忽略示例函数中的值或 NA

问题描述

我在 R 中有一个矩阵，我想从每一行中抽取一个随机样本.我的一些数据在 NA 中，但是在进行随机样本时，我不希望 NA 成为采样的一个选项.我将如何做到这一点?

I have a matrix in R that I would like to take a single random sample from each row. Some of my data is in NA, but when taking the random sample I do not want the NA to be an option for the sampling. How would I accomplish this?

例如

a <- matrix (c(rep(5, 10), rep(10, 10), rep(NA, 5)), ncol=5, nrow=5)
a
     [,1] [,2] [,3] [,4] [,5]
[1,]    5    5   10   10   NA
[2,]    5    5   10   10   NA
[3,]    5    5   10   10   NA
[4,]    5    5   10   10   NA
[5,]    5    5   10   10   NA

当我将样本函数应用于这个矩阵以输出另一个矩阵时，我得到

When I apply the sample function to this matrix to output another matrix I get

b <- matrix(apply(a, 1, sample, size=1), ncol=1)
b

     [,1]
[1,]   NA
[2,]   NA
[3,]   10
[4,]   10
[5,]    5

相反，我不希望 NA 能够成为输出并希望输出类似于:

Instead I do not want the NA to be capable of being the output and want the output to be something like:

b
     [,1]
[1,]   10
[2,]   10
[3,]   10
[4,]    5
[5,]   10

推荐答案

可能有更好的方法，但示例似乎没有任何与 NA 相关的参数，因此我只是编写了一个匿名函数来处理 NA.

There might be a better way but sample doesn't appear to have any parameters related to NAs so instead I just wrote an anonymous function to deal with the NAs.

apply(a, 1, function(x){sample(x[!is.na(x)], size = 1)})

基本上做你想做的事.如果你真的想要矩阵输出，你可以做

essentially does what you want. If you really want the matrix output you could do

b <- matrix(apply(a, 1, function(x){sample(x[!is.na(x)], size = 1)}), ncol = 1)

您没有要求这样做，但我提出的解决方案在某些情况下确实会失败(主要是如果一行仅包含 NA.

---

You didn't ask for this but my proposed solution does fail in certain cases (mainly if a row contains ONLY NAs.

a <- matrix (c(rep(5, 10), rep(10, 10), rep(NA, 5)), ncol=5, nrow=5)
# My solution works fine with your example data
apply(a, 1, function(x){sample(x[!is.na(x)], size = 1)})

# What happens if a row contains only NAs
a[1,] <- NA

# Now it doesn't work
apply(a, 1, function(x){sample(x[!is.na(x)], size = 1)})

# We can rewrite the function to deal with that case
mysample <- function(x, ...){
    if(all(is.na(x))){
        return(NA)
    }
    return(sample(x[!is.na(x)], ...))
}

# Using the new function things work.
apply(a, 1, mysample, size = 1)

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