为了实施共享策略（描述为here），适应度功能需要依赖于人口中的其他个体。
def shared_fitness(individual, population):
#compute
return result
如何在遗传算法要使用的工具箱中注册此功能？为了做到这一点：
toolbox = base.Toolbox()
toolbox.register('evaluate', shared_fitness, population=pop)
我首先必须定义总体pop。但是在算法时代，我希望使用当前总体而不是初始总体来评估适应度。
我如何实现依赖于人口中其他个体的适应度功能？
请注意，在99％的时间内，我们的人口将是某种列表（或其他容器对象）。当我们将这些对象传递给函数时，我们传递的是指针而不是值。这意味着我们对总体所做的任何更改都会影响评估函数，该函数具有总体的指针。为了进行健全性测试，我使用了N-Queens example from DEAP，并对评估功能进行了小幅更改-仅打印了当前的前五名人口成员。运行此命令时，即使评估功能收到“初始填充”作为输入，您也可以看到输出发生了变化。
如果由于某种原因而不是通过指针传递总体而不是指针，也许始终包含当前总体的全局变量可能会有所帮助，尽管这当然不太理想。
# This file is part of DEAP.
#
# DEAP is free software: you can redistribute it and/or modify
# it under the terms of the GNU Lesser General Public License as
# published by the Free Software Foundation, either version 3 of
# the License, or (at your option) any later version.
#
# DEAP is distributed in the hope that it will be useful,
# but WITHOUT ANY WARRANTY; without even the implied warranty of
# MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
# GNU Lesser General Public License for more details.
#
# You should have received a copy of the GNU Lesser General Public
# License along with DEAP. If not, see <http://www.gnu.org/licenses/>.
import random
import numpy
from deap import algorithms
from deap import base
from deap import creator
from deap import tools
# Problem parameter
NB_QUEENS = 20
INDIV_COUNT = 0
def evalNQueens(individual, population):
global INDIV_COUNT
"""Evaluation function for the n-queens problem.
The problem is to determine a configuration of n queens
on a nxn chessboard such that no queen can be taken by
one another. In this version, each queens is assigned
to one column, and only one queen can be on each line.
The evaluation function therefore only counts the number
of conflicts along the diagonals.
"""
size = len(individual)
# Count the number of conflicts with other queens.
# The conflicts can only be diagonal, count on each diagonal line
left_diagonal = [0] * (2 * size - 1)
right_diagonal = [0] * (2 * size - 1)
# Sum the number of queens on each diagonal:
for i in range(size):
left_diagonal[i + individual[i]] += 1
right_diagonal[size - 1 - i + individual[i]] += 1
# Count the number of conflicts on each diagonal
sum_ = 0
for i in range(2 * size - 1):
if left_diagonal[i] > 1:
sum_ += left_diagonal[i] - 1
if right_diagonal[i] > 1:
sum_ += right_diagonal[i] - 1
if INDIV_COUNT % len(population) == 0:
print(f'top 5 individuals @ generation {int(INDIV_COUNT / 300)}: {population[:5]}')
INDIV_COUNT += 1
return sum_,
creator.create("FitnessMin", base.Fitness, weights=(-1.0,))
creator.create("Individual", list, fitness=creator.FitnessMin)
# Since there is only one queen per line,
# individual are represented by a permutation
toolbox = base.Toolbox()
toolbox.register("permutation", random.sample, range(NB_QUEENS), NB_QUEENS)
# Structure initializers
# An individual is a list that represents the position of each queen.
# Only the line is stored, the column is the index of the number in the list.
toolbox.register("individual", tools.initIterate, creator.Individual, toolbox.permutation)
toolbox.register("population", tools.initRepeat, list, toolbox.individual)
toolbox.register("mate", tools.cxPartialyMatched)
toolbox.register("mutate", tools.mutShuffleIndexes, indpb=2.0 / NB_QUEENS)
toolbox.register("select", tools.selTournament, tournsize=3)
def main(seed=0):
random.seed(seed)
pop = toolbox.population(n=300)
toolbox.register("evaluate", evalNQueens, population=pop)
hof = tools.HallOfFame(1)
stats = tools.Statistics(lambda ind: ind.fitness.values)
stats.register("Avg", numpy.mean)
stats.register("Std", numpy.std)
stats.register("Min", numpy.min)
stats.register("Max", numpy.max)
algorithms.eaSimple(pop, toolbox, cxpb=0.5, mutpb=0.2, ngen=100, stats=stats,
halloffame=hof, verbose=True)
return pop, stats, hof
if __name__ == "__main__":
main()