Leaving the Nest

Question

Given a non-flat list of integers, output a list of lists containing the integers in each nesting level, starting with the least-nested level, with the values in their original order in the input list when read left-to-right. If two or more lists are at the same nesting level in the input list, they should be combined into a single list in the output. The output should not contain any empty lists - nesting levels that contain only lists should be skipped entirely.

You may assume that the integers are all in the (inclusive) range [-100, 100]. There is no maximum length or nesting depth for the lists. There will be no empty lists in the input - every nesting level will contain at least one integer or list.

The input and output must be in your language's native list/array/enumerable/iterable/etc. format, or in any reasonable, unambiguous format if your language lacks a sequence type.

Examples

[1, 2, [3, [4, 5], 6, [7, [8], 9]]] => [[1, 2], [3, 6], [4, 5, 7, 9], [8]]

[3, 1, [12, [14, [18], 2], 1], [[4]], 5] => [[3, 1, 5], [12, 1], [14, 2, 4], [18]]

[2, 1, [[5]], 6] => [[2, 1, 6], [5]]

[[54, [43, 76, [[[-19]]]], 20], 12] => [[12], [54, 20], [43, 76], [-19]]

[[[50]], [[50]]] => [[50, 50]]

Answer 1

Pyth, 17

 us-GeaYsI#GQ)S#Y

The leading space is important. This filters the list on whether the values are invariant on the s function, then removes these values from the list and flatten it one level. The values are also stored in Y and when we print we remove the empty values by filtering if the sorted value of the list is truthy.

Test Suite

Alternatively, a 15 byte answer with a dubious output format:

 us-GpWJsI#GJQ)

Test Suite

Expansion:

 us-GeaYsI#GQ)S#Y     ##   Q = eval(input)
 u          Q)        ##   reduce to fixed point, starting with G = Q
        sI#G          ##   get the values that are not lists from G
                      ##   this works because s<int> = <int> but s<list> = flatter list
      aY              ##   append the list of these values to Y
     e                ##   flatten the list
   -G                 ##   remove the values in the list from G
              S#Y     ##   remove empty lists from Y

Answer 2

Mathematica, 56 54 52 bytes

_{-2 bytes due to Alephalpha.}

_{-2 bytes due to CatsAreFluffy.}

Cases[#,_?AtomQ,{i}]~Table~{i,Depth@#}/.{}->Nothing&

Actually deletes empty levels.

Answer 3

Python 2, 78 bytes

f=lambda l:l and zip(*[[x]for x in l if[]>x])+f(sum([x for x in l if[]<x],[]))

Answer 4

Retina, 79

I know the Retina experts will golf this more, but here's a start:

{([^{}]+)}(,?)([^{}]*)
$3$2<$1>
)`[>}],?[<{]
,
(\d),+[<{]+
$1},{
<+
{
,*[>}]+
}

Try it online.

Answer 5

Mathematica 55 64 62 bytes

#~Select~AtomQ/.{}->Nothing&/@Table[Level[#,{k}],{k,Depth@#}]&

---

%&[{1, 2, {3, {4, 5}, 6, {7, {8}, 9}}}]

{{1, 2}, {3, 6}, {4, 5, 7, 9}, {8}}

Answer 6

JavaScript, 112 80 bytes

F=(a,b=[],c=0)=>a.map(d=>d!==+d?F(d,b,c+1):b[c]=[...b[c]||[],d])&&b.filter(d=>d)

Thanks Neil for helping shave off 32 bytes.

Answer 7

Jelly, 24 bytes

fFW®;©ṛ¹ḟF;/µŒḊ’$¡W®Tị¤;

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If newline-separated lists were allowed, this could be golfed down to 14 bytes.

fFṄ¹¹?ḟ@¹;/µ¹¿

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Answer 8

Jelly, 6 bytes

ŒJẈĠịF

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Jelly has substantially improved in the past 5 years

How it works

ŒJẈĠịF - Main link. Takes a list L on the left
ŒJ     - Multi-dimensional indices
  Ẉ    - Get the length of each element
         This yields the depth (starting at 1 for flat elements) of each element
   Ġ   - Group the indices of this list by the values
     F - Flatten L
    ị  - Index into

Huh? The way that ŒJẈĠ works is kinda confusing without an example. Let's take L = [3, 1, [12, [14, [18], 2], 1], [[4]], 5] and work through it

ŒJ is one of a series of multi-dimensional commands that reflect the behaviour of their one-dimensional versions. Specifically, this is the "multi-dimensional indices" command, a multi-dimensional version of J, the "indices" command.

Picture L arranged like this:

3 1 12 14 18 2 1 4 5
    12 14 18 2 1 4
       14 18 2   4
          18

That is, each element is repeated down a number of times equal to it's depth. ŒJ doesn't quite leave the elements like this (it changes the specific values, not the shape), but it's enough. Ẉ then calculates the length of each i.e. how many rows each element goes down:

1 1 2 3 4 3 2 3 1

Ġ the takes this list as it's argument. First, it's generates this list's indices: [1, 2, 3, 4, 5, 6, 7, 8, 9]. We then group these indices by the values of the argument. So because the values at indices 3 and 7 are equal (they're both 2), [3, 7] is grouped. The final result of this is [[1, 2, 9], [3, 7], [4, 6, 8], [5]] as the groups are sorted by the values they're grouped by.

