如果您担心实际速度，则一定要使用numpy（因为聪明的算法调整可能比使用数组操作获得的收益要小得多）。这是三个都计算相同功能的解决方案。该
is_pareto_efficient_dumb解决方案在大多数情况下较慢，但随着成本增加而变得更快，在许多点上，该
is_pareto_efficient_simple解决方案都比哑解决方案有效得多，并且最终
is_pareto_efficient函数可读性较差，但最快（所以所有这些都是帕累托高效的！）。
import numpy as np# Very slow for many datapoints. Fastest for many costs, most readabledef is_pareto_efficient_dumb(costs): """ Find the pareto-efficient points :param costs: An (n_points, n_costs) array :return: A (n_points, ) boolean array, indicating whether each point is Pareto efficient """ is_efficient = np.ones(costs.shape[0], dtype = bool) for i, c in enumerate(costs): is_efficient[i] = np.all(np.any(costs[:i]>c, axis=1)) and np.all(np.any(costs[i+1:]>c, axis=1)) return is_efficient# Fairly fast for many datapoints, less fast for many costs, somewhat readabledef is_pareto_efficient_simple(costs): """ Find the pareto-efficient points :param costs: An (n_points, n_costs) array :return: A (n_points, ) boolean array, indicating whether each point is Pareto efficient """ is_efficient = np.ones(costs.shape[0], dtype = bool) for i, c in enumerate(costs): if is_efficient[i]: is_efficient[is_efficient] = np.any(costs[is_efficient]<c, axis=1) # Keep any point with a lower cost is_efficient[i] = True # And keep self return is_efficient# Faster than is_pareto_efficient_simple, but less readable.def is_pareto_efficient(costs, return_mask = True): """ Find the pareto-efficient points :param costs: An (n_points, n_costs) array :param return_mask: True to return a mask :return: An array of indices of pareto-efficient points. If return_mask is True, this will be an (n_points, ) boolean array Otherwise it will be a (n_efficient_points, ) integer array of indices. """ is_efficient = np.arange(costs.shape[0]) n_points = costs.shape[0] next_point_index = 0 # Next index in the is_efficient array to search for while next_point_index<len(costs): nondominated_point_mask = np.any(costs<costs[next_point_index], axis=1) nondominated_point_mask[next_point_index] = True is_efficient = is_efficient[nondominated_point_mask] # Remove dominated points costs = costs[nondominated_point_mask] next_point_index = np.sum(nondominated_point_mask[:next_point_index])+1 if return_mask: is_efficient_mask = np.zeros(n_points, dtype = bool) is_efficient_mask[is_efficient] = True return is_efficient_mask else: return is_efficient
分析测试（使用从正态分布中得出的点）：
含10,000个样本，有2个成本：
is_pareto_efficient_dumb: Elapsed time is 1.586sis_pareto_efficient_simple: Elapsed time is 0.009653sis_pareto_efficient: Elapsed time is 0.005479s
拥有1,000,000个样本，有2个成本：
is_pareto_efficient_dumb: Really, really, slowis_pareto_efficient_simple: Elapsed time is 1.174sis_pareto_efficient: Elapsed time is 0.4033s
使用10,000个样本，需要15个费用：
is_pareto_efficient_dumb: Elapsed time is 4.019sis_pareto_efficient_simple: Elapsed time is 6.466sis_pareto_efficient: Elapsed time is 6.41s
请注意，如果您担心效率问题，可以通过预先对数据重新排序来进一步提高2倍左右的速度，请参见此处。
