如何在数组中搜索字符串的一部分？

Question

我想搜索整个字符串或字符串的一部分是否是数组的一部分。在 PHP 中如何实现这一点？

另外，我如何在其中使用 metaphone ？

示例：

array1={'India','USA','China'};
array2={'India is in east','United States of America is USA','Made in China'}

如果我在 array2 中搜索 array1，则：

“印度”应匹配“印度位于东部”，对于美国和印度也是如此。中国。

Answer 1

$array1 = array('India','USA','China');
$array2 = array('India is in east','United States of America is USA','Made in China');
$found = array();

foreach ($array1 as $key => $value) {
    // Thanks to @Andrea for this suggestion:
    $found[$value] = preg_grep("/$value/", $array2);
    // Alternative:
    //$found = $found + preg_grep("/$value/", $array2);
}

print_r($found);

结果：

Array
(
    [0] => India is in east
    [1] => United States of America is USA
    [2] => Made in China
)

使用 Metaphone 比较棘手。您必须确定什么构成匹配。一种方法是使用要比较的两个值的 Methaphone 结果之间的编辑距离。

更新：请参阅@Andrea 的 解决方案可实现更合理的每个单词变音位比较。

这是一个粗略的例子：

$meta1 = array_map(
    create_function( '$v', 'return array(metaphone($v) => $v);' ),
    $array1
);

$meta2 = array_map(
    create_function( '$v', 'return array(metaphone($v) => $v);' ),
    $array2
);

$threshold = 3;

foreach ($meta2 as $key2 => $value2) {

    $k2 = key($value2);
    $v2 = $value2[$k2];

    foreach ($meta1 as $key1 => $value1) {

        $k1  = key($value1);
        $v1  = $value1[$k1];
        $lev = levenshtein($k2, $k1);

        if( strpos($v2, $v1) !== false || levenshtein($k2, $k1) <= $threshold ) {
            array_push( $found, $v2 );
        }
    }
}

......但它需要工作。如果阈值太高，它会产生重复项。您可能更喜欢分两次进行比赛。一个用于查找简单匹配，如我的第一个代码示例中所示，然后另一个用于与 Metaphone 匹配（如果第一个没有返回匹配项）。

Answer 2

变音位盒也可以遵循 Mike 针对严格盒提出的相同结构。

我认为不需要额外的相似性函数，因为变音位的目的应该是为我们提供一个发音相同的单词所共有的键。

$array1 = array('India','USA','China');
$array2 = array(
    'Indiuh is in east',
    'United States of America is USA',
    'Gandhi was born in India',
    'Made in China'
);
$found = array();
foreach ($array1 as $key => $value) {
    $found[$value] = preg_grep('/\b'.$value.'\b/i', $array2);
}

var_export($found);

echo "\n\n";

function meta( $sentence )
{
    return implode(' ', array_map('metaphone', explode(' ', $sentence)));
}

$array2meta = array_map('meta', $array2);
foreach ($array1 as $key => $value) {
    $valuemeta = meta($value);
    $foundmeta[$value] = preg_grep('/\b'.$valuemeta.'\b/', $array2meta);
    $foundmeta[$value] = array_intersect_key($array2, $foundmeta[$value]);
}

var_export($foundmeta);

上面的代码打印出：

array (
  'India' => 
  array (
    2 => 'Gandhi was born in India',
  ),
  'USA' => 
  array (
    1 => 'United States of America is USA',
  ),
  'China' => 
  array (
    3 => 'Made in China',
  ),
)

array (
  'India' => 
  array (
    0 => 'Indiuh is in east',
    2 => 'Gandhi was born in India',
  ),
  'USA' => 
  array (
    1 => 'United States of America is USA',
  ),
  'China' => 
  array (
    3 => 'Made in China',
  ),
)

Answer 3

$a1 = array('India','USA','China');
$a2 = array('India is in east','United States of America is USA','Made in China');


foreach ( $a2 as $a )
{
  foreach( $a1 as $b  )
  {
    if ( strpos( $a, $b ) > -1 )
    {
      echo $a . " contains " . $b . "\n";
    }
  }
}