Flatten a list: Difference between revisions
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There is no nesting of lists or other data structures in TI-89 BASIC, short of using variable names as pointers. |
There is no nesting of lists or other data structures in TI-89 BASIC, short of using variable names as pointers. |
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=={{header|Trith}}== |
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<lang trith>[[1] 2 [[3 4] 5] [[[]]] [[[6]]] 7 8 []] flatten</lang> |
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=={{header|VBScript}}== |
=={{header|VBScript}}== |
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Revision as of 16:44, 18 May 2010
You are encouraged to solve this task according to the task description, using any language you may know.
Write a function to flatten the nesting in an arbitrary list of values. Your program should work on the equivalent of this list:
[[1], 2, [[3,4], 5], [[[]]], [[[6]]], 7, 8, []]
Where the correct result would be the list:
[1, 2, 3, 4, 5, 6, 7, 8]
C.f. Tree traversal
ActionScript
<lang ActionScript>function flatten(input:Array):Array { var output:Array = new Array(); for (var i:uint = 0; i < input.length; i++) {
//typeof returns "object" when applied to arrays. This line recursively evaluates nested arrays,
// although it may break if the array contains objects that are not arrays.
if (typeof input[i]=="object") { output=output.concat(flatten(input[i])); } else { output.push(input[i]); } } return output; } </lang>
Aikido
<lang aikido> function flatten (list, result) {
foreach item list {
if (typeof(item) == "vector") {
flatten (item, result)
} else {
result.append (item)
}
}
}
var l = [[1], 2, [[3,4], 5], [[[]]], [[[6]]], 7, 8, []] var newl = [] flatten (l, newl)
// print out the nicely formatted result list print ('[') var comma = "" foreach item newl {
print (comma + item) comma = ", "
} println("]")
</lang> Output:
[1, 2, 3, 4, 5, 6, 7, 8]
C++
<lang cpp>#include <list>
- include <boost/any.hpp>
typedef std::list<boost::any> anylist;
void flatten(std::list<boost::any>& list) {
typedef anylist::iterator iterator;
iterator current = list.begin();
while (current != list.end())
{
if (current->type() == typeid(anylist))
{
iterator next = current;
++next;
list.splice(next, boost::any_cast<anylist&>(*current));
next = current;
++next;
list.erase(current);
current = next;
}
else
++current;
}
}</lang>
Use example:
Since C++ currently doesn't have nice syntax for initializing lists, this includes a simple parser to create lists of integers and sublists. Also, there's no standard way to output this type of list, so some output code is added as well. <lang cpp>#include <cctype>
- include <iostream>
// ******************* // * the list parser * // *******************
void skipwhite(char const** s) {
while (**s && std::isspace((unsigned char)**s))
{
++*s;
}
}
anylist create_anylist_i(char const** s) {
anylist result;
skipwhite(s);
if (**s != '[')
throw "Not a list";
++*s;
while (true)
{
skipwhite(s);
if (!**s)
throw "Error";
else if (**s == ']')
{
++*s;
return result;
}
else if (**s == '[')
result.push_back(create_anylist_i(s));
else if (std::isdigit((unsigned char)**s))
{
int i = 0;
while (std::isdigit((unsigned char)**s))
{
i = 10*i + (**s - '0');
++*s;
}
result.push_back(i);
}
else
throw "Error";
skipwhite(s);
if (**s != ',' && **s != ']')
throw "Error";
if (**s == ',')
++*s;
}
}
anylist create_anylist(char const* i) {
return create_anylist_i(&i);
}
// ************************* // * printing nested lists * // *************************
void print_list(anylist const& list);
void print_item(boost::any const& a) {
if (a.type() == typeid(int)) std::cout << boost::any_cast<int>(a); else if (a.type() == typeid(anylist)) print_list(boost::any_cast<anylist const&>(a)); else std::cout << "???";
}
void print_list(anylist const& list) {
std::cout << '[';
anylist::const_iterator iter = list.begin();
while (iter != list.end())
{
print_item(*iter);
++iter;
if (iter != list.end())
std::cout << ", ";
}
std::cout << ']';
}
// *************************** // * The actual test program * // ***************************
int main() {
anylist list =
create_anylist("[[1], 2, [[3,4], 5], [[[]]], [[[6]]], 7, 8, []]");
print_list(list);
std::cout << "\n";
flatten(list);
print_list(list);
std::cout << "\n";
}</lang> The output of this program is
[[1], 2, [[3, 4], 5], [[[]]], [[[6]]], 7, 8, []] [1, 2, 3, 4, 5, 6, 7, 8]
Clojure
The following returns a lazy sequence of the flattened data structure. <lang lisp>(defn flatten [coll]
(lazy-seq
(when-let [s (seq coll)]
(if (coll? (first s))
(concat (flatten (first s)) (flatten (next s)))
(cons (first s) (flatten (next s)))))))</lang>
An alternative approach (from clojure.contrib.seq-utils).
