Question

我正在构建一个模拟，根据用户定义的参数随机为R中的子数组分配字符标签。

我的代码如下

K <- 2        ### Number of subarrays 
K1 <- c(1:3)  ### labels in first subarray
K2 <- c(4:5)  ### labels in second subarray

N <- 10 
Hstar <- 5 

perms <- 10  ### rows in each subarray

specs <- 1:N
specs1 <- 1:(N/2) ### specs in subarray 1
specs2 <- ((N/2) + 1):N ### specs in subarray 2

pop <- array(dim = c(c(perms, N/K), K)) ### population subarrays

haps <- as.character(1:Hstar) ### character labels

probs <- rep(1/Hstar, Hstar)  ### label probabilities


### 'for' loop to randomly populate 'pop' with 'haps' according to 'probs'

for(j in 1:perms){
    for(i in 1:K){
        if(i == 1){
            pop[j, specs, i] <- sample(haps, size = N, replace = TRUE, prob = probs)
    }
    else{
        pop[j, specs1,  1] <- sample(haps[K1], size = N/2, replace = TRUE, prob = probs[K1])
        pop[j, specs2,  2] <- sample(haps[K2], size = N/2, replace = TRUE, prob = probs[K2])    
    }
  }
}

我想要做的是填充（按行，而不是列）＆＃39; pops＆＃39;，它由两个子阵列组成，带有字符标签（＆＃39; haps＆＃39;）。具体地，子阵列1仅需要包含来自K1的标签，并且子阵列2必须仅包含来自K2的标签。＆＃39;弹出＆＃39;尺寸为10×5×2（子阵列1中的值为50，子阵列2中的值为50）。不幸的是，R抛出了错误

Error in `[<-`(`*tmp*`, j, specs, i, value = c("4", "1", "3", "4", "1",  : 
subscript out of bounds

当嵌套＆＃39; for＆＃39;循环运行，我似乎无法理解为什么。我认为它与specs，specs1，specs2有关。基本上，来自＆＃39; specs＆＃39;的值。分为＆＃39; specs1＆＃39;和＆＃39; specs2＆＃39;。但是，错误表明问题在于pop [j，specs，i]，但是由于K = 2，程序的这一部分不应该受到影响......但事实确实如此。

有关如何解决问题的任何想法，以便程序运行任何K值？

如果需要进一步澄清，请与我们联系。

Answer 1

让我划分零件上的错误。下面的行错误地指定了分配维度。我注意到有一些不一致，因为你试图逐行循环（10次迭代），每行有5个元素（5列）。我怀疑你要按列循环，所以它应该是perms=5。

为了描述这个问题，如果你按每个元素调试代码，你会看到pop[j, specs, i]。您正在尝试引用pop[ 1 , 1:10 , 1]，并且您的子阵列具有维度10x5，这意味着您必须切换到pop[,1,1]（您不需要指定1:10）就整个专栏而言。）

pop[j, specs, i] <- sample(haps, size = N, replace = TRUE, prob = probs)


sample(haps, size = N, replace = TRUE, prob = probs)
# [1] "3" "1" "4" "3" "2" "1" "1" "1" "2" "2"
pop[j, specs, i]
# Error in pop[j, specs, i] : subscript out of bounds
pop[specs, j, i]
# [1] "5" "2" "1" "4" "3" "5" "1" "5" "5" "2"

pop[, j, i] <- sample(haps, size = N, replace = TRUE, prob = probs)
#      [,1] [,2] [,3] [,4] [,5]
# [1,] "5"  NA   NA   NA   NA  
# [2,] "1"  NA   NA   NA   NA  
# [3,] "4"  NA   NA   NA   NA  
# [4,] "1"  NA   NA   NA   NA  
# [5,] "1"  NA   NA   NA   NA  
# [6,] "2"  NA   NA   NA   NA  
# [7,] "5"  NA   NA   NA   NA  
# [8,] "5"  NA   NA   NA   NA  
# [9,] "3"  NA   NA   NA   NA  
#[10,] "3"  NA   NA   NA   NA

