用两个目标。最小成本和最大满意度用python写出NSGA2精英策略中的快速非支配排序函数和计算拥挤度函数dominatespopulationi populationj没写写出来

NSGA2是一种常用的多目标优化算法，其核心是快速非支配排序（Fast Non-dominated Sorting）和拥挤度计算（Crowding Distance Calculation）。下面是一个NSGA2的Python实现，包含快速非支配排序函数和拥挤度计算函数。

首先，我们定义一个目标函数类，包含两个目标：最小成本和最大满意度。

class ObjectiveFunction:
    def __init__(self, cost, satisfaction):
        self.cost = cost
        self.satisfaction = satisfaction

    def dominates(self, other):
        return self.cost <= other.cost and self.satisfaction >= other.satisfaction

其中，dominates函数判断当前目标函数是否支配另一个目标函数。

接下来，我们实现快速非支配排序函数：

def fast_non_dominated_sort(population):
    fronts = [[]]
    dominated = {}
    rank = {}

    for i, p in enumerate(population):
        dominated[i] = set()
        rank[i] = 0
        for j, q in enumerate(population):
            if p.dominates(q):
                dominated[i].add(j)
            elif q.dominates(p):
                rank[i] += 1
        if rank[i] == 0:
            fronts[0].append(i)

    i = 0
    while len(fronts[i]) > 0:
        next_front = []
        for p in fronts[i]:
            for q in dominated[p]:
                rank[q] -= 1
                if rank[q] == 0:
                    next_front.append(q)
        i += 1
        fronts.append(next_front)

    return fronts[:-1]

该函数将种群划分为多个前沿，每个前沿中的个体都是非支配的。该函数的输入为一个种群，输出为一个前沿列表，其中每个前沿都是一个个体的索引列表。

最后，我们实现拥挤度计算函数：

def crowding_distance(population, front):
    distances = [0] * len(front)

    for obj_func in ObjectiveFunction.__dict__.keys():
        if obj_func.startswith("__"):
            continue
        sorted_front = sorted(front, key=lambda p: getattr(population[p], obj_func))
        distances[0] = distances[-1] = float("inf")
        for i in range(1, len(front) - 1):
            distances[i] += getattr(population[sorted_front[i+1]], obj_func) - getattr(population[sorted_front[i-1]], obj_func)

    return distances

该函数计算每个个体的拥挤度。对于每个目标函数，将当前前沿按照该目标函数的值进行排序，然后计算每个个体与其相邻个体之间的距离之和作为拥挤度。

完整代码如下：

class ObjectiveFunction:
    def __init__(self, cost, satisfaction):
        self.cost = cost
        self.satisfaction = satisfaction

    def dominates(self, other):
        return self.cost <= other.cost and self.satisfaction >= other.satisfaction


def fast_non_dominated_sort(population):
    fronts = [[]]
    dominated = {}
    rank = {}

    for i, p in enumerate(population):
        dominated[i] = set()
        rank[i] = 0
        for j, q in enumerate(population):
            if p.dominates(q):
                dominated[i].add(j)
            elif q.dominates(p):
                rank[i] += 1
        if rank[i] == 0:
            fronts[0].append(i)

    i = 0
    while len(fronts[i]) > 0:
        next_front = []
        for p in fronts[i]:
            for q in dominated[p]:
                rank[q] -= 1
                if rank[q] == 0:
                    next_front.append(q)
        i += 1
        fronts.append(next_front)

    return fronts[:-1]


def crowding_distance(population, front):
    distances = [0] * len(front)

    for obj_func in ObjectiveFunction.__dict__.keys():
        if obj_func.startswith("__"):
            continue
        sorted_front = sorted(front, key=lambda p: getattr(population[p], obj_func))
        distances[0] = distances[-1] = float("inf")
        for i in range(1, len(front) - 1):
            distances[i] += getattr(population[sorted_front[i+1]], obj_func) - getattr(population[sorted_front[i-1]], obj_func)

    return distances
``