Finally, we index into the flattened L using this list of groups, which yields the intended output

Answer 9

Python, 108 99 bytes

This seems a bit long to me, but I couldn't make a one-liner shorter, and if I try using or instead of if, I get empty lists in the results.

def f(L):
    o=[];i=[];j=[]
    for x in L:[i,j][[]<x]+=[x]
    if i:o+=[i]
    if j:o+=f(sum(j,[]))
    return o

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Edit: Saved 9 bytes thanks to Stack Overflow

Answer 10

Python 3, 109 bytes

As ever, stupid Python 2 features like comparing ints and lists mean that Python 3 comes out behind. Oh well...

def d(s):
 o=[]
 while s:
  l,*n=[],
  for i in s:
   try:n+=i
   except:l+=[i]
  if l:o+=[l]
  s=n
 return o

Answer 11

Perl, 63 bytes

{$o[$x++]=[@t]if@t=grep{!ref}@i;(@i=map{@$_}grep{ref}@i)&&redo}

Input is expected in @i, output produced in @o. (I hope this is acceptable).

Example:

@i=[[54, [43, 76, [[[-19]]]], 20], 12];                              # input

{$o[$x++]=[@t]if@t=grep{!ref}@i;(@i=map{@$_}grep{ref}@i)&&redo}      # the snippet

use Data::Dumper;                                                    # print output
$Data::Dumper::Indent=0;  # keep everything on one line
$Data::Dumper::Terse=1;   # don't print $VAR =
print Dumper(\@o);

Output:

[[12],[54,20],[43,76],[-19]]

Answer 12

Clojure, 119 bytes

(116 with seq? and input as lists, a trivial modification)

(defn f([v](vals(apply merge-with concat(sorted-map)(flatten(f 0 v)))))([l v](map #(if(number? %){l[%]}(f(inc l)%))v)))

Better intended:

(defn f([v]  (vals(apply merge-with concat(sorted-map)(flatten(f 0 v)))))
       ([l v](map #(if(number? %){l[%]}(f(inc l)%))v)))

When called with two arguments (the current level and a collection) it either creates an one-element unordered-map like {level: value}, or calls f recursively if a non-number (presumambly a collection) is seen.

These mini-maps are then merged into a single sorted-map and key collisions are handled by concat function. vals returns map's values from first level to the last.

If a number is the only one at its level then it remains a vec, others are converted to lists by concat.

(f [[54, [43, 76, [[[-19]]]], 20], 12])
([12] (54 20) (43 76) [-19])

If input was a list instead of vec then number? could be replaced by seq?, oddly vector isn't seq? but it is sequential?. But I'm too lazy to implement that version, re-do examples etc.

Answer 13

Racket 259 bytes

(let((ol'())(m 0))(let p((l l)(n 0))(cond[(empty? l)][(list?(car l))(set! m(+ 1 n))
(p(car l)(+ 1 n))(p(cdr l)n)][(set! ol(cons(list n(car l))ol))(p(cdr l)n )]))
(for/list((i(+ 1 m)))(flatten(map(λ(x)(cdr x))(filter(λ(x)(= i(list-ref x 0)))(reverse ol))))))

Ungolfed:

(define (f l)
  (define ol '())
  (define maxn 0)
  (let loop ((l l)              ; in this loop each item is added with its level
             (n 0))
    (cond
      [(empty? l)]
      [(list? (first l))
       (set! maxn (add1 n))
       (loop (first l) (add1 n))
       (loop (rest l) n)]
      [else
       (set! ol (cons (list n (first l)) ol))
       (loop (rest l) n )]))

  ; now ol is '((0 1) (0 2) (1 3) (2 4) (2 5) (1 6) (2 7) (3 8) (2 9)) 

  (for/list ((i (add1 maxn)))   ; here similar levels are combined
    (flatten
     (map (λ (x) (rest x))      ; level numbers are removed
          (filter (λ (x) (= i(list-ref x 0)))
                  (reverse ol))))))

Testing:

(f '[1 2 [3 [4 5] 6 [7 [8] 9]]])

Output:

'((1 2) (3 6) (4 5 7 9) (8))

Answer 14

MATL, 37 bytes

j']['!=dYsXKu"GK@=)'[\[\],]'32YXUn?1M

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Works with current release (13.0.0) of the language/compiler.

This produces the output as lines of space-separated values, where each line corresponds to the same nesting level, and different nesting levels are separated by newlines.

j            % read input as string (row array of chars)
']['!        % 2x1 array containing ']'  and '['
=            % test for equality, all combinations
d            % row array obtained as first row minus second row
Ys           % cumulative sum. Each number is a nesting level
XK           % copy to clibdoard K
u            % unique values: all existing nesting levels
"            % for each nesting level
  G          %   push input
  K          %   push array that indicates nesting level of each input character
  @          %   push level corresponding to this iteration
  =          %   true for characters corresponding to that nesting level
  )          %   pick those characters
  '[\[\],]'  %   characters to be replaced
  32         %   space
  YX         %   regexp replacement
  U          %   only numbers and spaces remain: convert string to array of numbers
  n?         %   if non-empty
    1M       %     push that array of numbers again
             %   end if implicitly
             % end for each implicitly

Answer 15

Stax, 18 bytes

é╔óó)₧ú↔u.≤∟╦Ç╟·○`

Run and debug it

I rhought I'd be able to beat pyth, but ¯\_(○-○)_/¯