类似于我的问题，即使用示例列表作为模板，以便在没有环绕的情况下从较大的列表中进行采样，我怎么知道这样做允许环绕？
因此，如果我有一个字母向量：
> all <- letters
> all
[1] "a" "b" "c" "d" "e" "f" "g" "h" "i" "j" "k" "l" "m" "n" "o" "p" "q" "r" "s" "t" "u" "v" "w" "x" "y" "z"
然后我从字母中定义一个参考样本，如下所示：
> refSample <- c("j","l","m","s")
其中元素之间的间距为 2（第 1 至 2）、1（第 2 至 3）和 6（第 3 至 4），那么我如何从元素之间具有相同环绕间距的all中选择 n 个样本以refSample？例如，"a","c","d","j"、"q" "s" "t" "z"和"r" "t" "u" "a"将是有效的样本，但"a","c","d","k"则不是。
同样，为函数参数化是最好的。
我会把它作为一个练习，但这里是——
all <- letters
refSample <- c("j","l","m","s")
pick_matches <- function(n, ref, full, wrap = FALSE) {
iref <- match(ref,full)
spaces <- diff(iref)
tot_space <- sum(spaces)
N <- length( full ) - 1
max_start <- N - tot_space*(1-wrap)
starts <- sample(0:max_start, n, replace = TRUE)
return( sapply( starts, function(s) full[ 1 + cumsum(c(s, spaces)) %% (N+1) ] ) )
}
> set.seed(1)
> pick_matches(5, refSample, all, wrap = FALSE)
[,1] [,2] [,3] [,4] [,5]
[1,] "e" "g" "j" "p" "d"
[2,] "g" "i" "l" "r" "f"
[3,] "h" "j" "m" "s" "g"
[4,] "n" "p" "s" "y" "m"
> pick_matches(5, refSample, all, wrap = TRUE)
[,1] [,2] [,3] [,4] [,5]
[1,] "x" "y" "r" "q" "b"
[2,] "z" "a" "t" "s" "d"
[3,] "a" "b" "u" "t" "e"
[4,] "g" "h" "a" "z" "k"