与Using a sample list as a template for sampling from a larger list without wraparound处的问题类似,我怎么知道这样可以实现环绕?
因此,如果我有一个字母矢量:
> 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"
然后我用字母定义参考样本如下:
> refSample <- c("j","l","m","s")
其中元素之间的间距是2(第1到第2),1(第2到第3)和第6(第3到第4),我怎样才能从{{1}中选择 n 样本它的元素与all
之间有相同的环绕间距?例如,refSample
,"a","c","d","j"
和"q" "s" "t" "z"
将是有效样本,但"r" "t" "u" "a"
则不会。
同样,为函数参数化是最好的。
答案 0 :(得分:3)
我会把它留作练习但是这里 -
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"