忽略示例函数中的值或NA

Question

我在R中有一个矩阵,我想从每一行中取一个随机样本.我的一些数据是在NA中,但在采用随机样本时,我不希望NA作为采样的选项.我怎么做到这一点？

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此矩阵以输出另一个矩阵时,我得到了

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能够作为输出,并希望输出类似于:

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

Answer 1

可能有更好的方法,但样本似乎没有任何与NA相关的参数,所以我只是写了一个匿名函数来处理NA.

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

基本上做你想要的.如果你真的想要矩阵输出,你可以做到

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

编辑:您没有要求这个,但我提议的解决方案在某些情况下确实失败(主要是如果一行只包含NA.

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)