2

假设我有:

chromosome_1 <- c('0010000001010000')

chromosome_2 <- c('0100000001001010')

如何实施步骤 3-5

  1. 评估
    • NC1 = 没有。1 的在chromosome_1
    • NC2 = 没有。1 的在chromosome_2
    • 米=min(NC1, NC2)
  2. 生成一个随机NC整数range(1, M)
  3. 从等位基因为“1”的基因中随机选择NC基因位置,chromosome_1并形成s1这些选定位置的一组索引。

    从等位基因为“1”的基因中随机选择NC基因位置,chromosome_2并形成s2这些选定位置的一组索引。

  4. s = union(s1, s2) 认为 s = 2, 3, 10, 15

  5. 对于每个i索引s

    交换染色体等位基因chromosome_1chromosome_2基因位置i

下图说明了结果:

在此处输入图像描述

我真的很感激任何帮助!

4

2 回答 2

1

您可以尝试使用GA 包

在手册(第 5 页)中,有一个示例。

ga(type = c("binary", "real-valued", "permutation"),
fitness, ...,
min, max, nBits,
population = gaControl(type)$population,
selection = gaControl(type)$selection,
crossover = gaControl(type)$crossover,
mutation = gaControl(type)$mutation,
popSize = 50,
pcrossover = 0.8,
pmutation = 0.1,
elitism = base::max(1, round(popSize*0.05)),
updatePop = FALSE,
postFitness = NULL,
maxiter = 100,
run = maxiter,
maxFitness = Inf,
names = NULL,
suggestions = NULL,
optim = FALSE,
optimArgs = list(method = "L-BFGS-B",
poptim = 0.05,
pressel = 0.5,
control = list(fnscale = -1, maxit = 100)),
keepBest = FALSE,
parallel = FALSE,
monitor = if(interactive())
{ if(is.RStudio()) gaMonitor else gaMonitor2 }
else FALSE,
seed = NULL)

例如,填充、选择、交叉、变异和监控算子分配新功能。在我的研究中,我使用了自己的突变和监控功能。例如;

myga <- ga(type = "binary",
fitness, ...,
min, max, nBits,
mutation = myMutationFunction
popSize = 50,
pcrossover = 0.8,
pmutation = 0.1,
maxiter = 100,
run = maxiter,
monitor = myMonitorFunction

myMutationFunction <- function (x) {
#...
}

myMonitorFunction <- function (x) {
#...
}

因此,您只需定义自己的函数并将函数名称赋予 ga 函数。为了便于参考,您可以查看默认功能。您可以在默认函数中看到必要的参数和返回值。

于 2017-01-21T10:41:49.750 回答
0

可能不是最简单的解决方案,但它有效

set.seed(12345)

## Step 1
a <- c(0,0,1,0,0,0,0,0,0,1,0,1,0,0,0,0)
b <- c(0,1,0,0,0,0,0,0,0,1,0,0,1,0,1,0)
m <- min(sum(a==1), sum(b==1))

## Step 2
random_int <- sample(1:m, 1)

## Step 3
random_a <- sample(which(a == 1), random_int)
random_b <- sample(which(b == 1), random_int)
#all <- sort(union(random_a, random_b))

## Step 4
## for demo purpose (assume it as the random output)
all <- c(2,3,10,15)     

temp_a <- a[all]
temp_b <- b[all]

## Step 5
##crossover
b[all] <- temp_a
a[all] <- temp_b

## Output
> a
 [1] 0 1 0 0 0 0 0 0 0 1 0 1 0 0 1 0
> b
 [1] 0 0 1 0 0 0 0 0 0 1 0 0 1 0 0 0
于 2017-02-02T05:40:22.587 回答