同样的问题出现在else部分，在那里我可以看到同样的错误。下面正确一个

pop[specs1 , j,  2] <- sample(haps[K1], size = N/2, replace = TRUE, prob = probs[K1])
pop[specs2 , j,  2] <- sample(haps[K2], size = N/2, replace = TRUE, prob = probs[K2])

无论如何，有更好的方法来完成这项任务：

pop[,,1] <- 
  apply(
    pop[,,1], 2, 
    function(x) sample(haps, size = N, replace = TRUE, prob = probs) )

pop[specs1,,2] <- 
  apply(
    pop[specs1,,2], 2, function(x)
     sample(haps[K1], size = N/2, replace = TRUE, prob = probs[K1]) )

pop[specs2,,2] <- 
  apply(
    pop[specs2,,2], 2, function(x)
      sample(haps[K2], size = N/2, replace = TRUE, prob = probs[K2]) )

Answer 2

R语言在矢量化方面非常有效。您可以使用此功能来防止使用for循环。

为了使代码有效，我需要纠正一些错误：

specs指的是数组的第一维而不是第二维。
我认为specs1和specs2是指第二个子阵列（在您的示例中为i=2）。我修改了以后。

要填充数组，我会生成与您要填充的数组对应的大小样本。我使用了length和dim。数组由列填充，即第一列，每行，第二列等......

K <- 2        ### Number of subarrays 
K1 <- c(1:3)  ### labels in first subarray
K2 <- c(4:5)  ### labels in second subarray

N <- 10 
Hstar <- 5 

perms <- 10  ### rows in each subarray

specs <- 1:N
specs1 <- 1:(N/2) ### specs in subarray 1
specs2 <- ((N/2) + 1):N ### specs in subarray 2

pop <- array(dim = c(c(perms, N/K), K)) ### population subarrays
haps <- as.character(1:Hstar) ### character labels
probs <- rep(1/Hstar, Hstar)  ### label probabilities

pop[specs, , 1] <- sample(haps, size = length(specs) * dim(pop)[2], replace = TRUE, prob = probs)
pop[specs1, , 2] <- sample(haps[K1], size = length(specs1) * dim(pop)[2], replace = TRUE, prob = probs[K1])
pop[specs2, , 2] <- sample(haps[K2], size = length(specs2) * dim(pop)[2], replace = TRUE, prob = probs[K2])

pop
#> , , 1
#> 
#>       [,1] [,2] [,3] [,4] [,5]
#>  [1,] "4"  "3"  "2"  "3"  "2" 
#>  [2,] "5"  "4"  "3"  "1"  "4" 
#>  [3,] "1"  "3"  "4"  "3"  "5" 
#>  [4,] "3"  "3"  "5"  "5"  "3" 
#>  [5,] "2"  "4"  "3"  "4"  "4" 
#>  [6,] "3"  "3"  "2"  "4"  "1" 
#>  [7,] "5"  "1"  "4"  "4"  "1" 
#>  [8,] "4"  "3"  "2"  "3"  "2" 
#>  [9,] "3"  "2"  "3"  "3"  "1" 
#> [10,] "3"  "4"  "1"  "4"  "2" 
#> 
#> , , 2
#> 
#>       [,1] [,2] [,3] [,4] [,5]
#>  [1,] "3"  "3"  "2"  "1"  "3" 
#>  [2,] "2"  "2"  "2"  "2"  "2" 
#>  [3,] "2"  "2"  "2"  "2"  "1" 
#>  [4,] "2"  "3"  "2"  "3"  "1" 
#>  [5,] "1"  "2"  "2"  "3"  "2" 
#>  [6,] "5"  "5"  "5"  "4"  "5" 
#>  [7,] "4"  "5"  "4"  "5"  "5" 
#>  [8,] "5"  "5"  "4"  "5"  "5" 
#>  [9,] "4"  "5"  "5"  "4"  "4" 
#> [10,] "5"  "4"  "5"  "5"  "4"

我认为你建立在参数化的基础上允许使用任何K值。

在R中填充子阵列[＆lt; -` * tmp *＆＃39; ：下标超出范围

2 个答案: