# Create chunks from an array

Your task is to write a program which given an array and a number, you need to split the array into chunks with size is number.

## Rules

Your program will receive an array A , as well as a positive integer n. The array should then be split into chunks of length n, if the length of the string isn't divisible by n any leftover at the end should be considered its own chunk.

• If n is greater than length of array A, you will need to return array A, for example: if n = 4 and array A = [1,2,3], you should return [1,2,3]

• The array can contain any type rather than number.

• You should not change order (or direction) of any item from left to right. For example if n = 2 and A= [1,2,3]. Any result rather than [[1,2],[3]] will be invalid.

## Test Cases

n   A               Output

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


This is , so you the shortest bytes of each language will be the winner.

• If n is greater than the length of A we need to return A‽ Are you sure you don't mean [A]? – Adám Mar 6 at 13:24
• @chaugiang I still think a too large n should return [A], e.g [[1,2,3]]. What if n is exactly the length of A? – Adám Mar 6 at 13:42
• @chaugiang Adam is correct imo. The return value should be consistent. – Jonah Mar 6 at 16:02
• @chaugiang Can n ever equal 1? – DJMcMayhem Mar 6 at 19:41
• In a strongly typed language, it's simply impossible to return A rather than [A] , which would exclude an awful lot of languages. – dfeuer Mar 6 at 22:48

# 05AB1E, 1 byte

ô


Builtins ftw. :)

# JavaScript (ES6), 36 bytes

Takes input as (n)(array).

n=>g=a=>a+a&&[a.splice(0,n),...g(a)]


Try it online!

### Commented

n =>                  // n = chunk size
g = a =>            // g = recursive function taking the array a[]
a + a             // if a[] is empty, stop recursion and return an empty string
&&                // otherwise, return an array made of:
[ a.splice(0, n), //   the next chunk
...g(a)         //   followed by the result of a recursive call

• Now that is a neat and clean solution, and I learned about recursive anonymous functions too! – Joe the Person Mar 7 at 17:44

# APL (Dyalog Unicode), 12 bytesSBCS

⊢⊂⍨(⍴⊢)⍴1↑⍨⊣


Big thanks to Adám for basically doing basically all the golfing (and for basically all the APL knowledge I have currently >_>).

### Explanation

 ⊂⍨           Partitioned enclose (commuted, i.e. left and right switched) - for each ⍵ in left, ⍺ in right, if ⍺ = 0, create a new sub-array, push ⍵ to latest sub-array
⊢             Right argument of entire expression
⍴      Reshape - Change size of right into dimensions specified by left
(⍴ )       Shape of (here, there is only one dimension - length)
⊢        Right argument of entire expression
↑⍨   Take (commuted) - takes ⍺ elements from left where ⍺ is right. Extra elements (zeroes here) are automatically added
1     1
⊣  Left argument of entire expression


### Execution

Arguments 2, 1 2 3 4 5 6 7. Note that APL arrays are of the form a b c, with optional surrounding parentheses.

           ⊣  2
1     1
↑⍨   1↑2 = 1 0
⊢        1 2 3 4 5 6 7
(⍴ )       ⍴1 2 3 4 5 6 7 = 7
⍴      7⍴1 0 = 1 0 1 0 1 0 1
⊢             1 2 3 4 5 6 7
⊂⍨           1 0 1 0 1 0 1⊂1 2 3 4 5 6 7 = (1 2)(3 4)(5 6)(7)


Try it online!

• Congratulations on your first APL answer. And nicely explained too! Here, have an APL pie: 🥧 – Adám Mar 6 at 13:08

# Python 3, 61 bytes

lambda A,n:[A,[A[x:x+n]for x in range(0,len(A),n)]][n<len(A)]


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Modifies Henry T's existing Python 3 solution to produce valid output for n >= len(A).
Posting as its own answer due to lack of commenting privileges.

# Prolog (SWI), 9084 61 bytes

### Code:

[]*_*[].
L*N*[P|R]:-length(P,N),append(P,T,L),T*N*R;P=L,R=[].


The input format might be a bit weird, but it is:

A * n * Result.


