Question

我有一个像：

这样的数组

arr = array("*" , "$" , "and" , "or" , "!" ,"/");

和另一个字符串如：

string = "this * is very beautiful but $ is more important in life.";

我正在寻找以最低成本找到此字符串中数组成员的最有效方法。此外，我需要在结果中有一个数组，可以显示字符串中存在哪些成员。

最简单的方法是使用for循环，但我相信应该有更有效的方法在PHP中执行此操作。

Answer 1

$arr=array("*" , "$" , "#" , "!");

$r = '~[' . preg_quote(implode('', $arr)) . ']~';

$str = "this * is very beautiful but $ is more important in life.";

preg_match_all($r, $str, $matches);

echo 'The following chars were found: ' . implode(', ', $matches[0]);

Answer 2

您可以使用array_insersect

$string = "this * is very beautiful but $ is more important in life.";
$arr=array("*" , "$" , "#" , "!");

$str = str_split($string);

$out = array_intersect($arr, $str);

print_r($out);

此代码将生成以下输出

数组（[0] =＆gt; * [1] =＆gt; $）

Answer 3

如果您正在寻找最有效的方法，以下代码的结果是：

预浸料：1.03257489204

array_intersect：2.62625193596

strpos：0.814728021622

看起来像循环数组并使用strpos进行匹配是最有效的方法。

        $arr=array("*" , "$" , "#" , "!");

        $string="this * is very beautiful but $ is more important in life.";



        $time = microtime(true);

        for ($i=0; $i<100000; $i++){

            $r = '~[' . preg_quote(implode('', $arr)) . ']~';

            $str = "this * is very beautiful but $ is more important in life.";

            preg_match_all($r, $str, $matches);

        }

        echo "preg:". (microtime(true)-$time)."\n";

        $time = microtime(true);

        for ($i=0; $i<100000; $i++){

            $str = str_split($string);

            $out = array_intersect($arr, $str);

        }


        echo "array_intersect:". (microtime(true)-$time)."\n";


        $time = microtime(true);

        for ($i=0; $i<100000; $i++){

            $res = array();
            foreach($arr as $a){
                if(strpos($string, $a) !== false){
                    $res[] = $a;
                }
            }

        }

        echo "strpos:". (microtime(true)-$time)."\n";

有效检查字符串中的数组项

3 个答案: