Flatten the Array!

Question

In this challenge, your task is to create a program which takes in a nested array and returns a single-dimensional flattened array. For Example [10,20,[30,[40]],50] should output [10,20,30,40,50].

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Input

The input will be a nested array (eg. [10,20,[[[10]]]]). It will contain only Integers (both negative and positive), Strings and Arrays. You can take the input as function argument, STDIN or whatever suits your language. You can assume that the input array won't have an empty array.

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Output

The output will be a flatted single-dimensional array with the same elements of same type as in the nested array and in the SAME order.

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Test Cases

[10,20,30] -> [10,20,30]
[[10]] -> [10]
[["Hi"],[[10]]] -> ["Hi",10]
[[[20],["Hi"],"Hi",20]] -> [20,"Hi","Hi",20]
[[["[]"],"[]"]] -> ["[]","[]"]

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Feel free to ask for any clarification by using comments. This is code-golf, so the shortest code in bytes wins!

Note: If your language contains a built-in for this, then you must NOT use it.

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Edit

Please also include a link to a website where your code can be executed.

Answer 1

K, 3 bytes

,//

This is a fairly common idiom. "Join over converge".

try it here with oK.

How it works:

Join (,) fuses together atoms or lists to produce a list. Over (/) takes a verb (in this case join) and applies it between each element of a list, left to right. Thus, the compound ,/ will flatten all the top level elements of the list. The symbol / actually has different meanings depending on the valence (number of arguments) of the verb with which it is compounded. When we provide ,/ as the verb, the final / acts as "converge"- it repeatedly applies ,/ to the input until it stops changing. Some other languages call a feature like this a "fixed point combinator". By repeatedly fusing bottom level lists, you will eventually arrive at a single flat list, and none of the operations will perturb the order of elements. This seems to solve the problem.

Answer 2

JavaScript (ES6), 35 bytes

Inspired by @user81655's answer:

f=a=>a.map?[].concat(...a.map(f)):a

Answer 3

Mathematica, 16 14 bytes

{##&@@#&//@#}&

An unnamed function which takes and returns a list, e.g.:

{##&@@#&//@#}& @ {{{20}, {"Hi"}, "Hi", 20}}
(* {20, "Hi", "Hi", 20} *)

Explanation

Syntactic sugar party!

To understand how this works, note that every expression in Mathematica is either an atom (e.g. numbers, strings, symbols) or a compound expression of the form f[a, b, c, ...], where f, a, b, c are themselves arbitrary expressions. Here, f is called the head of the expression. Everything else on top of that is just syntactic sugar. E.g. {a, b, c} is just List[a, b, c].

We start with //@ which maps a functions over all levels of a list. For instance:

f //@ {{{20}, {"Hi"}, "Hi", 20}}
(* f[{f[{f[{f[20]}], f[{f["Hi"]}], f["Hi"], f[20]}]}] *)

Note that this maps f over atoms as well as compound expressions. What we're now looking for is a way to get rid of the list heads and keep everything else.

The Apply function is normally used to feed the elements of a list as separate arguments to a function, but its actual definition is more general and simply replaces the head of an expression. E.g. Apply[g, f[a, b]] gives g[a, b].

Now there's a special "head" called Sequence that simply vanishes. E.g. {a, Sequence[b, c], d} just evaluates to {a, b, c, d}. The idea for flattening the list is to replace the heads of all inner lists with Sequence so that they get splatted into their surrounding list. So what we want is to Apply the head Sequence to the lists. Conveniently if we Apply something to an atom, it just leaves the atom unchanged, so we don't have to distinguish between types of expressions at all.

Finally, there's one small issue: f is also applied to the outermost level, so that it also removes the outermost List, which we don't want. The shortest way to counter that is simply to wrap the result in a list again, such that the surrounding Sequence can safely vanish.

Note that there's neither Apply nor Sequence in the code. @@ is an operator form of Apply and ##& is a standard golfing trick to shorten the long built-in name Sequence. So ungolfing everything a bit, we get something like:

flatten[list_] := { MapAll[Apply[Sequence], list] }

For more details on how and why the ##& works, see the section on "Sequences of arguments" in my answer for the Mathematica tips.

Answer 4

Python 2, 43 bytes

f=lambda l:[l]*(l*0!=[])or sum(map(f,l),[])

On a list, recurses on the elements and concatenates the results. On a string or number, encases in a singleton list.

Unfortunately, Python 2's ordering for types int < list < string sandwiches list between the others, requiring two inequalities to check. So, instead, l*0 is checked against the empty list [], otherwise giving 0 or "".

