测试一个数组是否是另一个数组的子集

How can I determine if one array is a subset of another (all elements in the first are present in the second)?

如何确定一个数组是否是另一个数组的子集（第一个数组中的所有元素都存在于第二个数组中）？

 $s1 = "string1>string2>string3>string4>string5>string6>";
 $arr1 = explode(">", $s1);
 $s2 = "string1>string4>string5";
 $arr2 = explode(">", $s2);

 $isSubset = /* ??? */

6 个解决方案

#1

If you start from strings, you could check strstr($fullString,$subsetStr);. But that'll only work when all chars have the same order: 'abcd','cd' will work, but 'abcd','ad' won't.

如果从字符串开始，可以检查strstr（$ fullString，$ subsetStr）;.但这只会在所有字符具有相同顺序时起作用：'abcd'，'cd'可以工作，但'abcd'，'ad'不会。

But instead of writing your own, custom, function you should know that PHP has TONS of array functions, so its neigh on impossible that there isn't a std function that can do what you need it to do. In this case, I'd suggest array_diff:

但是你不应该编写自己的自定义函数，而应该知道PHP有大量的数组函数，因此不可能没有std函数可以做你需要它做的事情。在这种情况下，我建议使用array_diff：

$srcString = explode('>','string1>string2>string3>string4>string5');
$subset = explode('>','string3>string2>string5');
$isSubset = array_diff($subset,$srcString);
//if (empty($isSubset)) --> cf comments: somewhat safer branch:
if (!$isSubset)
{
    echo 'Subset';
    return true;
}
else
{
    echo 'Nope, substrings: '.implode(', ',$isSubset).' Didn\'t match';
    return false;
}

#2

if (array_intersect($array1, $array2) == $array1) {
    // $array1 is a subset of $array2
}

#3

Simple: use array subtraction.

简单：使用数组减法。

On array subtraction, you will know whether or not one array is a subset of the other.

在数组减法中，您将知道一个数组是否是另一个数组的子集。

Example:

例：

if (!array_diff($array1, $array2)) {
    // $array1 is a subset of $array2
}

Reference: array_diff

参考：array_diff

You can use array_intersect also.

您也可以使用array_intersect。

#4

I would create an associated array of the larger array, then iterate through the smaller array, looking for a non collision, if you find one, return false.

我将创建一个较大数组的关联数组，然后遍历较小的数组，寻找非碰撞，如果找到一个，则返回false。

function isSubset($arr1,$arr2){
    $map = Array();
    for ($i=0;$i<count($arr1);$i++){
      $map[$arr[$i]]=true;
    }
    for ($i=0;$i<count($arr2);$i++){
       if (!isset($map[$arr2[$i]])){
          return false;
       }
    }
    return true;

#5

$s1 = "1>2>3>4>5>6>7";

$arr1 = explode(">",$s1);

$s2 = "1>2>3";

$arr2 = explode(">",$s2); 

if(isSub($arr1,$arr2)){

         echo 'true';

}else{

         echo 'false';
}

function isSub($a1,$a2){

    $num2 = count($a2);
    $sub  = $num2;

    for($i = 0;$i < $num2 ;$i++){
        if(in_array($a2[$i],$a1)){
            $sub--;
        }
    }
    return ($sub==0)? true:false;
}

#6

Simple function which will return true if array is exact subset otherwise false. Solution is applicable for two dimensional array as well.

简单函数，如果数组是精确子集，则返回true，否则返回false。解决方案也适用于二维阵列。

 function is_array_subset($superArr, $subArr) {
        foreach ($subArr as $key => $value) {
            //check if keys not set in super array OR values are unequal in both array.
            if (!isset($superArr[$key]) || $superArr[$key] != $value) {
                return false;
            }
        }
        return true;
    }

#1