<lang lisp>(defn flatten [x]
(filter (complement sequential?)
(rest (tree-seq sequential? seq x))))</lang>
Common Lisp
<lang lisp>(defun flatten (tree)
(let ((result '()))
(labels ((scan (item)
(if (listp item)
(map nil #'scan item)
(push item result))))
(scan tree))
(nreverse result)))</lang>
While this is a common homework problem in Common Lisp, it should be noted that it is typically a mistake to use it in a realistic program; it is almost always easier to rewrite the program such that it does not generate the deeply-nested lists in the first place. In particular, note that flattening means that none of the values stored in the nested-list-structure can be a list (a cons or nil) itself. For example, producing and flattening a tree of boolean values would not do the right thing, since false = nil, a list.
D
There are many ways to implement this in D. This version minimizes heap activity using a tagged union. A similar object-oriented version too is possible. D v.2. <lang d>import std.stdio: writeln; import std.conv: to;
struct TreeList(T) {
bool isArr = true; // contains an arr on default
union {
TreeList!T[] arr;
T data;
}
static TreeList!T opCall(Types...)(Types items) {
TreeList!T result;
foreach (i, el; items)
static if (is(Types[i] == T)) {
TreeList!T item;
item.isArr = false;
item.data = el;
result.arr ~= item;
} else
result.arr ~= el;
return result; }
// this is better as a general free function
TreeList!T flatten() {
TreeList!T result;
if (this.isArr)
foreach (el; this.arr)
result.arr ~= el.flatten().arr;
else
result.arr ~= this;
return result; }
string toString() {
if (isArr)
return to!string(arr, "[", ", ", "]");
else
return to!string(data);
}
}
void main() {
alias TreeList!(int) L; // 12 bytes if T is int auto list = L(L(1), 2, L(L(3,4), 5), L(L(L())), L(L(L(6))), 7, 8, L()); writeln(list); writeln(list.flatten());
}</lang> Output:
[[1], 2, [[3, 4], 5], [[[]]], [[[6]]], 7, 8, []] [1, 2, 3, 4, 5, 6, 7, 8]
E
<lang e>def flatten(nested) {
def flat := [].diverge()
def recur(x) {
switch (x) {
match list :List { for elem in list { recur(elem) } }
match other { flat.push(other) }
}
}
recur(nested)
return flat.snapshot()
}</lang>
<lang e>? flatten([[1], 2, [[3,4], 5], [[[]]], [[[6]]], 7, 8, []])
- value: [1, 2, 3, 4, 5, 6, 7, 8]</lang>
Erlang
There's a standard function (lists:flatten/2) that does it more efficiently, but this is the cleanest implementation you could have; <lang Erlang>flatten([]) -> []; flatten([H|T]) -> flatten(H) ++ flatten(T); flatten(H) -> [H].</lang>
Factor
USE: sequences.deep
( scratchpad ) { { 1 } 2 { { 3 4 } 5 } { { { } } } { { { 6 } } } 7 8 { } } flatten .
{ 1 2 3 4 5 6 7 8 }
Groovy
List.flatten() is a Groovy built-in that returns a flattened copy of the source list:
<lang groovy>assert [[1], 2, [[3,4], 5], [[[]]], [[[6]]], 7, 8, []].flatten() == [1, 2, 3, 4, 5, 6, 7, 8]</lang>
Haskell
In Haskell we have to interpret this structure as an algebraic data type.
