom[x:y+1]
        sis = copy.copy(mom)
        sis[x:y+1] = dad[x:y+1]
        for parent, child in zip([dad, mom], [bro, sis]):
            for i in range(x, y+1):
                if parent[i] not in child[x:y+1]:
                    spot = i
                    while x <= spot <= y:
                        print(child[spot])
                        spot = parent.index(child[spot])
                    child[spot] = parent[i]
        return [bro, sis]
    else:
        return [mom, dad]

以上来自基于 Python 的 inpyred 库。我正在尝试为车辆路线问题制定一个算法，其中上述列表是建议解决方案的一个示例。

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