如何在 Python 中使用 DEAP 绘制多目标帕累托边界
How to plot multi-objectives pareto frontier with DEAP in Python
对于我正在从事的项目,我在 DEAP 中设置了 3 个不同的目标作为优化目标,an evolution framework based on Python。
使用NSGA-II等算法可以解决多目标问题。有没有办法生成pareto前沿面来可视化结果呢
按照这个 link 中的一个方法(不是我自己的)来计算 Pareto 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()
,你会注意到最后一部分是对 Pareto 点应用三角剖分并将其绘制为三角形表面。结果是这样的(红色形状是帕累托前沿):
编辑:你也可能想看看this(虽然它似乎是二维空间)。
您可能还想看看这个 Link,它实现了一种更有效的方法来通过块嵌套循环 (BNL) 求解二维帕累托边界。这比上面的暴力法快30倍。
对于我正在从事的项目,我在 DEAP 中设置了 3 个不同的目标作为优化目标,an evolution framework based on Python。
使用NSGA-II等算法可以解决多目标问题。有没有办法生成pareto前沿面来可视化结果呢
按照这个 link 中的一个方法(不是我自己的)来计算 Pareto 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()
,你会注意到最后一部分是对 Pareto 点应用三角剖分并将其绘制为三角形表面。结果是这样的(红色形状是帕累托前沿):
编辑:你也可能想看看this(虽然它似乎是二维空间)。
您可能还想看看这个 Link,它实现了一种更有效的方法来通过块嵌套循环 (BNL) 求解二维帕累托边界。这比上面的暴力法快30倍。