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
x <- c(NA, 4, 3, NA, NA, 2, 1, 1, NA ); y <- letters[x]; z <- factor(y)
gives examples with missing data, repeated values, that are not already sorted, and are in different classes (integer, character, factor).
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
$`4`
[1] 7 6 2 4 3 3 8 11 6 6 1 8 4
chunk2 <- function(x,n) split(x, cut(seq_along(x), n, labels = FALSE))
labels = FALSE
speed up twice, and using cut()
is 4 times faster than using ceiling(seq_along(x) / n
on my data.
split()
approaches. @verbarmour's answer below is the fastest overall. It is blazing fast because it doesn't have to work with factor, nor does it need to sort. That answer deserves a lot more upvotes.
A simplified version:
n = 3
split(x, sort(x%%n))
NB: This will only work on numeric vectors.
split(x, sort(1:length(x) %% n))
split()
is the slowest part of this code (because it calls as.factor
). So maybe consider using a data.frame and do something like data$group <- sort(1:length(data) %% n)
, then use the group column in the rest of your code.
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
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
x
, y
, or z
defined in this comment. In particular, it sorts the results, which may or may not be okay, depending on the application.
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.
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
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.
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
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.
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
Here's another variant.
NOTE: with this sample you're specifying the CHUNK SIZE in the second parameter
all chunks are uniform, except for the last; 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,|
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
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])
}))
}
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
Credit to @Sebastian for this function
chunk <- function(x,y){
split(x, factor(sort(rank(row.names(x))%%y)))
}
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.
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.
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.
no applicable method for 'extract_' applied to an object of class "c('integer', 'numeric')
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
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))
Success story sharing
n
chunks of equal size. This gets you an unknown number of chunks of sizen
. I had the same problem and used the solutions from @mathheadinclouds.n-chunks
I usedmax <- length(d)%/%n
. I used this with a vector of 31 strings and obtained a list of 3 vectors of 10 sentences and one of 1 sentence.