输出对象的类别因输入数据不同而不同

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

这篇关于输出对象的类别因输入数据不同而不同的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！