Flatten a list
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
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>
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 can be a list, and therefore they can not be nil.
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>
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>
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>
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 java>import java.util.LinkedList; import java.util.List;
public class Flatten {
public static List flatten(List list){
List retVal = new LinkedList();
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 test = new LinkedList();
LinkedList one = new LinkedList();
one.add(1);
LinkedList tff = new LinkedList(); LinkedList tf = new LinkedList(); tf.add(3); tf.add(4); tff.add(tf); tff.add(5);
LinkedList empty1 = new LinkedList(); LinkedList empty2 = new LinkedList(); LinkedList empty3 = new LinkedList(); empty2.add(empty3); empty1.add(empty2);
LinkedList six = new LinkedList(); LinkedList inner1 = new LinkedList(); LinkedList inner2 = new LinkedList(); inner2.add(6); inner1.add(inner2); six.add(inner1);
test.add(one); test.add(2); test.add(tff); test.add(empty1); test.add(six); test.add(7); test.add(8); test.add(new LinkedList());
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, []]; var f = a.flatten();
f.each(function(e){document.write("
")});
</script></lang> output:
1 number 2 number 3 number 4 number 5 number 6 number 7 number 8 number
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>
Perl
<lang perl>sub flatten {
map { ref eq 'ARRAY' ? flatten(@$_) : $_ } @_
}
my @lst = ([1], 2, [[3,4], 5], [[[]]], [[[6]]], 7, 8, []); print flatten(@lst), "\n";</lang>
PHP
<lang php>/* Note: This code is only for PHP 4.
It won't work on PHP 5 due to the change in behavior of array_merge(). */
while (array_filter($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($lst); ?></lang>
Python
Recursive
Function flatten is recursive and preserves its argument:
<lang python>>>> def flatten(lst):
return sum( ([x] if type(x) is not list else flatten(x)
for x in lst), [] )
>>> lst = [[1], 2, [[3,4], 5], [[[]]], [[[6]]], 7, 8, []] >>> flatten(lst) [1, 2, 3, 4, 5, 6, 7, 8]</lang>
Non-recursive
Function flat is iterative and flattens the list in-place. It follows the Python idiom of returning None when acting in-place: <lang python>>>> def flat(lst):
i=0
while i<len(lst):
while True:
try:
lst[i:i+1] = lst[i]
except (TypeError, IndexError):
break
i += 1
>>> lst = [[1], 2, [[3,4], 5], [[[]]], [[[6]]], 7, 8, []] >>> flat(lst) >>> lst [1, 2, 3, 4, 5, 6, 7, 8]</lang>
R
<lang R> l <- list(list(1), 2, list(list(3, 4), 5), list(list(list())), list(list(list(6))), 7, 8, list()) unlist(l) </lang>
Ruby
flatten is a built-in method of Arrays
<lang ruby>flat = [[1], 2, [[3,4], 5], [[[]]], [[[6]]], 7, 8, []].flatten
p flat # => [1, 2, 3, 4, 5, 6, 7, 8]</lang>
Scheme
<lang scheme>> (define (flatten x)
(cond ((null? x) '())
((not (pair? x)) (list x))
(else (append (flatten (car x))
(flatten (cdr x))))))
> (flatten '((1) 2 ((3 4) 5) ((())) (((6))) 7 8 ())) (1 2 3 4 5 6 7 8)</lang>
Slate
<lang slate> s@(Sequence traits) flatten [
[| :out | s flattenOn: out] writingAs: s
].
s@(Sequence traits) flattenOn: w@(WriteStream traits) [
s do: [| :value |
(value is: s)
ifTrue: [value flattenOn: w]
ifFalse: [w nextPut: value]].
].</lang>
Smalltalk
<lang smalltalk>OrderedCollection extend [
flatten [ |f|
f := OrderedCollection new.
self do: [ :i |
i isNumber
ifTrue: [ f add: i ]
ifFalse: [ |t|
t := (OrderedCollection withAll: i) flatten.
f addAll: t
]
].
^ f
]
].
|list|
list := OrderedCollection
withAll: { {1} . 2 . { {3 . 4} . 5 } .
{{{}}} . {{{6}}} . 7 . 8 . {} }.
(list flatten) printNl.</lang>
Tcl
<lang tcl>proc flatten list {
for {set old {}} {$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.
TI-89 BASIC
There is no nesting of lists or other data structures in TI-89 BASIC, short of using variable names as pointers.