如何在PHP中迭代多个具有多个数组值的字符串？

Question

*来自revo答案的更新问题

这是一个工作脚本,带有一组更好的示例字符串来显示我的意图 -

$strings[] = 'seventy five yards out';
$strings[] = 'sixty yards out';
$strings[] = 'one hundred fifty yards out';

$inputString = 'seventy two yards out';
$inputWords = str_word_count($inputString, 1);

$foundWords = [];

foreach ($strings as $key => $string) {
    $stringWords = str_word_count($string, 1);
    $wordsCount = array_count_values($stringWords);
    $commonWords = array_intersect($inputWords, array_keys($wordsCount));
    if (count($commonWords) > 0) {
        foreach ($commonWords as $commonWord) {
            $foundWords[$key][$commonWord] = $wordsCount[$commonWord];
        }
    }
}

print_r($foundWords);

如何将它打印出"七十五码外",因为它实际上最接近文本？我正在考虑将字数除以得到一个百分比,但现在认为现在可能有效了..

Answer 1

关键是str_word_count()对每个提供的字符串分别执行 a 。通过这种方式，我们可以将其转换为数组，并且根据您的需要处理数组要简单得多。

array_count_values()计算数组的值，从而得出单词出现的次数。

$strings[] = 'seventy five yards out';
$strings[] = 'sixty yards out';
$strings[] = 'one hundred fifty yards out';

$inputString = 'seventy two yards out';
$inputWords = str_word_count($inputString, 1);

$probabilities = [];

foreach ($strings as $key => $string) {
    $stringWords = str_word_count($string, 1);
    $wordsCount = array_count_values($stringWords);
    $commonWords = array_intersect($inputWords, array_keys($wordsCount));
    if (count($commonWords) > 0) {
        foreach ($commonWords as $commonWord) {
            if (!isset($probabilities[$key])) $probabilities[$key] = 0;
            $probabilities[$key] += $wordsCount[$commonWord];
        }
        $probabilities[$key] /= count($stringWords);
    }
}
arsort($probabilities);
echo $strings[key($probabilities)];

输出：

seventy five yards out

概率print_r($probabilities);：

Array
(
    [0] => 0.75
    [1] => 0.66666666666667
    [2] => 0.4
)

现场演示