Checking if array is multidimensional or not?

Question

1. What is the most efficient way to check if an array is a flat array of primitive values or if it is a multidimensional array?
2. Is there any way to do this without actually looping through an array and running is_array() on each of its elements?

Answer 1

This function will return int number of array dimensions (stolen from here).

function countdim($array)
{
   if (is_array(reset($array))) 
     $return = countdim(reset($array)) + 1;
   else
     $return = 1;
 
   return $return;
}

Answer 2

After PHP 7 you could simply do:

public function is_multi(array $array):bool
{
    return is_array($array[array_key_first($array)]);
}

Answer 3

$is_multi_array = array_reduce(array_keys($arr), function ($carry, $key) use ($arr) { return $carry && is_array($arr[$key]); }, true);

Here is a nice one liner. It iterates over every key to check if the value at that key is an array. This will ensure true

Answer 4

Its as simple as

$isMulti = !empty(array_filter($array, function($e) {
                    return is_array($e);
                }));

Answer 5

In my case. I stuck in vary strange condition.
1st case = array("data"=> "name");
2nd case = array("data"=> array("name"=>"username","fname"=>"fname"));
But if data has array instead of value then sizeof() or count() function not work for this condition. Then i create custom function to check.
If first index of array have value then it return "only value"
But if index have array instead of value then it return "has array"
I use this way

 function is_multi($a) {
        foreach ($a as $v) {
          if (is_array($v)) 
          {
            return "has array";
            break;
          }
          break;
        }
        return 'only value';
    }

Special thanks to Vinko Vrsalovic

Answer 6

Don't use COUNT_RECURSIVE

click this site for know why

use rsort and then use isset

function is_multi_array( $arr ) {
rsort( $arr );
return isset( $arr[0] ) && is_array( $arr[0] );
}
//Usage
var_dump( is_multi_array( $some_array ) );

Answer 7

Even this works

is_array(current($array));

If false its a single dimension array if true its a multi dimension array.

current will give you the first element of your array and check if the first element is an array or not by is_array function.

Answer 8

I think this one is classy (props to another user I don't know his username):

static public function isMulti($array)
{
    $result = array_unique(array_map("gettype",$array));

    return count($result) == 1 && array_shift($result) == "array";
}

Answer 9

You can also do a simple check like this:

$array = array('yo'=>'dream', 'mydear'=> array('anotherYo'=>'dream'));
$array1 = array('yo'=>'dream', 'mydear'=> 'not_array');

function is_multi_dimensional($array){
    $flag = 0;
    while(list($k,$value)=each($array)){
        if(is_array($value))
            $flag = 1;
    }
    return $flag;
}
echo is_multi_dimensional($array); // returns 1
echo is_multi_dimensional($array1); // returns 0

Answer 10

Use count() twice; one time in default mode and one time in recursive mode. If the values match, the array is not multidimensional, as a multidimensional array would have a higher recursive count.

if (count($array) == count($array, COUNT_RECURSIVE)) 
{
  echo 'array is not multidimensional';
}
else
{
  echo 'array is multidimensional';
}

This option second value mode was added in PHP 4.2.0. From the PHP Docs:

If the optional mode parameter is set to COUNT_RECURSIVE (or 1), count() will recursively count the array. This is particularly useful for counting all the elements of a multidimensional array. count() does not detect infinite recursion.

However this method does not detect array(array()).

Answer 11

I think you will find that this function is the simplest, most efficient, and fastest way.

function isMultiArray($a){
    foreach($a as $v) if(is_array($v)) return TRUE;
    return FALSE;
}

You can test it like this:

$a = array(1 => 'a',2 => 'b',3 => array(1,2,3));
$b = array(1 => 'a',2 => 'b');

echo isMultiArray($a) ? 'is multi':'is not multi';
echo '<br />';
echo isMultiArray($b) ? 'is multi':'is not multi';

Answer 12

For PHP 4.2.0 or newer:

function is_multi($array) {
    return (count($array) != count($array, 1));
}

Answer 13

The short answer is no you can't do it without at least looping implicitly if the 'second dimension' could be anywhere. If it has to be in the first item, you'd just do

is_array($arr[0]);

But, the most efficient general way I could find is to use a foreach loop on the array, shortcircuiting whenever a hit is found (at least the implicit loop is better than the straight for()):

$ more multi.php
<?php

$a = array(1 => 'a',2 => 'b',3 => array(1,2,3));
$b = array(1 => 'a',2 => 'b');
$c = array(1 => 'a',2 => 'b','foo' => array(1,array(2)));

function is_multi($a) {
    $rv = array_filter($a,'is_array');
    if(count($rv)>0) return true;
    return false;
}

function is_multi2($a) {
    foreach ($a as $v) {
        if (is_array($v)) return true;
    }
    return false;
}

function is_multi3($a) {
    $c = count($a);
    for ($i=0;$i<$c;$i++) {
        if (is_array($a[$i])) return true;
    }
    return false;
}
$iters = 500000;
$time = microtime(true);
for ($i = 0; $i < $iters; $i++) {
    is_multi($a);
    is_multi($b);
    is_multi($c);
}
$end = microtime(true);
echo "is_multi  took ".($end-$time)." seconds in $iters times\n";

$time = microtime(true);
for ($i = 0; $i < $iters; $i++) {
    is_multi2($a);
    is_multi2($b);
    is_multi2($c);
}
$end = microtime(true);
echo "is_multi2 took ".($end-$time)." seconds in $iters times\n";
$time = microtime(true);
for ($i = 0; $i < $iters; $i++) {
    is_multi3($a);
    is_multi3($b);
    is_multi3($c);
}
$end = microtime(true);
echo "is_multi3 took ".($end-$time)." seconds in $iters times\n";
?>

$ php multi.php
is_multi  took 7.53565130424 seconds in 500000 times
is_multi2 took 4.56964588165 seconds in 500000 times
is_multi3 took 9.01706600189 seconds in 500000 times

Implicit looping, but we can't shortcircuit as soon as a match is found...

$ more multi.php
<?php

$a = array(1 => 'a',2 => 'b',3 => array(1,2,3));
$b = array(1 => 'a',2 => 'b');

function is_multi($a) {
    $rv = array_filter($a,'is_array');
    if(count($rv)>0) return true;
    return false;
}

var_dump(is_multi($a));
var_dump(is_multi($b));
?>

$ php multi.php
bool(true)
bool(false)

Answer 14

You could look check is_array() on the first element, under the assumption that if the first element of an array is an array, then the rest of them are too.