Split a vector into chunks

Question

I have to split a vector into n chunks of equal size in R. I couldn't find any base function to do that. Also Google didn't get me anywhere. Here is what I came up with so far;

x <- 1:10
n <- 3
chunk <- function(x,n) split(x, factor(sort(rank(x)%%n)))
chunk(x,n)
$`0`
[1] 1 2 3

$`1`
[1] 4 5 6 7

$`2`
[1]  8  9 10

Answer 1

A one-liner splitting d into chunks of size 20:

split(d, ceiling(seq_along(d)/20))

More details: I think all you need is seq_along(), split() and ceiling():

> d <- rpois(73,5)
> d
 [1]  3  1 11  4  1  2  3  2  4 10 10  2  7  4  6  6  2  1  1  2  3  8  3 10  7  4
[27]  3  4  4  1  1  7  2  4  6  0  5  7  4  6  8  4  7 12  4  6  8  4  2  7  6  5
[53]  4  5  4  5  5  8  7  7  7  6  2  4  3  3  8 11  6  6  1  8  4
> max <- 20
> x <- seq_along(d)
> d1 <- split(d, ceiling(x/max))
> d1
$`1`
 [1]  3  1 11  4  1  2  3  2  4 10 10  2  7  4  6  6  2  1  1  2

$`2`
 [1]  3  8  3 10  7  4  3  4  4  1  1  7  2  4  6  0  5  7  4  6

$`3`
 [1]  8  4  7 12  4  6  8  4  2  7  6  5  4  5  4  5  5  8  7  7

$`4`
 [1]  7  6  2  4  3  3  8 11  6  6  1  8  4

Answer 2

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

Answer 3

A simplified version:

n = 3
split(x, sort(x%%n))

NB: This will only work on numeric vectors.

Answer 4

Using base R's rep_len:

x <- 1:10
n <- 3

split(x, rep_len(1:n, length(x)))
# $`1`
# [1]  1  4  7 10
# 
# $`2`
# [1] 2 5 8
# 
# $`3`
# [1] 3 6 9

And as already mentioned if you want sorted indices, simply:

split(x, sort(rep_len(1:n, length(x))))
# $`1`
# [1] 1 2 3 4
# 
# $`2`
# [1] 5 6 7
# 
# $`3`
# [1]  8  9 10

Answer 5

Try the ggplot2 function, cut_number:

library(ggplot2)
x <- 1:10
n <- 3
cut_number(x, n) # labels = FALSE if you just want an integer result
#>  [1] [1,4]  [1,4]  [1,4]  [1,4]  (4,7]  (4,7]  (4,7]  (7,10] (7,10] (7,10]
#> Levels: [1,4] (4,7] (7,10]

# if you want it split into a list:
split(x, cut_number(x, n))
#> $`[1,4]`
#> [1] 1 2 3 4
#> 
#> $`(4,7]`
#> [1] 5 6 7
#> 
#> $`(7,10]`
#> [1]  8  9 10

Answer 6

If you don't like split() and you don't like matrix() (with its dangling NAs), there's this:

chunk <- function(x, n) (mapply(function(a, b) (x[a:b]), seq.int(from=1, to=length(x), by=n), pmin(seq.int(from=1, to=length(x), by=n)+(n-1), length(x)), SIMPLIFY=FALSE))

Like split(), it returns a list, but it doesn't waste time or space with labels, so it may be more performant.

Answer 7

This will split it differently to what you have, but is still quite a nice list structure I think:

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])
}

Which will give you the following, depending on how you want it formatted:

> 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

Running a couple of timings using these settings:

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

Then we have the following results:

> 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 

Note: Changing as.factor() to as.character() made my function twice as fast.

Answer 8

A few more variants to the pile...

> x <- 1:10
> n <- 3

Note, that you don't need to use the factor function here, but you still want to sort o/w your first vector would be 1 2 3 10:

> chunk <- function(x, n) split(x, sort(rank(x) %% n))
> chunk(x,n)
$`0`
[1] 1 2 3
$`1`
[1] 4 5 6 7
$`2`
[1]  8  9 10

Or you can assign character indices, vice the numbers in left ticks above:

> my.chunk <- function(x, n) split(x, sort(rep(letters[1:n], each=n, len=length(x))))
> my.chunk(x, n)
$a
[1] 1 2 3 4
$b
[1] 5 6 7
$c
[1]  8  9 10

Or you can use plainword names stored in a vector. Note that using sort to get consecutive values in x alphabetizes the labels:

> my.other.chunk <- function(x, n) split(x, sort(rep(c("tom", "dick", "harry"), each=n, len=length(x))))
> my.other.chunk(x, n)
$dick
[1] 1 2 3
$harry
[1] 4 5 6
$tom
[1]  7  8  9 10

Answer 9

Yet another possibility is the splitIndices function from package parallel:

library(parallel)
splitIndices(20, 3)

Gives:

[[1]]
[1] 1 2 3 4 5 6 7

[[2]]
[1]  8  9 10 11 12 13

[[3]]
[1] 14 15 16 17 18 19 20

NB: this works only with numeric values though. If you want to split a character vector, you would need to do some indexing: lapply(splitIndices(20, 3), \(x) letters[1:20][x])

Answer 10

You could combine the split/cut, as suggested by mdsummer, with quantile to create even groups:

split(x,cut(x,quantile(x,(0:n)/n), include.lowest=TRUE, labels=FALSE))

This gives the same result for your example, but not for skewed variables.

Answer 11

split(x,matrix(1:n,n,length(x))[1:length(x)])

perhaps this is more clear, but the same idea:
split(x,rep(1:n, ceiling(length(x)/n),length.out = length(x)))

if you want it ordered,throw a sort around it

Answer 12

Here's another variant.

