views:

127

answers:

2

I'm looking for an elegant way to turn this array:

Array (
  [foo] => 1
  [bar] => 1
  [zim] => 3
  [dib] => 6
  [gir] => 1
  [gaz] => 3
)

Into this array:

Array (
  [1] => Array ( foo, bar, gir ),
  [3] => Array ( zim, gaz ),
  [6] => Array ( dib )
)

Note:, there is no relationship between the keys or values. They are completely arbitrary and used as examples only. The resulting array should be an associative array grouped by the values of the input array.

Thanks!

+10  A: 
$input = array(
  'foo' => 1,
  'bar' => 1,
  'zim' => 3,
  'dib' => 6,
  'gir' => 1,
  'gaz' => 3
)

$output = array();
foreach ( $input as $k => $v ) {
  if ( !isset($output[$v]) ) {
    $output[$v] = array();
  }

  $output[$v][] = $k;
}
hsz
+3  A: 

I think this will do it just fine:

foreach ($arr1 as $k => $val) $arr2[$val][] = $k;

where $arr1 is the original array outputting the new array to $arr2.

animuson
I thought this would've generated a PHP warning, but nope! Concise and sweet FTW!
macek
this indeed doesn't generate a warning, but relying on a php bug feels quite dirty. hsz's code is cleaner.
stereofrog
@stereofrog, can you detail the "bug"?
macek
@macek: no warning when using an undefined variable is clearly a bug
stereofrog
@stereofrog, I suppose that's a fair argument. You're right, too. I shouldn't sacrifice reliable functionality just for the sake of trimming a line or two out.
macek
@stereofrog: I could have sworn I've seen this type of example in the manual pages. Cause if what your saying is true my code is totally f`ed` up.
Gutzofter
I never define variables first, what a waste of a line of code. That's why I use PHP, because you don't have too.
animuson