我已经使用了array_keys($obj) !== range(0, count($obj) - 1)和array_values($arr) !== $arr(它们是彼此的对偶，尽管第二个比第一个更便宜)，但对于非常大的数组都失败了。

这是因为array_keys和array_values都是非常昂贵的操作(因为它们构建了一个大小与原始数组大致相同的全新数组)。

下面的函数比上面提供的方法更健壮:

function array_type( $obj ){
    $last_key = -1;
    $type = 'index';
    foreach( $obj as $key => $val ){
        if( !is_int( $key ) || $key < 0 ){
            return 'assoc';
        }
        if( $key !== $last_key + 1 ){
            $type = 'sparse';
        }
        $last_key = $key;
    }
    return $type;
}

还要注意，如果你不关心区分稀疏数组和关联数组，你可以简单地从两个if块中返回'assoc'。

最后，虽然这可能看起来没有本页上的许多“解决方案”那么“优雅”，但实际上它的效率要高得多。几乎任何关联数组都会立即被检测到。只有索引数组才会被彻底检查，上面列出的方法不仅会彻底检查索引数组，还会复制它们。

我认为下面两个函数是最好的方法去检查'如果一个数组是结合或数字'。由于'numeric'可能只表示数字键或顺序数字键，下面列出了两个函数来检查这两种情况:

function is_indexed_array(&$arr) {
  for (reset($arr); is_int(key($arr)); next($arr));
  return is_null(key($arr));
}

function is_sequential_array(&$arr, $base = 0) {
  for (reset($arr), $base = (int) $base; key($arr) === $base++; next($arr));
  return is_null(key($arr));
}

The first function checks if each key is an integer value. The second function checks if each key is an integer value and in addition checks if all keys are sequential starting at $base, which defaults to 0 and thus can be omitted if you do not need to specify another base value. key($my_array) returns null if the read pointer is moved past the end of the array, which is what ends the for loop and makes the statement after the for loop return true if all keys were integer. If not, the loop ends prematurely because a key is of type string, and the statement after the for loop will return false. The latter function in addition adds one to $base after each compare, to be able to check if the next key is of the correct value. The strict compare makes it also check if the key is of type integer. The $base = (int) $base part in the first section of the for loop can be left out when $base is omitted or if you make sure it is only called using an integer. But since I can't be sure for everybody, I left it in. The statement is executed only once, anyway. I think these are the most efficient solutions:

Memory wise: No copying of data or key ranges. Doing an array_values or array_keys may seem shorter (less code) but keep in mind what goes on in the background once you make that call. Yes there are more (visible) statements than in some other solutions, but that is not what counts, is it? Time wise: Besides the fact that copying/extracting data and/or keys also takes time, this solution is more efficient than doing a foreach. Again a foreach may seem more efficient to some because it is shorter in notation, but in the background foreach also calls reset, key and next to do it's looping. But in addition it also calls valid to check the end condition, which is avoided here due to the combination with the integer check.

请记住，数组键只能是整数或字符串，严格的数字字符串，如“1”(但不是“01”)将被转换为整数。如果您希望数组是顺序的，这使得检查整数键成为除了计数之外唯一需要的操作。当然，如果is_indexed_array返回false，则可以将该数组视为关联数组。我说“见过”，因为事实上它们都是。

或者你可以用这个:

Arr::isAssoc($array)

它将检查数组是否包含任何非数字键或:

Arr:isAssoc($array, true)

检查数组是否严格顺序(包含自动生成的int键0到n-1)

使用这个库。

简单和性能友好的解决方案，只检查第一个键。

function isAssoc($arr = NULL)
{
    if ($arr && is_array($arr))
    {
        foreach ($arr as $key => $val)
        {
            if (is_numeric($key)) { return true; }

            break;
        }
    }

    return false;
}

简单地说，你可以通过以下步骤检查数组是否关联

使用array_keys()将数组的所有键转换为一个数组 使用array_filter()和从数组中过滤非数字键 is_numeric () 比较过滤数组和实际数组中的元素数量，如果两个数组中的元素数量不相等，则为关联数组。

以上步骤的功能如下所示。

 function isAssociative(array $array)
    {
        return count(array_filter(array_keys($array), function($v){return is_numeric($v);})) !== count($array));
    }

大多数答案的时间/空间复杂度都不是最佳的，或者语义正在发生变化。所以，这里有另一个最快和功能最正确的解决方案:

function is_sequential_array(Array &$a) {
    $n = count($a);
    for($i=0; $i<$n; $i++) {
        if(!array_key_exists($i, $a)) {
            return false;
        }
    }
    return true;
}

与其他答案相比，这个答案有以下优点:

O(1)的空间复杂度(这里很多答案使用O(n)个空间!) 不对键应用相等性(这是一种不必要且代价高昂的操作) 将输入数组视为不可变(许多答案通过应用突变函数隐式创建了一个副本) 使用函数array_key_exists代替isset(记住，isset额外检查'is not null'，从而改变语义) 最坏情况时间复杂度为O(n)(这里许多答案的最佳情况时间复杂度为O(n))