ř由索引

Question

創建的分割化矢量的列表有了這樣例子的代碼：

set.seed(11) 
x <- sample(letters) 
x 
[1] "h" "a" "m" "y" "b" "u" "v" "f" "p" "c" "q" "g" "x" "l" "i" "o" "e" "r" "t" "d" "z" "k" "s" "j" "w" "n" 

並設置有值的這種載體

y <- c(4, 13, 20) 

我想通過分裂x矢量向上y矢量作爲「切片」索引。並將結果分組爲列表。期望的輸出：

z <- list(c("h", "a", "m", "y"),c("b", "u", "v", "f", "p", "c", "q", "g", "x"), c("l", "i", "o", "e", "r", "t", "d"), c("z", "k", "s", "j", "w", "n")) 
z 
[[1]] 
[1] "h" "a" "m" "y" 

[[2]] 
[1] "b" "u" "v" "f" "p" "c" "q" "g" "x" 

[[3]] 
[1] "l" "i" "o" "e" "r" "t" "d" 

[[4]] 
[1] "z" "k" "s" "j" "w" "n" 

Answer 1

我們可以創建的「0」等於「x」的長度的矢量，使用y作爲數字索引與1替換元件在v1，cumsum結果，並使用該作爲分組矢量分裂「X」

v1 <- numeric(length(x)) 
v1[y+1] <- 1 
split(x,cumsum(v1)) 

或者，我們可以這樣做的結果cumsum從得到的分組矢量tabulate

split(x,cumsum(tabulate(y+1, length(x)))) 

或者使用match

split(x,cumsum(c(TRUE,!is.na(match(seq_along(x), y)[-length(x)])))) 

或者%in%

split(x,cumsum(seq_along(x) %in% (y+1))) 

Answer 2

爲了好玩，另闢蹊徑，開創分裂載體可以是使用cut：

split(x, cut(seq_along(x), c(-Inf, y, Inf))) 
# $`(-Inf,4]` 
# [1] "h" "a" "m" "y" 
# 
# $`(4,13]` 
# [1] "b" "u" "v" "f" "p" "c" "q" "g" "x" 
# 
# $`(13,20]` 
# [1] "l" "i" "o" "e" "r" "t" "d" 
# 
# $`(20, Inf]` 
# [1] "z" "k" "s" "j" "w" "n" 

它甚至可以告訴你哪個組數據屬於:-)

---

通過擴展，這也意味着findInterval將工作：

split(x, findInterval(seq_along(x), y+1)) 

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在這兩種情況下，我們在尋找從1倉中的值到所述輸入矢量「x」的長度落入，端點由「y」定義的地方。