Answer 5

Ruby, 43 42 34 bytes

Recursive solution. Now with exception handling! (might as well credit @akostadinov for inspiring the change though)

f=->a{a.map(&f).inject:+rescue[a]}

IDEOne link

Answer 6

JavaScript (ES6), 41 bytes

f=a=>[].concat(...a.map(v=>v.pop?f(v):v))

<textarea id="input" rows="6" cols="40">[[[20],["Hi"],"Hi",20]]</textarea><br /><button onclick="result.textContent=JSON.stringify(f(eval(input.value)))">Go</button><pre id="result"></pre>

Answer 7

Perl 6, 24 bytes

{gather {$_».&{.take}}}

Explanation:

{ # has $_ as an implicit parameter

  gather {

    $_\ # the parameter from the outer block
    »\  # for each single value in the structure
    .&( # call the following block as if it was a method
      { # this block has its own $_ for a parameter
        .take # call the .take method implicitly on $_
      }
    )
  }
}

Test:

#! /usr/bin/env perl6

use v6.c;
use Test;

my &flatten = {gather {$_».&{.take}}}

my @tests = (
  [10,20,30], [10,20,30],
  [[10,],], [10,],
  [["Hi",],[[10,],],], ["Hi",10],
  [[["[]",],"[]"],], ["[]","[]"],
);

plan @tests / 2;

for @tests -> $input, $expected {
  # is-deeply cares about the exact type of its inputs
  # so we have to coerce the Seq into an Array
  is-deeply flatten($input).Array, $expected, $input.perl;
}

1..4
ok 1 - $[10, 20, 30]
ok 2 - $[[10],]
ok 3 - $[["Hi"], [[10],]]
ok 4 - $[[["[]"], "[]"],]

Answer 8

Haskell, 43 bytes

data D a=L a|N[D a]
f(L x)=[x]
f(N l)=f=<<l

Haskell has neither nested lists with different depths of the sublists nor mixed types for the list elements. For nesting I define a custom data type D which is either a leaf L that holds some element or a node N which is a list of Ds. For the mixed elements I use the predefined data type Either which combines two types into one, here Either String Integer. The new type D and the flatten function f are fully polymorphic in the type of the leaf elements, so I don't have to take extra care of anything regarding Either.

Usage example: f (N[N[L(Right 20)], N[L(Left "Hi")], L(Left "Hi") , L(Right 20)]) -> [Right 20,Left "Hi",Left "Hi",Right 20].

Answer 9

JavaScript (Firefox 30-57), 43 bytes

f=a=>a.map?[for(b of a)for(c of f(b))c]:[a]

Just because I could even avoid using concat.

Answer 10

Pyth, 7 6 5 bytes

us+]Y

Try it online: Demonstration or Test Suite

But of course, there is also a build-in function, that handles the task in just 2 bytes: .n (Test Suite)

Answer 11

Perl, 34 29 bytes

Functions.

If needs to flatten to list like my @a = f(@a), 29 bytes:

sub f{map{ref()?f(@$_):$_}@_}

Test it on Ideone

If needs to flatten to array ref like my $a = f($a), 34 bytes:

sub f{[map{ref()?@{f(@$_)}:$_}@_]}

Test it on Ideone.

Perl 5.22.0+, 27 bytes

Thanks to hobbs.

If needs to flatten to list like my @a = f(@a), 27 bytes:

sub f{map{ref?f(@$_):$_}@_}

Test it on JDoodle

If needs to flatten to array ref like my $a = f($a), 32 bytes:

sub f{[map{ref?@{f(@$_)}:$_}@_]}

Test it on JDoodle.

Answer 12

Julia, 29 bytes

f(x,y=vcat(x...))=x==y?x:f(y)

This is recursive splatting into a concatenate function until a reaching a fix point. Example

julia> f([1,[2,[3,[4,[5,[6]]]]]])
6-element Array{Int64,1}:
 1
 2
 3
 4
 5
 6

Answer 13

Retina, 30 bytes

1>`("(\\.|[^"])+")|[][]
$1
$
]

Try it online! (The first line is only used to run multiple test cases at once.)

Retina has no concept of arrays, string literals or numbers, so I decided to go with a "common" input format of [...,...] style arrays and "-delimited strings, where \ can be used inside the strings to escape any character (in particular " and \ itself).

The program itself simply matches either a full string or a square bracket, and replaces them with $1 which keeps strings and removes square brackets. The limit 1> skips the first match so that we don't remove the leading [. However, this does remove the trailing ], so we add it back in in a separate stage.

Answer 14

Pyke, 11 bytes

.F~]+=])K~]

Try it here!

