*更新来自雷沃回答的问题

这里有一个工作脚本，它提供了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);

我怎样才能打印出'七十五码'呢，因为它实际上最接近文字？ 我正在考虑将字数统计分配到一个百分比，但现在认为现在可能会有效。

---

关键是分别对每个提供的字符串执行str_word_count() 。 通过这种方式，我们正在转换为数组，处理数组对于您所期望的更加简单。

array_count_values()计算导致出现单词出现次数的数组值。

$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

---

像这样的东西应该工作：

<?php

$g = 'the weather is nice'; // strings to loop through
$n = 'the water is blue';
$b = 'that was a bad movie';

$t = 'hows the weather';  // example input
$test = (str_word_count($t, 1)); // breaks out each word into array

// Comparisons
$comps = array();
// Array sums
$sums = array();
// Search each variable that's been set, as long as it's less that 't'
// A "for" loop will accept letters in addition to numbers, so we'll start with the
// letter "a" and loop through each letter up to "s" (which is one less than "t")
for ($inc = 'a'; $inc < 't'; $inc++) {
  // Now, a variable assigned as $$inc will translate into $a, $b, $c ... $s
  // and if $a, $b, $c, etc, are set...
  if (isset($$inc)) {
    // ... assign them to the $comps array with a key of $$inc
    $comps[$$inc] = str_word_count($$inc, 1);

    // For example, when the "for" loop reaches "f", nothing will be added to the
    // $comps array because $f is not set above.

    // But when it gets to "g" it'll find that $g HAS been set, and that it has a
    // value of "the weather is nice". At this point the $comps array will now look
    // like this:
    // $comps['the weather is nice'] = array('the', 'weather', 'is', 'nice');

    // If you'd like to see this in action (since it might sound a little confusing),
    // remove the # from the beginning of each of the following lines that start with #
    // (there should be 10 total):

    #print "<pre>The loop has reached the letter <b>{$inc}</b> for the value of ";
    #print "<b>$inc</b> and has found that <b>${$inc}</b> HAS been set in the code.n";
    #print "Adding another dollar sign to <b>$inc</b> has had the following effects:n";
    #print "- <b>$inc</b> now looks like <b>$$inc</b> (from within the written part of the code)n";
    #print "- <b>$$inc</b> translates into <b>${$inc}</b> (the variable that is acually being evaluated)n";
    #print "- <b>${$inc}</b> evaluates to <b>{$$inc}</b>n</pre>";
  }
  #else {
  #  print "<pre>The loop has reached the letter <b>{$inc}</b> for the value of <b>$inc</b>";
  #  print " and has found that <b>${$inc}</b> has NOT been set in the code, so it's being skipped.n";
  #}
}
// Avoid errors by checking if empty or not
if (!empty($comps)) {
  foreach ($comps as $key => $comp) {
    // Find intersections, if any
    $candidates[$key] = array_intersect($test, $comp);
    // Count the intersections
    $counts[$key] = array_count_values($candidates[$key]);
    // Add up the intersections
    $sums[$key] = array_sum($counts[$key]);
  }
}
$winner = '';
if (!empty($sums)) {
  // Reverse sort $sums, putting the highest value first
  arsort($sums);
  // Flip $sums so we can extract the key
  $flipped = array_flip($sums);
  // Extract the first key off of $sums
  $winner = array_shift($flipped);
}

print $winner;

---

首先，你的问题是要求出现的次数。 但是，当你明显走得更远时，我觉得我应该争取另一种解决方案。

similar_text()函数！

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

$inputString = 'seventy two yards out';

$p = 0;
$k = null;
foreach ($strings as $key => $string) {
    similar_text($inputString, $string, $percent);
    if ($percent > $p) {
        $p = $percent;
        $k = $key;
    }
}

echo !is_null($k) ? $strings[$k] : "";

输出：

seventy five yards out

现场演示

链接地址: http://www.djcxy.com/p/59107.html

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