这将以不同的方式划分它，但我认为这仍然是一个相当不错的列表结构:

chunk.2 <- function(x, n, force.number.of.groups = TRUE, len = length(x), groups = trunc(len/n), overflow = len%%n) { 
  if(force.number.of.groups) {
    f1 <- as.character(sort(rep(1:n, groups)))
    f <- as.character(c(f1, rep(n, overflow)))
  } else {
    f1 <- as.character(sort(rep(1:groups, n)))
    f <- as.character(c(f1, rep("overflow", overflow)))
  }
  
  g <- split(x, f)
  
  if(force.number.of.groups) {
    g.names <- names(g)
    g.names.ordered <- as.character(sort(as.numeric(g.names)))
  } else {
    g.names <- names(g[-length(g)])
    g.names.ordered <- as.character(sort(as.numeric(g.names)))
    g.names.ordered <- c(g.names.ordered, "overflow")
  }
  
  return(g[g.names.ordered])
}

这将给你以下，取决于你想要它的格式:

> x <- 1:10; n <- 3
> chunk.2(x, n, force.number.of.groups = FALSE)
$`1`
[1] 1 2 3

$`2`
[1] 4 5 6

$`3`
[1] 7 8 9

$overflow
[1] 10

> chunk.2(x, n, force.number.of.groups = TRUE)
$`1`
[1] 1 2 3

$`2`
[1] 4 5 6

$`3`
[1]  7  8  9 10

使用这些设置运行几个计时:

set.seed(42)
x <- rnorm(1:1e7)
n <- 3

然后我们得到以下结果:

> system.time(chunk(x, n)) # your function 
   user  system elapsed 
 29.500   0.620  30.125 

> system.time(chunk.2(x, n, force.number.of.groups = TRUE))
   user  system elapsed 
  5.360   0.300   5.663 

注意:将as.factor()更改为as.character()使我的函数速度提高了两倍。

如果你不喜欢split()并且你不介意NAs填充你的短尾巴:

chunk <- function(x, n) { if((length(x)%%n)==0) {return(matrix(x, nrow=n))} else {return(matrix(append(x, rep(NA, n-(length(x)%%n))), nrow=n))} }

返回矩阵([，1:ncol])的列是您正在寻找的机器人。

它分成大小为⌊n/k⌋+1或⌊n/k⌋的块，并且不使用O(n log n)排序。

get_chunk_id<-function(n, k){
    r <- n %% k
    s <- n %/% k
    i<-seq_len(n)
    1 + ifelse (i <= r * (s+1), (i-1) %/% (s+1), r + ((i - r * (s+1)-1) %/% s))
}

split(1:10, get_chunk_id(10,3))

这里还有另一个，允许你控制你想要的结果是否有序:

split_to_chunks <- function(x, n, keep.order=TRUE){
  if(keep.order){
    return(split(x, sort(rep(1:n, length.out = length(x)))))
  }else{
    return(split(x, rep(1:n, length.out = length(x))))
  }
}

split_to_chunks(x = 1:11, n = 3)
$`1`
[1] 1 2 3 4

$`2`
[1] 5 6 7 8

$`3`
[1]  9 10 11

split_to_chunks(x = 1:11, n = 3, keep.order=FALSE)

$`1`
[1]  1  4  7 10

$`2`
[1]  2  5  8 11

$`3`
[1] 3 6 9

chunk2 <- function(x,n) split(x, cut(seq_along(x), n, labels = FALSE)) 

简单的函数通过简单地使用索引来分割一个向量-不需要过于复杂

vsplit <- function(v, n) {
    l = length(v)
    r = l/n
    return(lapply(1:n, function(i) {
        s = max(1, round(r*(i-1))+1)
        e = min(l, round(r*i))
        return(v[s:e])
    }))
}