NOTE: with this sample you're specifying the CHUNK SIZE in the second parameter

1. all chunks are uniform, except for the last;
2. the last will at worst be smaller, never bigger than the chunk size.

chunk <- function(x,n)
{
    f <- sort(rep(1:(trunc(length(x)/n)+1),n))[1:length(x)]
    return(split(x,f))
}

#Test
n<-c(1,2,3,4,5,6,7,8,9,10,11)

c<-chunk(n,5)

q<-lapply(c, function(r) cat(r,sep=",",collapse="|") )
#output
1,2,3,4,5,|6,7,8,9,10,|11,|

Answer 13

I needed the same function and have read the previous solutions, however i also needed to have the unbalanced chunk to be at the end i.e if i have 10 elements to split them into vectors of 3 each, then my result should have vectors with 3,3,4 elements respectively. So i used the following (i left the code unoptimised for readability, otherwise no need to have many variables):

chunk <- function(x,n){
  numOfVectors <- floor(length(x)/n)
  elementsPerVector <- c(rep(n,numOfVectors-1),n+length(x) %% n)
  elemDistPerVector <- rep(1:numOfVectors,elementsPerVector)
  split(x,factor(elemDistPerVector))
}
set.seed(1)
x <- rnorm(10)
n <- 3
chunk(x,n)
$`1`
[1] -0.6264538  0.1836433 -0.8356286

$`2`
[1]  1.5952808  0.3295078 -0.8204684

$`3`
[1]  0.4874291  0.7383247  0.5757814 -0.3053884

Answer 14

Simple function for splitting a vector by simply using indexes - no need to over complicate this

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])
    }))
}

Answer 15

Sorry if this answer comes so late, but maybe it can be useful for someone else. Actually there is a very useful solution to this problem, explained at the end of ?split.

> testVector <- c(1:10) #I want to divide it into 5 parts
> VectorList <- split(testVector, 1:5)
> VectorList
$`1`
[1] 1 6

$`2`
[1] 2 7

$`3`
[1] 3 8

$`4`
[1] 4 9

$`5`
[1]  5 10

Answer 16

Credit to @Sebastian for this function

chunk <- function(x,y){
         split(x, factor(sort(rank(row.names(x))%%y)))
         }

Answer 17

If you don't like split() and you don't mind NAs padding out your short tail:

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))} }

The columns of the returned matrix ([,1:ncol]) are the droids you are looking for.

Answer 18

I need a function that takes the argument of a data.table (in quotes) and another argument that is the upper limit on the number of rows in the subsets of that original data.table. This function produces whatever number of data.tables that upper limit allows for:

library(data.table)    
split_dt <- function(x,y) 
    {
    for(i in seq(from=1,to=nrow(get(x)),by=y)) 
        {df_ <<- get(x)[i:(i + y)];
            assign(paste0("df_",i),df_,inherits=TRUE)}
    rm(df_,inherits=TRUE)
    }

This function gives me a series of data.tables named df_[number] with the starting row from the original data.table in the name. The last data.table can be short and filled with NAs so you have to subset that back to whatever data is left. This type of function is useful because certain GIS software have limits on how many address pins you can import, for example. So slicing up data.tables into smaller chunks may not be recommended, but it may not be avoidable.

Answer 19

I have come up with this solution:

require(magrittr)
create.chunks <- function(x, elements.per.chunk){
    # plain R version
    # split(x, rep(seq_along(x), each = elements.per.chunk)[seq_along(x)])
    # magrittr version - because that's what people use now
    x %>% seq_along %>% rep(., each = elements.per.chunk) %>% extract(seq_along(x)) %>% split(x, .) 
}
create.chunks(letters[1:10], 3)
$`1`
[1] "a" "b" "c"

$`2`
[1] "d" "e" "f"

$`3`
[1] "g" "h" "i"

$`4`
[1] "j"

The key is to use the seq(each = chunk.size) parameter so make it work. Using seq_along acts like rank(x) in my previous solution, but is actually able to produce the correct result with duplicated entries.

Answer 20

As there were many good answers, I benchmarked the solutions that worked properly:

Below, the code:

require(microbenchmark)
require(ggplot2)
require(parallel)

x <- rpois(75,5)
result <- microbenchmark(times = 1000,
                         "Harlan" = split(x, ceiling(seq_along(x)/(length(x)/5))),
                         "mathheadinclouds" = split(x, cut(seq_along(x), 5, labels = FALSE)),
                         "Richard DiSalvo on zhan2383" = split(x, sort(1:length(x) %% 5)),
                         "FXQuantTrader" = split(x, sort(rep_len(1:5, length(x)))),
                         "verbamour" = {mapply(function(a, b) (x[a:b]), seq.int(from=1, to=length(x), by=floor(length(x)/5)),
                                               pmin(seq.int(from=1, to=length(x), by=floor(length(x)/5))+floor((length(x)/5-1)), length(x)),
                                               SIMPLIFY=FALSE)},
                         "Tony Breyal" = chunk.2(x, 5),
                         "Richard Herron"  = split(x, sort(rep(letters[1:5], each=5, len=length(x)))),
                         "Philip Michaelsen" = vsplit(x, 5),
                         "Iyar Lin" =  split_to_chunks(x, 5)
)

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))))
  }
}

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])
  }))
}

Answer 21

Here's yet another one, allowing you to control if you want the result ordered or not:

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

Answer 22

Not sure if this answers OP's question, but I think the %% can be useful here

df # some data.frame
N_CHUNKS <- 10
I_VEC <- 1:nrow(df)
df_split <- split(df, sort(I_VEC %% N_CHUNKS))

Answer 23

This splits into chunks of size ⌊n/k⌋+1 or ⌊n/k⌋ and does not use the O(n log n) sort.

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))