如何在PHP中迭代多个具有多个数组值的字符串?

时间:2016-07-23 02:34:19

标签: php arrays string loops preg-grep

*来自revo的回答

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);

如何将它打印出来?七十五码外出现'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..

3 个答案:

答案 0 :(得分:2)

这样的事情应该有效:

<?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;

答案 1 :(得分:2)

关键是分别对每个提供的字符串执行<?xml version="1.0" encoding="utf-8"?> <LinearLayout xmlns:android="http://schemas.android.com/apk/res/android" android:orientation="vertical" android:layout_width="match_parent" android:layout_height="match_parent"> <EditText android:layout_width="match_parent" android:layout_height="wrap_content" android:layout_marginTop="40dp" android:hint="Enter something" /> </LinearLayout> 。这样我们就可以转换成数组,处理数组对于你想要的东西来说简单得多。

str_word_count()计算数组的值,这会导致出现多个单词。

array_count_values()

输出:

$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)];

概率seventy five yards out

print_r($probabilities);

<强> Live demo

答案 2 :(得分:0)

首先,你的问题是询问发生次数。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()功能!

$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

<强> Live demo