Explanation:

.F~]+=])    - Deep for loop
  ~]        -    contents of `]` ([] by default)
    +       -  ^+i
     =]     - `]` = ^
        K~] - Output value
        K   - Remove the output from the for loop
         ~] - Return the contents of `]`

Or 7 bytes after a bugfix

M?+]K~]

Try it here!

Explanation:

M?+]    - Deep map
 ?+]    -  `]` = `]`+i
    K~] - Output value
    K   - Remove the output from the for loop
     ~] - Return the contents of `]`

Or even 2 bytes if printing to stdout is allowed (This might come under built-ins)

M
<newline required>

Try it here!

This deeply applies the print_newline function to every non-sequence item in the input and recurses for sequence items.

Answer 15

Java (v8) 390 276 bytes

public static Object[] f(final Object[]a) {
    List<Object>r=new ArrayList<>();boolean t=false;int n=0;
    for(final Object p:a)
        if(t=p instanceof Object[]){for(final Object q:(Object[])p) r.add(q);}
        else r.add(p);
    return(t)?f(r.toArray()):r.toArray();
}  

Just for completeness and all that. :) Can't say Java's code-efficient.

Answer 16

ANSI C, 193 bytes

#define b break;
#define c case
#define p putch(_);
char f;main(_){switch(_){c 1:putch(91);b c 34:f^=1;p b c 91:f&&p b c 93:f&&p b c 10:c 13:putch(93);return;default:p}_=getch();main(_);}

:-/, any suggestions? Btw, I did try to find an online source to compile this but the WL is strict for this code to compile. It will work for VS and gcc otherwise.

Answer 17

JavaScript 20 bytes

a=>(a+[]).split(',')

The array + array is equal to array.toString

Answer 18

C#, 48 bytes

()=>{$"[{i.Replace("[","").Replace("]","")}]";};

Thought I'd post it also since nobody has given a C# solution yet. Suggestions welcome!

Answer 19

Python, 57 bytes

f=lambda a:sum([list==type(x)and f(x)or[x]for x in a],[])

Try it online: Python 2, Python 3

Thanks to Kevin Lau for the list==type(x) trick.

Answer 20

Ruby

there is builtin flatten method.

You can run here: http://www.tutorialspoint.com/execute_ruby_online.php

One 43 bytes, but thought to share:

f=->a{a.inject([]){|r,e|r+(f[e]rescue[e])}}

One 45 bytes that is more efficient than the previous and the other ruby answer:

f=->a{a.map{|e|Array===e ?f[e]:[e]}.inject:+}

here's benchmark:

require 'benchmark'
n=10^9
arr=[[[20],[[[[[[[[123]]]]]]]],"ads",[[[[[[[4]]]]]]],5,[[[[[[[[[[6]]]]]]]]]],7,8,[[[[[[[[[[9]]]]]]]]]],[[[[[[[[[[0]]]]]]]]]],[[[[[[[[[[[["Hi"]]]]]]]]]]]],[[[[[["Hi"]]]]]],[[[[[20]]]]]]]
Benchmark.bm do |x|
  x.report { f=->a{a.map(&f).inject:+rescue[a]}; f[arr] }
  x.report { f=->a{a.map{|e|e!=[*e]?[e]:f[e]}.inject:+}; f[arr] }
  x.report { f=->a{a.inject([]){|r,e|r+(f[e]rescue[e])}}; f[arr] }
  x.report { f=->a{a.map{|e|Array===e ?f[e]:[e]}.inject:+}; f[arr] }
end

result:

       user     system      total        real
   0.010000   0.000000   0.010000 (  0.000432)
   0.000000   0.000000   0.000000 (  0.000303)
   0.000000   0.000000   0.000000 (  0.000486)
   0.000000   0.000000   0.000000 (  0.000228)

Answer 21

Perl, 39 34 + 1 (-p flag) 35 bytes

Oneliner. Inspired by Martin Büttner.

#!perl -p
s/("(\\.|[^"])+"|]$|^\[)|[][]/$1/g

Test it on Ideone.

Answer 22

Clojure, 68 bytes

(def f #(if(some vector? %)(f(mapcat(fn[z](if(vector? z)z[z]))%))%))

mapcat first applies function to each element and then concats results. So every time it concats one 'nesting level' is lost. Concat does not work on not sequences so elements have to be wrapped into vector if they're not vector.

You can try it here: http://www.tryclj.com

(f [[[20],["Hi"],"Hi",20]])
(f [[["[]"],"[]"]])

Answer 23

PHP, 73 Bytes

<?array_walk_recursive($_GET,function($i)use(&$r){$r[]=$i;});print_r($r);

Answer 24

Attache, 14 bytes

{Reap[Sow@>_]}

Try it online!