For example, for the input:

n = 2
A = [1, 2, 3, 4, 5, 6]

You would need to use [1, 2, 3, 4, 5, 6] * 2 * Result..

Try it online!

### Ungolfed version:

divide([], _, []).
divide(List, N, [Prefix | Result]) :-
length(Prefix, N), append(Prefix, Remaining, List), divide(Remaining, N, Result)
; Prefix = List, Result = [].


# PHP, 15 bytes

$f=array_chunk;  requires PHP 7. Call with $f(ARRAY, N).

• I don't think you're required to give another name to a builtin, so this just scores 11, doesn't it? – Neil Mar 6 at 16:12
• @Neil I thought that might be a forbidden loophole; but you may be right. – Titus Mar 8 at 15:22

{*.batch($_)}  Try it online! Curried function wrapping the batch built-in. # Clean, 54 bytes import StdEnv$n l=[l%(i,i+n-1)\\i<-[0,n..length l-1]]


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# Python 2, 39 bytes

i,j=input()
while j:print j[:i];j=j[i:]


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Assumes that 1 chunk per line is acceptable output.

• 36 bytes as a recursive lambda function – ovs Mar 6 at 14:45
• @ovs - Very nice and also different enough for you to post as your own answer if you wish. – ElPedro Mar 6 at 14:55

# Brainfuck, 71 bytes

,[>+>+<<-]>>>,[<[>.,<-]>>>++++[<++++++++>-]<.[-]<<<[<+>>+<-]<[->+<]>>>]


Dunno if this counts or not... input format:

<character whose ascii is n>AAAAAAAAAAAAA
For example, in the input:
1234567890123492034
n is 32 since the ASCII value of space is 32


Takes the input and puts in a space every time n characters pass

Explanation (no commas because that would break the program):

, take n
[>+>+<<-] copy into next two cells (destroys original)
>>>, take first of A into next cell
[ while that input exists
<[>.,<-] if n is nonzero output take next of A subtract one from n
>>>++++[<++++++++>-]<.[-]< n is zero so put a space
<<[<+>>+<-] copy the old n into surrounding cells
<[->+<] move from first cell to second
>>>] take input, do again

• Remove the spaces for 71 characters – MilkyWay90 Mar 7 at 1:31
• lol, I thought I removed all of them but I didn't notice those, thanks! – vityavv Mar 7 at 1:45
• Try reorganizing the cells such that the cells you use more are more accessible (for example, if the input cell (the one where you use , more) is used more it could be put an a cell which is easier to access than if it was placed in other cells) or use a bruteforcer. I am not skilled in golfing in BF so these suggestions may not be helpful. – MilkyWay90 Mar 7 at 3:06
• So far I have n n n A space as my cell setup, if you can think of a better way... – vityavv Mar 8 at 1:01
• Could A space n n n ... work (or space A n n n...)? – MilkyWay90 Mar 8 at 2:50

# Python 3, 46 chars

lambda A,n:[A[:n],*(f(A[n:],n)if A[n:]else[])]


-1 thanks to @Collin Phillips.

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# CJam, 3 bytes

{/}


This is an anonymous block that takes an array of numbers and a number from the stack, and replaces them by an array of arrays.

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# Brachylog, 2 bytes

ġ₎


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• I read the statement, This came to mind immediately =). #builtinsFTW – Kroppeb Mar 9 at 17:13

# Elixir, 16 bytes

Enum.chunk_every


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# Charcoal, 1 byte

⪪


Try it online! Charcoal's default I/O makes it difficult to demonstrate using anything except strings. If you want a full program that takes numeric lists and outputs formatted lists then this can be done as follows:

Ｅ⪪ＡＮ⪫ι,


Try it online! Link is to verbose version of code. Explanation:

  Ａ      Input array
⪪       Split into chunks of
Ｎ     Input number
Ｅ       Map over chunks
ι  Current chunk
⪫   Joined with
, Literal ,
Implicitly print each chunk on its own line


# C# (Visual C# Interactive Compiler), 7877 43 bytes

a=>b=>{int i=0;return a.GroupBy(_=>i++/b);}


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I think we should be able to just write int i; because 0 is the default of int. I let it to avoid the error: error CS0165: Use of unassigned local variable 'i'.

