Question

我有这样的问题：我需要找到不超过最大重量的最佳项目组合。对于这个问题，我使用了遗传算法。

这是我的数据

dataset <- data.frame(name = paste0("x",1:11),
                  Weight = c(2.14083022,7.32592911,0.50945094,4.94405846,12.02631340,14.59102403,0.07583312,0.36318323,10.64413370,3.54882187,1.79507759),
                  stringsAsFactors = F)

这是我的成本函数：

max_weight = 10
fitness_function <- function(x){
   current_weight <- x %*% dataset$Weight
          if ( current_weight > max_weight){
      return(0)
   } else {
      return( -1* current_weight)
   }
}

然后我尝试了两个包：genalg和GA

genalg

    ga_genalg <- rbga.bin(size = 11,
                          popSize = 100, 
                          mutationChance = .1,
                          evalFunc = fitness_function)

好的，结果是：

cat(summary(ga_genalg))
GA Settings
  Type                  = binary chromosome
  Population size       = 100
  Number of Generations = 100
  Elitism               = 20
  Mutation Chance       = 0.1

Search Domain
  Var 1 = [,]
  Var 0 = [,]

GA Results
  Best Solution : 0 1 1 0 0 0 0 1 0 0 1

我检查了最佳解决方案，看起来不错：

genalg_best_solution = c(0,1,1,0,0,0,0,1,0,0,1)
dataset$Weight %*% genalg_best_solution 
         [,1]
[1,] 9.993641

PS。有人知道如何在没有输入和正则表达式的情况下获得这个最佳解决方案向量吗？

GA

ga_GA <- ga(type = "binary", fitness = fitness_function, popSize = 100, pmutation = .1, nBits = 11)
ga_best_solution = ga_GA@solution 
dim(ga_best_solution)
[1] 73 11

解决方案是具有73行的矩阵。此外，ga_GA@bestSol还会返回list()

这个套餐中我最好的解决方案在哪里？或者我需要检查所有73行并找到最佳（我已经尝试并获得73个零）？

PPS。第二个问题解决方案：GA最大化功能和genalg最小化功能= /。 有人知道如何从genalg包中提取最佳解决方案吗？

Answer 1

这里有很多问题。我的观点是，GA可以为您提供更简单的输出：最佳解决方案和健身分数。

你是GA的最大化健康分数，而genalg最小化 - 我创建了第二个适应度函数，它没有返回健身值乘以-1。这导致两者都有相同的解决方案。

另外，我没有得到你为ga（）输出提供的尺寸。就我而言，这只是一行，包含11个二进制值：

library(GA)
library(genalg)

dataset <- data.frame(name = paste0("x",1:11),
  Weight = c(
    2.14083022,7.32592911,0.50945094,4.94405846,
    12.02631340,14.59102403,0.07583312,0.36318323,
    10.64413370,3.54882187,1.79507759
  ),
  stringsAsFactors = F
)

max_weight = 10


# genalg ------------------------------------------------------------------

# fitness function for genalg 
fitness_function <- function(x){
   current_weight <- x %*% dataset$Weight
   if ( current_weight > max_weight){
      return(0)
   } else {
      return(-current_weight)
   }
}


ga_genalg <- rbga.bin(size = 11,
  popSize = 100, 
  mutationChance = .1,
  evalFunc = fitness_function
)
tail(ga_genalg$best, 1) # best fitness
summary(ga_genalg, echo=TRUE)

plot(ga_genalg) # plot

# helper function from ?rbga.bin
monitor <- function(obj) {
    minEval = min(obj$evaluations);
    filter = obj$evaluations == minEval;
    bestObjectCount = sum(rep(1, obj$popSize)[filter]);
    # ok, deal with the situation that more than one object is best
    if (bestObjectCount > 1) {
        bestSolution = obj$population[filter,][1,];
    } else {
        bestSolution = obj$population[filter,];
    }
    outputBest = paste(obj$iter, " #selected=", sum(bestSolution),
                       " Best (Error=", minEval, "): ", sep="");
    for (var in 1:length(bestSolution)) {
        outputBest = paste(outputBest,
            bestSolution[var], " ",
            sep="");
    }
    outputBest = paste(outputBest, "\n", sep="");

    cat(outputBest);
}

monitor(ga_genalg)
# 100 #selected=4 Best (Error=-9.99364087): 0 1 1 0 0 0 0 1 0 0 1 



# GA ----------------------------------------------------------------------

# fitness function for GA (maximizes fitness)
fitness_function2 <- function(x){
   current_weight <- x %*% dataset$Weight
   if ( current_weight > max_weight){
      return(0)
   } else {
      return(current_weight)
   }
}

ga_GA <- ga(type = "binary", fitness = fitness_function2, popSize = 100, pmutation = .1, nBits = 11)
ga_GA@solution 
#     x1 x2 x3 x4 x5 x6 x7 x8 x9 x10 x11
# [1,]  0  1  1  0  0  0  0  1  0   0   1
dim(ga_best_solution)
# [1]  1 11

ga_GA@fitnessValue
# [1] 9.993641

R中的遗传算法

1 个答案: