36
\$\begingroup\$

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].


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.


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.


Test Cases

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


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

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


Edit

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

\$\endgroup\$
17
  • 7
    \$\begingroup\$ Some languages treat strings as arrays, is [["Hi"],[[10]]] -> ["H","i",10] ok? \$\endgroup\$
    – Adám
    May 18, 2016 at 5:30
  • 4
    \$\begingroup\$ @Mego I was surprised too to find out that there was an unflatten question but no flatten question on PPCG. \$\endgroup\$
    – Arjun
    May 18, 2016 at 6:13
  • 3
    \$\begingroup\$ What if your language only supports subarrays of the same size? (E.g. Java?) What if the type of each element must be the same? (E.g. Java, C++ etc.?) Also, please add e.g. ["[",[["[",],'[',"['['"]] as a test case. \$\endgroup\$
    – flawr
    May 18, 2016 at 11:30
  • 4
    \$\begingroup\$ @flawr That test case only makes sense for languages that support bot ' and " as delimiters. (But I agree that a test case involving [, ], " and \ inside a string would be useful.) \$\endgroup\$ May 18, 2016 at 11:40
  • 5
    \$\begingroup\$ The test cases also exclude languages which do not support these kinds of arrays with multiple types, or with another notation for array literals. \$\endgroup\$
    – flawr
    May 18, 2016 at 11:42

36 Answers 36

42
\$\begingroup\$

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.

\$\endgroup\$
6
  • 1
    \$\begingroup\$ All right, thanks for the explanation! Have your well-earned +1. \$\endgroup\$
    – Value Ink
    May 18, 2016 at 5:52
  • \$\begingroup\$ a similar solution to a different problem. \$\endgroup\$
    – JohnE
    May 18, 2016 at 5:53
  • 1
    \$\begingroup\$ I came up with the same algorithm (but not in this language). +1 for picking the right language to implement it in! \$\endgroup\$
    – Cyoce
    May 18, 2016 at 5:56
  • \$\begingroup\$ @Cyoce If your language has equivalents to the three operators used here, it's an extremely natural solution. By all means post your variation. \$\endgroup\$
    – JohnE
    May 18, 2016 at 5:58
  • 1
    \$\begingroup\$ @JohnE Long story, I'm deriving a language from algorithms I come up with, so the language isn't finished (and thus implemented) yet. \$\endgroup\$
    – Cyoce
    May 18, 2016 at 6:02
39
\$\begingroup\$

JavaScript (ES6), 35 bytes

Inspired by @user81655's answer:

f=a=>a.map?[].concat(...a.map(f)):a
\$\endgroup\$
7
  • 3
    \$\begingroup\$ Very clever! +1 for [ab]using JS's weird way of handling missing keys! \$\endgroup\$
    – Cyoce
    May 18, 2016 at 7:03
  • \$\begingroup\$ I can beat that. \$\endgroup\$ May 19, 2016 at 16:37
  • \$\begingroup\$ @BaldBantha: We're looking forward for your answer :-) \$\endgroup\$
    – Bergi
    May 19, 2016 at 16:39
  • 3
    \$\begingroup\$ Crap NVM My 33 byte solution fails on one of the test cases. NOOOO \$\endgroup\$ May 19, 2016 at 16:39
  • 2
    \$\begingroup\$ @BaldBantha, join-split will fail on commas inside of strings. \$\endgroup\$
    – Qwertiy
    May 20, 2016 at 1:18
19
\$\begingroup\$

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.

\$\endgroup\$
3
  • \$\begingroup\$ First time I've seen //@. Very useful to know about! \$\endgroup\$
    – DavidC
    May 18, 2016 at 15:03
  • \$\begingroup\$ //@ captures a neat pattern. Reminds me a bit of some of the recursive combinators in Joy. Do you have a link to a good reference to any related functions in Mathematica? I'm very interested in ways of factoring explicit recursion out of programs. \$\endgroup\$
    – JohnE
    May 18, 2016 at 16:15
  • 1
    \$\begingroup\$ @JohnE Well, here's the docs. You could also look at things like Map, MapAt, Apply, as well as Replace and related functions. In general though there's a lot of function which take an optional levelspec parameter (see my original 16-byte solution), which lets you apply the function at multiple/all levels at once. \$\endgroup\$ May 18, 2016 at 16:25
13
\$\begingroup\$

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 "".

\$\endgroup\$
1
  • 1
    \$\begingroup\$ f=lambda s:[s]["">s>f:]or sum(map(f,s),[]) saves 1. \$\endgroup\$ Jun 6 at 9:14
10
\$\begingroup\$

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

\$\endgroup\$
5
  • \$\begingroup\$ kudos for shortness, awesome \$\endgroup\$ May 18, 2016 at 21:03
  • \$\begingroup\$ I didn't know you could use rescue like that \$\endgroup\$
    – Cyoce
    May 19, 2016 at 1:47
  • 1
    \$\begingroup\$ @Cyoce I think it's because Ruby technically doesn't have a try block, so you use begin instead to differentiate the parts you want to be catching for and the parts that you don't. So since you're catching for the entire rest of the block before it, you technically don't need it? The rest is just trimmed whitespace, since Ruby interprets the line as ...inject(:+) rescue [a] \$\endgroup\$
    – Value Ink
    May 19, 2016 at 3:16
  • 1
    \$\begingroup\$ @KevinLau-notKenny, no, rescue on same line is different, just rescuing that line. e.g. a = raise("haha") rescue 1 would assign 1 to a. It' \$\endgroup\$ May 19, 2016 at 6:45
  • \$\begingroup\$ @KevinLau-notKenny There's an inline rescue, like there's an inline if and while. \$\endgroup\$
    – Nic
    May 21, 2016 at 16:23
8
\$\begingroup\$

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>

\$\endgroup\$
0
8
\$\begingroup\$

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 - $[[["[]"], "[]"],]
\$\endgroup\$
7
\$\begingroup\$

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].

\$\endgroup\$
6
\$\begingroup\$

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)

\$\endgroup\$
3
  • \$\begingroup\$ Just 3 away from the current winner! +1 \$\endgroup\$
    – Arjun
    May 18, 2016 at 13:11
  • \$\begingroup\$ @Sting: Golfed away another byte. Forgot that Pyth append the last character G implicitly, if I don't write it. \$\endgroup\$
    – Jakube
    May 18, 2016 at 13:25
  • \$\begingroup\$ Congratulations! \$\endgroup\$
    – Arjun
    May 18, 2016 at 13:30
6
\$\begingroup\$

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.

\$\endgroup\$
4
  • \$\begingroup\$ Isn't it ECMAScript 6 not Firefox 30+? \$\endgroup\$ May 19, 2016 at 18:49
  • 1
    \$\begingroup\$ @SolomonUcko No, [for(of)] is only available in Firefox 30+. It was proposed for ES7 but later dropped. \$\endgroup\$
    – Neil
    May 19, 2016 at 19:29
  • 1
    \$\begingroup\$ thanks for explaining! Mostly, i just thought it was for(__ in __) \$\endgroup\$ May 19, 2016 at 21:57
  • \$\begingroup\$ @SolomonUcko [for(in)] was an alternative experimental syntax which gave you the keys of the object. \$\endgroup\$
    – Neil
    May 19, 2016 at 23:02
5
\$\begingroup\$

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.

\$\endgroup\$
7
  • \$\begingroup\$ I haven't tested it, but think ?@{f@$_}: should work instead of ?@{f(@$_)}:, saving two bytes. \$\endgroup\$
    – msh210
    May 18, 2016 at 21:29
  • 1
    \$\begingroup\$ @msh210 No, it’s not working. The compiler doesn’t khow that f is a function because f not yet declared. sub f{}sub f{... f@$_ ...} working. \$\endgroup\$ May 19, 2016 at 9:52
  • \$\begingroup\$ 1. ref doesn't need the parens to work, saving 2 bytes. 2. As far as I can see, sub f{map{ref?f(@$_):$_}@_} is within the rules and saves another 5. f takes an array (nonref) as a list, so it can return the same. \$\endgroup\$
    – hobbs
    May 20, 2016 at 0:15
  • \$\begingroup\$ @hobbs 1. If no parentheses with ref then the compiler assumes that ? is starting ?PATTERN? operation like ref(?PATTERN?). So the compiler searches second ? and throws error. \$\endgroup\$ May 20, 2016 at 5:27
  • \$\begingroup\$ @DenisIbaev ah. ?PATTERN? was removed in 5.22.0 (m?PATTERN? still works) and I'm testing on a recent version. So you can gain those two bytes by specifying 5.22+. \$\endgroup\$
    – hobbs
    May 20, 2016 at 5:34
4
\$\begingroup\$

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
\$\endgroup\$
4
\$\begingroup\$

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]

\$\endgroup\$
3
\$\begingroup\$

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.

\$\endgroup\$
3
\$\begingroup\$

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.

\$\endgroup\$
1
  • \$\begingroup\$ Just 4 away from K! +1 \$\endgroup\$
    – Arjun
    May 18, 2016 at 13:11
3
\$\begingroup\$

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.

\$\endgroup\$
10
  • 3
    \$\begingroup\$ Hello, and welcome to PPCG! This question is code-golf, so please try to minimize your code. Thanks! \$\endgroup\$ May 18, 2016 at 20:06
  • 3
    \$\begingroup\$ Remove all the unnecessary spaces, tabs, and newlines. Change oaf to o, and change flatten to f. \$\endgroup\$ May 18, 2016 at 22:01
  • 2
    \$\begingroup\$ You don't need the finals, the whole thing can be a lambda, you don't need public static... \$\endgroup\$ May 19, 2016 at 6:41
  • 1
    \$\begingroup\$ you could save couple characters if you use generics instead object \$\endgroup\$
    – user902383
    May 19, 2016 at 9:16
  • 1
    \$\begingroup\$ you could also save 2 bytes if you replace false with 1>2, and additional 2 bytes you could get if you declare n but not define (compiler automatically define it as 0) \$\endgroup\$
    – user902383
    May 19, 2016 at 9:17
2
\$\begingroup\$

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.

\$\endgroup\$
2
  • 2
    \$\begingroup\$ type(x)==list is shorter than isinstance(x,list). \$\endgroup\$
    – Value Ink
    May 18, 2016 at 5:37
  • 1
    \$\begingroup\$ “It will contain only Integers (both negative and positive), Strings and Arrays.” How about [`x`>'['and...? (That works in Python 2 only.) \$\endgroup\$
    – Lynn
    May 18, 2016 at 11:47
2
\$\begingroup\$

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)
\$\endgroup\$
6
  • 1
    \$\begingroup\$ Hello, and welcome to PPCG! Unfortunately, your answer is not valid, because of this rule: Note: If your language contains a built-in for this, then you must NOT use it. \$\endgroup\$ May 18, 2016 at 20:04
  • \$\begingroup\$ @NoOneIsHere, thanks, didn't know that \$\endgroup\$ May 18, 2016 at 20:22
  • 1
    \$\begingroup\$ How does my new update stack against time-wise against yours? Also, just like my new answer, you can remove the spaces around rescue \$\endgroup\$
    – Value Ink
    May 18, 2016 at 20:59
  • \$\begingroup\$ @KevinLau-notKenny updated, thanks! rescue looks to be rather slow btw, like try/catch in java \$\endgroup\$ May 18, 2016 at 21:08
  • 1
    \$\begingroup\$ Update your bytecount, too \$\endgroup\$
    – Value Ink
    May 18, 2016 at 21:11
2
\$\begingroup\$

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

Oneliner. Inspired by Martin Büttner.

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

Test it on Ideone.