<lang Haskell>import Data.Tree
-- [[1], 2, [[3,4], 5], [[[]]], [[[6]]], 7, 8, []] -- implemented as multiway tree:
-- Data.Tree represents trees where nodes have values too, unlike the trees in our problem. -- so we use a list as that value, where a node will have an empty list value, -- and a leaf will have a one-element list value and no subtrees list :: Tree [Int] list = Node [] [
Node [] [Node [1] []],
Node [2] [],
Node [] [
Node [] [ Node [3] [],Node [4] []],
Node [5] []
],
Node [] [Node [] [Node [] []]],
Node [] [Node [] [Node [6] []]],
Node [7] [],
Node [8] [],
Node [] []
]
flattenList = concat.flatten</lang> Flattening the list:
*Main> flattenList list [1,2,3,4,5,6,7,8]
Alternately: <lang haskell>data Tree a = Leaf a | Node [Tree a]
flatten :: Tree a -> [a] flatten (Leaf x) = [x] flatten (Node xs) = concatMap flatten xs
main = print $ flatten $ Node [Node [Leaf 1],
Leaf 2,
Node [Node [Leaf 3, Leaf 4], Leaf 5],
Node [Node [Node []]],
Node [Node [Node [Leaf 6]]],
Leaf 7,
Leaf 8,
Node []]
-- output: [1,2,3,4,5,6,7,8]</lang>
Yet another choice, custom data structure, efficient lazy flattening:
<lang haskell>data NestedList a = NList [NestedList a] | Entry a
flatten :: NestedList a -> [a] flatten nl = flatten' nl []
where -- By passing through a list which the results will be preprended to we allow efficient lazy evaluation flatten' :: NestedList a -> [a] -> [a] flatten' (Entry a) cont = a:cont flatten' (NList entries) cont = foldr flatten' cont entries
example :: NestedList Int example = NList [ NList [Entry 1], Entry 2, NList [NList [Entry 3, Entry 4], Entry 5], NList [NList [NList []]], NList [ NList [ NList [Entry 6]]], Entry 7, Entry 8, NList []]
main :: IO () main = print $ flatten example
-- output [1,2,3,4,5,6,7,8]</lang>
Icon and Unicon
Icon
The following procedure solves the task using a string representation of nested lists and cares not if the list is well formed or not. <lang Icon>link strings # for compress,deletec,pretrim
procedure sflatten(s) # uninteresting string solution return pretrim(trim(compress(deletec(s,'[ ]'),',') ,','),',') end</lang>
The solution uses several procedures from strings in the IPL
This procedure is more in the spirit of the task handling actual lists rather than representations. It uses a recursive approach using some of the built-in list manipulation functions and operators. <lang Icon>procedure flatten(L) # in the spirt of the problem a structure local l,x
l := [] every x := !L do
if type(x) == "list" then l |||:= flatten(x) else put(l,x)
return l end</lang>
Finally a demo routine to drive these and a helper to show how it works. <lang Icon>procedure main() write(sflatten(" [[1], 2, [[3,4], 5], [[[]]], [[[6]]], 7, 8, []]")) writelist(flatten( [[1], 2, [[3,4], 5], [[[]]], [[[6]]], 7, 8, []])) end
procedure writelist(L) writes("[") every writes(" ",image(!L)) write(" ]") return end</lang>
Unicon
This Icon solution works in Unicon.
