按照这个link(不是我自己的)中的方法来计算帕累托点，你可以这样做：def simple_cull(inputPoints, dominates):

paretoPoints = set()

candidateRowNr = 0

dominatedPoints = set()

while True:

candidateRow = inputPoints[candidateRowNr]

inputPoints.remove(candidateRow)

rowNr = 0

nonDominated = True

while len(inputPoints) != 0 and rowNr < len(inputPoints):

row = inputPoints[rowNr]

if dominates(candidateRow, row):

# If it is worse on all features remove the row from the array

inputPoints.remove(row)

dominatedPoints.add(tuple(row))

elif dominates(row, candidateRow):

nonDominated = False

dominatedPoints.add(tuple(candidateRow))

rowNr += 1

else:

rowNr += 1

if nonDominated:

# add the non-dominated point to the Pareto frontier

paretoPoints.add(tuple(candidateRow))

if len(inputPoints) == 0:

break

return paretoPoints, dominatedPoints

def dominates(row, candidateRow):

return sum([row[x] >= candidateRow[x] for x in range(len(row))]) == len(row)

import random

inputPoints = [[random.randint(70,100) for i in range(3)] for j in range(500)]

paretoPoints, dominatedPoints = simple_cull(inputPoints, dominates)

import matplotlib.pyplot as plt

from mpl_toolkits.mplot3d import Axes3D

fig = plt.figure()

ax = fig.add_subplot(111, projection='3d')

dp = np.array(list(dominatedPoints))

pp = np.array(list(paretoPoints))

print(pp.shape,dp.shape)

ax.scatter(dp[:,0],dp[:,1],dp[:,2])

ax.scatter(pp[:,0],pp[:,1],pp[:,2],color='red')

import matplotlib.tri as mtri

triang = mtri.Triangulation(pp[:,0],pp[:,1])

ax.plot_trisurf(triang,pp[:,2],color='red')

plt.show()

，您将注意到最后一部分正在对Pareto点应用三角剖分并将其打印为三角形曲面。结果如下(其中红色形状为帕累托前沿)：

编辑：另外，您可能想看看this(尽管它似乎是针对2D空间的)。在