在数组中查找相似的字符串

Finding similar strings in array

我需要利用 similar_text() 的值数组,如下所示:

$strings = ["lawyer" => 3, "business" => 3, "lawyers" => 1, "a" => 3];

我想做的是在上面的数组中找到几乎相同的词,即 lawyerlawyers,然后将它们的计数加到一个新数组中.

因此 lawyer 将是 4,因为 lawyers 将关联到 lawyer 的原始字符串。

请记住,这个数组只会是单数词,长度未指定,范围可以从 1>99

我不知道从哪里开始,所以我用 foreach 循环解决了这个问题,如下所示,但预期的输出与预期不符。

foreach ( $strings as $key_one => $count_one ) {
    foreach ( $strings as $key_two => $count_two ) {
        similar_text($key_two, $key_one, $percent);
        if ($percent > 80) {
            if(!isset($counts[$key_one])) {
                $counts[$key_one] = $count_one;
            } else {
                $counts[$key_one] += $count_two;
            }
        }
    }
}

注意: 此示例的匹配百分比为 80(作为 lawyer 的匹配& lawyers~92%)

这最终给了我类似于以下内容的内容:

Array
(
    [lawyer] => 4
    [business] => 3
    [a] => 3
    [lawyers] => 2
)

我要求的位置:

Array
(
    [lawyer] => 4
    [business] => 3
    [a] => 3
)

请注意我是如何要求它实际删除 lawyers 并将计数添加到 lawyer.

您可以随时使用

unset( $counts[$key_two] ) ;

你的困难在于,正如律师与律师相似,律师与律师也相似。所以他们俩的计数都被对方提高了。

试试这个:

foreach ( $strings as $key_one => &$count_one ) {
    if ($count_one == 0) continue; // skip it if we've already processed it
    if (!isset($counts[$key_one]) {
        $counts[$key_one] = $count_one;
        $count_one = 0;
    }
    foreach ( $strings as $key_two => &$count_two ) {
        similar_text($key_two, $key_one, $percent);
        if ($percent > 80) {
            $counts[$key_one] += $count_two;
            $count_two = 0;
        }
    }
}

这样做的缺点是您更改了可能不理想的原始 $strings 数组。这是另一种方法,在另一个散列中跟踪已处理的字符串:

$already = $counts = array(); // not really necessary, but nice to init
foreach ( $strings as $key_one => $count_one ) {
    if (isset($already[$key_one])) continue; // skip if already processed
    $counts[$key_one] = $count_one; // by definition this should be new
    foreach ( $strings as $key_two => $count_two ) {
        similar_text($key_two, $key_one, $percent);
        if ($percent > 80) {
            $counts[$key_one] += $count_two;
            $already[$key_two] = true;
        }
    }
}

我会推荐第二种解决方案。