Ioke
<lang ioke>iik> [[1], 2, [[3,4], 5], [[[]]], [[[6]]], 7, 8, []] flatten [[1], 2, [[3,4], 5], [[[]]], [[[6]]], 7, 8, []] flatten +> [1, 2, 3, 4, 5, 6, 7, 8]</lang>
J
Solution: <lang j>flatten =: [: ; <S:0</lang>
Example: <lang j> NB. create and display nested noun li
]li =. (<1) ; 2; ((<3;4); 5) ; ((<a:)) ; ((<(<6))) ; 7; 8; <a:
+---+-+-----------+----+-----+-+-+--+ |+-+|2|+-------+-+|+--+|+---+|7|8|++| ||1|| ||+-----+|5|||++|||+-+|| | |||| |+-+| |||+-+-+|| |||||||||6||| | |++| | | ||||3|4||| |||++|||+-+|| | | | | | |||+-+-+|| ||+--+|+---+| | | | | | ||+-----+| || | | | | | | | |+-------+-+| | | | | | +---+-+-----------+----+-----+-+-+--+
flatten li
1 2 3 4 5 6 7 8</lang>
Java
<lang java5>import java.util.LinkedList; import java.util.List; import java.util.Arrays;
public class Flatten {
public static List<Object> flatten(List<?> list){
List<Object> retVal = new LinkedList<Object>();
for(Object item:list){
if(item instanceof List){
retVal.addAll(flatten((List)item));
}else{
retVal.add(item);
}
}
return retVal;
}
public static void main(String[] args){
List<Object> test = Arrays.<Object>asList(
Arrays.<Object>asList(1),
2,
Arrays.<Object>asList(
Arrays.<Object>asList(3,4),
5
),
Arrays.<Object>asList(
Arrays.<Object>asList(
Arrays.<Object>asList()
)
),
Arrays.<Object>asList(
Arrays.<Object>asList(
Arrays.<Object>asList(6)
)
),
7,
8,
Arrays.<Object>asList()
);
System.out.println(test); System.out.println(flatten(test)); }
}</lang> Output:
[[1], 2, [[3, 4], 5], [[[]]], [[[6]]], 7, 8, []] [1, 2, 3, 4, 5, 6, 7, 8]
JavaScript
<lang javascript><script type="text/javascript" src="/path/to/prototype.js"></script> <script type="text/javascript">
var a = [[1], 2, [[3,4], 5], [[[]]], [[[6]]], 7, 8, []];
document.write("
before
"); a.each(function(e){document.write("
")});
var f = a.flatten();
document.write("
after
"); f.each(function(e){document.write("
")});
</script></lang> output:
before 1 object 2 number 3,4,5 object object 6 object 7 number 8 number object after 1 number 2 number 3 number 4 number 5 number 6 number 7 number 8 number
Lua
<lang lua>function flatten(list)
if type(list) ~= "table" then return {list} end
local flat_list = {}
for _, elem in ipairs(list) do
for _, val in ipairs(flatten(elem)) do
flat_list[#flat_list + 1] = val
end
end
return flat_list
end
test_list = {{1}, 2, {{3,4}, 5}, {{{}}}, {{{6}}}, 7, 8, {}}
print(table.concat(flatten(test_list), ","))</lang>
Logo
<lang logo>to flatten :l
if not list? :l [output :l] if empty? :l [output []] output sentence flatten first :l flatten butfirst :l
end
- using a template iterator (map combining results into a sentence)
to flatten :l
output map.se [ifelse or not list? ? empty? ? [?] [flatten ?]] :l
end
make "a [[1] 2 [[3 4] 5] [[[]]] [[[6]]] 7 8 []] show flatten :a</lang>
OCaml
<lang ocaml># let flatten = List.concat ;; val flatten : 'a list list -> 'a list = <fun>
- let li = [[1]; 2; [[3;4]; 5]; [[[]]]; [[[6]]]; 7; 8; []] ;;
^^^
Error: This expression has type int but is here used with type int list
- (* use another data which can be accepted by the type system *)
flatten [[1]; [2; 3; 4]; []; [5; 6]; [7]; [8]] ;;
- : int list = [1; 2; 3; 4; 5; 6; 7; 8]</lang>
Since OCaml is statically typed, it is not possible to have a value that could be both a list and a non-list. Instead, we can use an algebraic datatype:
<lang ocaml># type 'a tree = Leaf of 'a | Node of 'a tree list ;; type 'a tree = Leaf of 'a | Node of 'a tree list
- let rec flatten = function
Leaf x -> [x] | Node xs -> List.concat (List.map flatten xs) ;;
val flatten : 'a tree -> 'a list = <fun>
- flatten (Node [Node [Leaf 1]; Leaf 2; Node [Node [Leaf 3; Leaf 4]; Leaf 5]; Node [Node [Node []]]; Node [Node [Node [Leaf 6]]]; Leaf 7; Leaf 8; Node []]) ;;
- : int list = [1; 2; 3; 4; 5; 6; 7; 8]</lang>
Oz
Oz has a standard library function "Flatten": <lang oz>{Show {Flatten [[1] 2 [[3 4] 5] nil [[[6]]] 7 8 nil]}}</lang> A simple, non-optimized implementation could look like this: <lang oz>fun {Flatten2 Xs}
case Xs of nil then nil
[] X|Xr then
{Append {Flatten2 X} {Flatten2 Xr}}
else [Xs]
end
end </lang>
Perl
<lang perl>sub flatten {
map { ref eq 'ARRAY' ? flatten(@$_) : $_ } @_
}
my @lst = ([1], 2, [[3,4], 5], [[[]]], [[[6]]], 7, 8, []); print flatten(@lst), "\n";</lang>
Perl 6
<lang perl6>multi flatten (@a) { map { flatten $^x }, @a } multi flatten ($x) { $x }</lang>
PHP
<lang php>/* Note: This code is only for PHP 4.
