输出对象的类别因输入数据不同而不同 [英] Class of output object differs as input data differs

问题描述

我正在尝试为 n 次尝试中的每次尝试绘制可变数量的样本.在这个例子中 n = 8 因为 length(n.obs) == 8.绘制完所有样本后，我想将它们组合成一个 matrix.

I am trying to draw a variable number of samples for each of n attempts. In this example n = 8 because length(n.obs) == 8. Once all of the samples have been drawn I want to combine them into a matrix.

这是我的第一次尝试:

set.seed(1234)
n.obs <- c(2,1,2,2,2,2,2,2)
my.samples <- sapply(1:8, function(x) sample(1:4, size=n.obs[x], prob=c(0.1,0.2,0.3,0.4), replace=TRUE))
my.samples

这种方法产生一个list.

class(my.samples)
#[1] "list"

我使用:

max.len <- max(sapply(my.samples, length))
max.len
#[1] 2

输出 matrix 可以使用:

 corrected.list <- lapply(my.samples, function(x) {c(x, rep(NA, max.len - length(x)))})
 output.matrix <- do.call(rbind, corrected.list)
 output.matrix[is.na(output.matrix)] <- 0
 output.matrix
 #     [,1] [,2]
 #[1,]    4    3 
 #[2,]    3    0
 #[3,]    3    2
 #[4,]    3    4
 #[5,]    4    3
 #[6,]    3    3
 #[7,]    3    4
 #[8,]    1    4

上述方法似乎工作得很好，因为 n.obs 在 n.obs > 中包含多个值和至少一个 element1.但是，我希望代码足够灵活以处理以下每个 n.obs:

The above approach seems to work fine as along as n.obs includes multiple values and at least one element in n.obs > 1. However, I want the code to be flexible enough to handle each of the following n.obs:

上面的 sapply 语句返回一个 2 x 8 matrix 和以下 n.obs.

The above sapply statement returns a 2 x 8 matrix with the following n.obs.

set.seed(1234)
n.obs <- c(2,2,2,2,2,2,2,2)

上面的 sapply 语句返回一个带有以下 n.obs 的 integer.

The above sapply statement returns an integer with the following n.obs.

set.seed(3333)
n.obs <- c(1,1,1,1,1,1,1,1)

上面的 sapply 语句返回一个带有以下 n.obs 的 list.

The above sapply statement returns a list with the following n.obs.

n.obs <- c(0,0,0,0,0,0,0,0)

以下是上述三个 n.obs 中每一个的期望结果示例:

Here are example desired results for each of the above three n.obs:

desired.output <- matrix(c(4, 3,
                           3, 3,
                           2, 3,
                           4, 4,
                           3, 3,
                           3, 3,
                           4, 1,
                           4, 2), ncol = 2, byrow = TRUE)

desired.output <- matrix(c(2,
                           3,
                           4,
                           2,
                           3,
                           4,
                           4,
                           1), ncol = 1, byrow = TRUE)

desired.output <- matrix(c(0,
                           0,
                           0,
                           0,
                           0,
                           0,
                           0,
                           0), ncol = 1, byrow = TRUE)

如何概括代码，使其始终返回一个包含八行的 matrix，而不管用作输入的 n.obs 是什么?一种方法是使用一系列 if 语句来处理有问题的情况，但我认为可能有更简单、更有效的解决方案.

How can I generalize the code so that it always returns a matrix with eight rows regardless of the n.obs used as input? One way would be to use a series of if statements to handle problematic cases, but I thought there might be a simpler and more efficient solution.

推荐答案

我们可以写一个函数:

get_matrix <- function(n.obs) {

   nr <- length(n.obs)
   my.samples <- sapply(n.obs, function(x) 
                  sample(1:4, size=x, prob=c(0.1,0.2,0.3,0.4), replace=TRUE))
   max.len <- max(lengths(my.samples))
   mat <- matrix(c(sapply(my.samples, `[`, 1:max.len)), nrow = nr, byrow = TRUE)
   mat[is.na(mat)] <- 0
   mat
}

检查输出:

get_matrix(c(2,1,2,2,2,2,2,2))

#     [,1] [,2]
#[1,]    1    4
#[2,]    4    0
#[3,]    4    3
#[4,]    4    4
#[5,]    4    2
#[6,]    4    3
#[7,]    4    4
#[8,]    4    4

get_matrix(c(1,1,1,1,1,1,1,1))

#     [,1]
#[1,]    4
#[2,]    4
#[3,]    3
#[4,]    4
#[5,]    2
#[6,]    4
#[7,]    1
#[8,]    4

get_matrix(c(0,0,0,0,0,0,0,0))
#     [,1]
#[1,]    0
#[2,]    0
#[3,]    0
#[4,]    0
#[5,]    0
#[6,]    0
#[7,]    0
#[8,]    0

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