\$\endgroup\$
2
\$\begingroup\$

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 [[["[]"],"[]"]])
\$\endgroup\$
1
  • \$\begingroup\$ Nice first code-golf. +1 :) \$\endgroup\$
    – Arjun
    May 20, 2016 at 11:56
2
\$\begingroup\$

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.

\$\endgroup\$
3
  • 2
    \$\begingroup\$ Welcome to PPCG! \$\endgroup\$ May 15, 2017 at 14:35
  • 1
    \$\begingroup\$ Welcome to PPCG! Nice first golf. Good luck ahead! \$\endgroup\$
    – Arjun
    May 15, 2017 at 14:59
  • \$\begingroup\$ Thanks! It was an attempt to up my points so that I can get commenting privileges elsewhere. It appears things don't work like that the accounts are for different portals. :D I will see if some nifty features from c++ can be used. \$\endgroup\$
    – amritanshu
    May 15, 2017 at 15:07
2
\$\begingroup\$

JavaScript 20 bytes

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

The array + array is equal to array.toString

\$\endgroup\$
6
  • \$\begingroup\$ @WheatWizard thanks for the welcome and I am new to the site. actually a is an argument of the function. I will try to edit out the function now. \$\endgroup\$
    – i--
    Jul 19, 2017 at 15:19
  • \$\begingroup\$ I think now it's ok @WheatWizard. Please let me know if there is a problem with this \$\endgroup\$
    – i--
    Jul 19, 2017 at 15:21
  • 1
    \$\begingroup\$ Actually looking at the javaScript docs an anonymous function would definitely be shorter, you would only have to add a=> to the beginning of your code. \$\endgroup\$
    – Wheat Wizard
    Jul 19, 2017 at 15:26
  • 1
    \$\begingroup\$ Save 2 bytes with split`,` \$\endgroup\$
    – Arjun
    Jul 23, 2017 at 1:11
  • 3
    \$\begingroup\$ This doesn't handle strings with commas in them correctly \$\endgroup\$
    – Jo King
    Feb 16, 2019 at 10:17
2
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Haskell + free, 7 bytes

In order to represent a ragged list in Haskell we use a free monad of lists. Luckily for us provided with the free monad is retract, which does a lot of different things, but when given the particular type here it flattens the free monad of lists down to a single list.

retract

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But we can also make our own solution using other tools, like foldFree:

11 bytes

foldFree id

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However foldFree basically is just a more general version of retract, and we can do it without it:

21 bytes

iter(>>=id).fmap(:[])

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If we want to use none of the functions provided by free just the definition of the monad and prelude functions we can do:

29 bytes

f(Pure a)=[a]
f(Free a)=a>>=f

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1
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Racket, 63 bytes

(define(f l)(apply append(map(λ(x)(if(list? x)(f x)`(,x)))l)))
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1
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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.

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1
  • \$\begingroup\$ would it be not better when using Arrays.asList() on "array" and then go the foreach with lambda and end this with a Collector? \$\endgroup\$
    – Serverfrog
    Mar 23, 2017 at 14:21
1
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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>

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3
  • 3
    \$\begingroup\$ This doesn't seem to work when run in the console for input ["["]... I tried running (a=>eval([${a}]))(["["]) and got a SyntaxError \$\endgroup\$
    – jrich
    May 18, 2016 at 22:35
  • \$\begingroup\$ @jrich. You just get this error when you type character by character. If you copy and paste any valid array, it will work as expected. By the way, nice answer SpeedNinja, I would only change oninput event with a button click. \$\endgroup\$ May 24, 2016 at 20:58
  • 1
    \$\begingroup\$ This doesn't work for strings with commas in them \$\endgroup\$
    – Jo King
    Feb 16, 2019 at 10:18
1
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PHP, 73 Bytes

<?array_walk_recursive($_GET,function($i)use(&$r){$r[]=$i;});print_r($r);
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1
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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

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1
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Jelly, 4 bytes

;/ÐL

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Explanation

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

Built-in F would be one byte if allowed.

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Wolfram Language (Mathematica), 13 bytes

#~Level~{-1}&

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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).

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