It won't work on PHP 5 due to the change in behavior of array_merge(). */
while (array_filter($lst, 'is_array'))
$lst = call_user_func_array('array_merge', $lst);</lang>
Explanation: while $lst has any elements which are themselves arrays (i.e. $lst is not flat), we merge the elements all together (in PHP 4, array_merge() treated non-array arguments as if they were 1-element arrays; PHP 5 array_merge() no longer allows non-array arguments.), thus flattening the top level of any embedded arrays. Repeat this process until the array is flat.
Recursive
<lang php><?php function flatten($ary) {
$result = array();
foreach ($ary as $x) {
if (is_array($x))
// append flatten($x) onto $result
array_splice($result, count($result), 0, flatten($x));
else
$result[] = $x;
}
return $result;
}
$lst = array(array(1), 2, array(array(3, 4), 5), array(array(array())), array(array(array(6))), 7, 8, array()); var_dump(flatten($lst)); ?></lang>
Non-recursive
Function flat is iterative and flattens the array in-place. <lang php><?php function flat(&$ary) { // argument must be by reference or array will just be copied
for ($i = 0; $i < count($ary); $i++) {
while (is_array($ary[$i])) {
array_splice($ary, $i, 1, $ary[$i]);
}
}
}
$lst = array(array(1), 2, array(array(3, 4), 5), array(array(array())), array(array(array(6))), 7, 8, array()); flat($lst); var_dump($lst); ?></lang>
PicoLisp
<lang PicoLisp>(de flatten (X)
(make # Build a list
(recur (X) # recursively over 'X'
(if (atom X)
(link X) # Put atoms into the result
(mapc recurse X) ) ) ) ) # or recurse on sub-lists</lang>
or a more succint way using fish:
<lang PicoLisp>(de flatten (X)
(fish atom X) )</lang>
PL/I
<lang PL/I> list = translate (list, ' ', '[]' ); list = '[' || list || ']'; </lang>
Python
Recursive
Function flatten is recursive and preserves its argument:
<lang python>>>> def flatten(lst):
return sum( ([x] if isinstance(x, list) else flatten(x)
for x in lst), [] )
>>> lst = [[1], 2, [[3,4], 5], [[[]]], [[[6]]], 7, 8, []] >>> flatten(lst) [1, 2, 3, 4, 5, 6, 7, 8]</lang>
Non-recursive
Function flat is iterative and flattens the list in-place. It follows the Python idiom of returning None when acting in-place: <lang python>>>> def flat(lst):
i=0
while i<len(lst):
while True:
try:
lst[i:i+1] = lst[i]
except (TypeError, IndexError):
break
i += 1
>>> lst = [[1], 2, [[3,4], 5], [[[]]], [[[6]]], 7, 8, []] >>> flat(lst) >>> lst [1, 2, 3, 4, 5, 6, 7, 8]</lang>
Generative
This method shows a solution using Python generators.
flatten is a generator that yields the non-list values of its input in order.
The primary advantage of generators is that there are no temporary values needed.
For large input, this can reduce memory usage dramatically.
The generator is converted back to a list before printing.
<lang python>>>> def flatten(lst):
for x in lst:
if isinstance(x, list):
for x in flatten(x):
yield x
else:
yield x
>>>lst = [[1], 2, [[3,4], 5], [[[]]], [[[6]]], 7, 8, []] >>>print list(flatten(lst)) [1, 2, 3, 4, 5, 6, 7, 8]</lang>
R
<lang R>x <- list(list(1), 2, list(list(3, 4), 5), list(list(list())), list(list(list(6))), 7, 8, list())
unlist(x)</lang>
REBOL
<lang rebol> flatten: func [
"Flatten the block in place." block [any-block!]
