PHP：如果在多维数组中发现重复项，则合并相邻值

我有一些 PHP 变量集，我正在从中创建一个多维数组。现在，在该数组中，我想检查特定的键（[font]) 对于重复项。

如果发现重复项，则对应的相应值[lang] and [weight]应该合并。

这是我到目前为止所尝试的（这会取消设置/删除数组中的重复值）：

// Font [0]
$font_1 = "Poppins";
$font_1_l = "Hindi, English";
$font_1_w = "700, 700i";

// Font [1]
$font_2 = "Lora";
$font_2_l = "Vietnamese, Japanese";
$font_2_w = "200, 300, 400, 400i";

// Font [2]
$font_3 = "Noto Sans";
$font_3_l = "Punjabi, Latin, Hindi";
$font_3_w = "200, 200i, 300, 300i, 400, 500";

// Font [3]
$font_4 = "Lora";
$font_4_l = "Greek, Roman, Vietnamese";
$font_4_w = "400, 400i, 500, 500b";

// Array of all the values
$font_f = array( array( 'font' => $font_1, 'lang' => $font_1_l, 'weight' => $font_1_w ), array( 'font' => $font_2, 'lang' => $font_2_l, 'weight' => $font_2_w ), array( 'font' => $font_3, 'lang' => $font_3_l, 'weight' => $font_3_w ), array( 'font' => $font_4, 'lang' => $font_4_l, 'weight' => $font_4_w ) ); 

// Printing the array for testing
echo "<pre>";
print_r( array_map("unserialize", array_unique(array_map("serialize", $font_f))) );

// Removing duplicates
$font_f_copy = $font_f; // Copy of $font_f for modification
$fonts = array(); // To get unique fonts

for( $i=0; $i<count($font_f); $i++ ) {
  if ( in_array( $font_f[$i]['font'], $fonts ) ) {
    unset($font_f_copy[$i]);
  }
  else {
    $fonts[] = $font_f[$i]['font'];
  }
}

// Printing $font_f_copy for testing
print_r($font_f_copy);

Output:

Array
(
    [0] => Array
        (
            [font] => Poppins
            [lang] => Hindi, English
            [weight] => 700, 700i
        )

    [1] => Array
        (
            [font] => Lora
            [lang] => Vietnamese, Japanese
            [weight] => 200, 300, 400, 400i
        )

    [2] => Array
        (
            [font] => Noto Sans
            [lang] => Punjabi, Latin, Hindi
            [weight] => 200, 200i, 300, 300i, 400, 500
        )

    [3] => Array
        (
            [font] => Lora
            [lang] => Greek, Roman, Vietnamese
            [weight] => 400, 400i, 500, 500b
        )

)
Array
(
    [0] => Array
        (
            [font] => Poppins
            [lang] => Hindi, English
            [weight] => 700, 700i
        )

    [1] => Array
        (
            [font] => Lora
            [lang] => Vietnamese, Japanese
            [weight] => 200, 300, 400, 400i
        )

    [2] => Array
        (
            [font] => Noto Sans
            [lang] => Punjabi, Latin, Hindi
            [weight] => 200, 200i, 300, 300i, 400, 500
        )

)

正如您在上面的代码中看到的，Font [1] 和 Font [3] 将具有相同的值[font]即洛拉，所以[lang] and [weight]对于字体 [1] 应合并[lang] and [weight]分别为字体 [3]。

我想知道如何继续实现这一目标。

我个人会做类似的事情，而不是 for 循环：

$result = [];
foreach ($font_f as $f) {
     if (isset($result[$f["font"]])) {
         $result[$f["font"]] = [ 
               "font" => $f["font"], 
               "lang" => $f["lang"]." ,".$result[$f["font"]]["lang"],
               "weight" => $f["weight"]." ,".$result[$f["font"]]["weight"]
         ]; 
     } else {
        $result[$f["font"]] = $f;
     }
}