Fortunately, Attache has a "vectorization" operator, which applies a function at the atoms of a list. In this case, all we need to do is to set up a reaper with Reap and Sow all atoms of the input _ with @>. I think it's quite elegant.

Alternatives

15 bytes: Fixpoint{`'^^_}

16 bytes: Fixpoint!&Concat

17 bytes: {q:=[]q&Push@>_q}

17 bytes: Fixpoint[&Concat]

Answer 25

Elixir, 74 bytes

def d(l)do l|>Stream.flat_map(fn x->if is_list(x)do d(x)else[x]end end)end

First Elixir answer, so can probably be golfed a bit.

Try it online.

Explanation:

def d(l)do l|>            # Recursive method taking a list as input:
  Stream.flat_map(fn x->  #  Map over each item `x` of the input-list:
    if is_list(x)do       #   If `x` is a list itself:
      d(x)                #    Do a recursive call with `x`
    else                  #   Else:
      [x]                 #    Simply leave `x` unchanged
    end                   #   End of the if-else statements
  end)                    #  End of the map
end                       # End of the recursive method

Of course, if builtins were allowed, this could have been 25 bytes instead:

fn(l)->List.flatten(l)end

Try it online.

Answer 26

Jelly, 4 bytes

;/ÐL

Try it online!

Explanation

;/   | Reduce by concatenation
  ÐL | Loop until no further changes

Built-in F would be one byte if allowed.

Answer 27

Wolfram Language (Mathematica), 13 bytes

#~Level~{-1}&

Try it online!

Un-golfed: F[x_] := Level[x, {-1}]

picks out the elements of the structure at the last level of its tree form. I'm not sure this counts as "avoiding the builtin" (which would be Flatten).

Answer 28

Brachylog, 6 bytes

ċ↰ᵐc|g

Try it online!

ċ         The input is a list,
    |     and the output is
   c      the concatenation of
 ↰ᵐ       the results of this predicate on the list's elements.
    |     If the input is not a list, then it
     g    wrapped in a list is the output.

Answer 29

Racket, 63 bytes

(define(f l)(apply append(map(λ(x)(if(list? x)(f x)`(,x)))l)))

Answer 30

JavaScript, 17 Bytes

a=>eval(`[${a}]`)

Finally, JavaScript's type conversions can be put to some good use! Please note that this will actually output an array, but string conversion (putting it into HTML) causes it to become a comma separated list.

If comma separated lists are acceptable output, then the following is valid:

7 Bytes

a=>""+a

NOTE: Snippet is broken for some reason

var subject = 
  a=>eval(`[${a}]`)

<input oninput="try {output.innerHTML = subject(this.value)} catch(e) {output.innerHTML='Invaild Input'}" />
<div id="output"></div>

Answer 31

Java 8 165 chars

import java.util.*;<T>T[]f(T[]a){List<T>l=new ArrayList<>();for(T e:a)if(e instanceof Object[])Collections.addAll(l,f((T[])e));else l.add(e);return(T[])l.toArray();}

Ungolfed into a class:

public class Q80096 {

    public static <T> T[] flatten(T[] array) {
        List<T> flattenedList = new ArrayList<>();
        for (T element : array)
            if (element instanceof Object[])
                 Collections.addAll(flattenedList, flatten((T[]) element));
            else
                flattenedList.add(element);
        return (T[]) flattenedList.toArray();
    }
}

This answer is based on Jeremy Harton's approach. I used it changed it in some places and created a more golf-like version.

Answer 32

Lua, 123 bytes

function f(a)o,t={},function(s)for _,v in next,s do if type(v)=="table"then t(v)else o[#o+1]=v end end end t(a)return o end

I'm sure this is not the best answer, but I thought I'd post it since nobody has given a Lua solution yet. Suggestions welcome!

Try it online!

Answer 33

PowerShell, 35 bytes

filter f{if($_.Rank){$_|f}else{$_}}

Try it online!

Rank - Returns the number of dimensions in the array. Rank = 0 if an object is not array.

Answer 34

C# (Visual C# Interactive Compiler), 72 bytes

T[]f<T>(T[]a)=>a.SelectMany(x=>x is T[]?f(x as T[]):new[]{x}).ToArray();

Try it online!

Recursive function that takes an object[] as input. Yes, there is a type parameter, so you could pass an array of any type... But it only works correctly if all nested arrays are of the same type. That is a requirement that I have added :)

It would be more flexible to get this to work with an input of type IList or Array.