][
parse block [
any [block: any-block! (change/part block first block 1) :block | skip]
]
head block
] </lang>
Sample:
>> flatten [[1] 2 [[3 4] 5] [[[]]] [[[6]]] 7 8 []] == [1 2 3 4 5 6 7 8]
Ruby
flatten is a built-in method of Arrays
<lang ruby>flat = [[1], 2, [[3,4], 5], [[[]]], [[[6]]], 7, 8, []].flatten
p flat # => [1, 2, 3, 4, 5, 6, 7, 8]</lang>
Scala
<lang scala>def flatList(l: List[_]): List[Any] = l match {
case Nil => Nil case (head: List[_]) :: tail => flatList(head) ::: flatList(tail) case head :: tail => head :: flatList(tail)
}</lang>
Sample:
scala> List(List(1), 2, List(List(3, 4), 5), List(List(List())), List(List(List(6))), 7, 8, List()) res10: List[Any] = List(List(1), 2, List(List(3, 4), 5), List(List(List())), List(List(List(6))), 7, 8, List()) scala> flatList(res10) res12: List[Any] = List(1, 2, 3, 4, 5, 6, 7, 8)
Scheme
<lang scheme>> (define (flatten x)
(cond ((null? x) '())
((not (pair? x)) (list x))
(else (append (flatten (car x))
(flatten (cdr x))))))
> (flatten '((1) 2 ((3 4) 5) ((())) (((6))) 7 8 ())) (1 2 3 4 5 6 7 8)</lang>
Slate
<lang slate>s@(Sequence traits) flatten [
[| :out | s flattenOn: out] writingAs: s
].
s@(Sequence traits) flattenOn: w@(WriteStream traits) [
s do: [| :value |
(value is: s)
ifTrue: [value flattenOn: w]
ifFalse: [w nextPut: value]].
].</lang>
Smalltalk
<lang smalltalk>OrderedCollection extend [
flatten [ |f|
f := OrderedCollection new.
self do: [ :i |
i isNumber
ifTrue: [ f add: i ]
ifFalse: [ |t|
t := (OrderedCollection withAll: i) flatten.
f addAll: t
]
].
^ f
]
].
|list|
list := OrderedCollection
withAll: { {1} . 2 . { {3 . 4} . 5 } .
{{{}}} . {{{6}}} . 7 . 8 . {} }.
(list flatten) printNl.</lang>
Tcl
<lang tcl>proc flatten list {
for {set old {}} {$old ne $list} {} {
set old $list
set list [join $list]
}
return $list
}
puts [flatten {{1} 2 {{3 4} 5} {{{}}} {{{6}}} 7 8 {}}]
- ===> 1 2 3 4 5 6 7 8</lang>
Note that because lists are not syntactically distinct from strings, it is probably a mistake to use this procedure with real (especially non-numeric) data. Also note that there are no parentheses around the outside of the list when printed; this is just a feature of how Tcl regards lists, and the value is a proper list (it can be indexed into with lindex, iterated over with foreach, etc.)
TI-89 BASIC
There is no nesting of lists or other data structures in TI-89 BASIC, short of using variable names as pointers.
Trith
<lang trith>[[1] 2 [[3 4] 5] [[[]]] [[[6]]] 7 8 []] flatten</lang>
VBScript
Working on embedded arrays as that's about the closest we get to lists.
Implementation
<lang vb> class flattener dim separator
sub class_initialize separator = "," end sub
private function makeflat( a ) dim i dim res for i = lbound( a ) to ubound( a ) if isarray( a( i ) ) then res = res & makeflat( a( i ) ) else res = res & a( i ) & separator end if next makeflat = res end function
public function flatten( a ) dim res res = makeflat( a ) res = left( res, len( res ) - len(separator)) res = split( res, separator ) flatten = res end function
public property let itemSeparator( c ) separator = c end property end class </lang>
Invocation
<lang vb> dim flat set flat = new flattener flat.itemSeparator = "~" wscript.echo join( flat.flatten( array( array( 1 ),2,array(array(3,4),5),array(array(array())),array(array(array(6))),7,8,array())), "!") </lang>
Output
<lang vb> 1!2!3!4!5!6!7!8 </lang>