# F# (.NET Core), 15 bytes

Seq.chunkBySize


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Well F# has a builtin...

# J, 4 bytes

<\~-


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Takes the array as left arg and chunk size as right arg.

Uses a dyadic hook and the infix adverb with a negative argument, which does what we want by definition.

Note: The return type must be boxed because J only allows tables of equal sized items.

# Japt, 2 bytes

òV


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# PHP, 45 bytes

function f($a,$b){return array_chunk($a,$b);}


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• Would just array_chunk be a valid answer? – Arnauld Mar 6 at 12:19
• @Arnauld I dont know. Never golfed in php before although I use it at work. – Luis felipe De jesus Munoz Mar 6 at 12:21
• I'm not 100% sure either, but we can abuse the implicit conversion of undeclared variables to a string and do something like that. – Arnauld Mar 6 at 12:30
• (erratum: I meant undefined constants) – Arnauld Mar 6 at 12:38

# Java 10, 106 80 bytes

L->n->{for(int l=L.size(),i=0;i<l;)System.out.print(L.subList(i,(i+=n)<l?i:l));}


Prints the chunks without delimiter.

Try it online.

### 106 bytes:

L->n->{var r=new java.util.Stack();for(int l=L.size(),i=0;i<l;)r.add(L.subList(i,(i+=n)<l?i:l));return r;}


Actually returns a list of lists.

Try it online.

Explanation:

L->n->{                       // Method with List and integer parameters and List return-type
var r=new java.util.Stack();//  Create an empty List
for(int l=L.size(),         //  Determine the size of the input-List
i=0;i<l;)               //  Loop i in the range [0, size):
L.subList(i,            //    A sublist of the input-list in the range from i
Math.min(i+=n,l)));   //    to the minimum of: i + input-integer or the size
//    (and increase i by the input-integer at the same)
return r;}                  //  Return the List of Lists of integers as result


# K (oK), 10 bytes

{(0N,x)#y}


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# Ruby, 25 bytes

->n,a{[*a.each_slice(n)]}


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If we can return enumerators instead of arrays, then it becomes simply:

# Ruby, 21 bytes

->n,a{a.each_slice n}


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# PicoLisp, 75 74 bytes

(de f(n l)(if(>= n(length l))(list l)(cons(head n l)(f n(tail(- 0 n)l)))))


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# Coconut, 8 bytes

groupsof


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# V, 6 bytes

òÀf,r



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Hexdump:

00000000: f2c0 662c 720a                           ..f,r.


Explanation:

ò           " Until an error happens:
f         "   (f)ind the...
À          "     n'th...
,        "     ","
"   (If there are less than n commas after the cursor, throw an error)
r       "   Replace the char under the cursor with...
<cr>   "     A newline


# Clojure, 14 bytes

#(partition %)


builtins I guess

• Hi, welcome. The function should take two arguments: the array to be partitioned and the length of the chunk. Also what happens if the last chunk isn't "full" when using partition? – NikoNyrh Mar 7 at 17:05

import Data.Lists
chunksOf


Here's a more interesting version, with just a few more bytes (thanks to nimi for five bytes in each solution):

n![]=[]
n!x=take n x:n!drop n x


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• I think you can – aloisdg says Reinstate Monica Mar 6 at 22:27
• n!x=take n x:n!drop n x. Data.Lists provides also chunksOf. – nimi Mar 6 at 23:00

# PowerShell, 67 65 bytes

param($n,$a)$a|%{$b+=,$_ if($b.Count-ge$n){,$b;rv b}}
if($b){,$b}


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• You should be able to rv b (alias for Remove-Variable) instead of \$b=@() to save two bytes. – AdmBorkBork Mar 6 at 20:45

# Jelly, 1 byte

s


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While the printer makes it look like single-element splits are not wrapped into lists, they actually are.

• This night give a better output as to showing that single element arrays are still actually arrays. – Nick Kennedy Mar 6 at 13:33
• Er, is the downvote because I didn’t add @Nick Kennedy’s link? – Ven Mar 7 at 7:48
• certainly not from me – Nick Kennedy Mar 7 at 7:49