我有两个php函数来计算两个文本之间的关系。它们都使用单词包模型,但是check2()要快得多。无论如何,这两个函数都给出相同的结果。为什么?check1()使用一个包含所有单词的大字典数组-如单词包模型中所述。check2()不使用一个大数组,而是只包含一个文本的单词的数组。所以check2()应该不起作用,但它不起作用。为什么两个函数都给出相同的结果?

function check1($terms_in_article1, $terms_in_article2) {
    global $zeit_check1;
    $zeit_s = microtime(TRUE);
    $length1 = count($terms_in_article1); // number of words
    $length2 = count($terms_in_article2); // number of words
    $all_terms = array_merge($terms_in_article1, $terms_in_article2);
    $all_terms = array_unique($all_terms);
    foreach ($all_terms as $all_termsa) {
        $term_vector1[$all_termsa] = 0;
        $term_vector2[$all_termsa] = 0;
    }
    foreach ($terms_in_article1 as $terms_in_article1a) {
        $term_vector1[$terms_in_article1a]++;
    }
    foreach ($terms_in_article2 as $terms_in_article2a) {
        $term_vector2[$terms_in_article2a]++;
    }
    $score = 0;
    foreach ($all_terms as $all_termsa) {
        $score += $term_vector1[$all_termsa]*$term_vector2[$all_termsa];
    }
    $score = $score/($length1*$length2);
    $score *= 500; // for better readability
    $zeit_e = microtime(TRUE);
    $zeit_check1 += ($zeit_e-$zeit_s);
    return $score;
}
function check2($terms_in_article1, $terms_in_article2) {
    global $zeit_check2;
    $zeit_s = microtime(TRUE);
    $length1 = count($terms_in_article1); // number of words
    $length2 = count($terms_in_article2); // number of words
    $score_table = array();
    foreach($terms_in_article1 as $term){
        if(!isset($score_table[$term])) $score_table[$term] = 0;
        $score_table[$term] += 1;
    }
    $score_table2 = array();
    foreach($terms_in_article2 as $term){
        if(isset($score_table[$term])){
            if(!isset($score_table2[$term])) $score_table2[$term] = 0;
            $score_table2[$term] += 1;
        }
    }
    $score = 0;
    foreach($score_table2 as $key => $entry){
        $score += $score_table[$key] * $entry;
    }
    $score = $score/($length1*$length2);
    $score *= 500;
    $zeit_e = microtime(TRUE);
    $zeit_check2 += ($zeit_e-$zeit_s);
    return $score;
}

我希望你能帮助我。提前谢谢!

最佳答案

这两个函数实现了几乎相同的算法,但是第一个函数以简单的方式实现,第二个函数则更聪明一些,跳过了一部分不必要的工作。
支票1如下:

// loop length(words1) times
for each word in words1:
    freq1[word]++

// loop length(words2) times
for each word in words2:
    freq2[word]++

// loop length(union(words1, words2)) times
for each word in union(words1, words2):
    score += freq1[word] * freq2[word]

但是记住:当你用零乘某物时,你会得到零。
这意味着,计算不在两个集合中的单词的频率是浪费时间-我们将频率乘以零,这将不会增加分数。
支票2考虑了这一点:
// loop length(words1) times
for each word in words1:
    freq1[word]++

// loop length(words2) times
for each word in words2:
    if freq1[word] > 0:
        freq2[word]++

// loop length(intersection(words1, words2)) times
for each word in freq2:
    score += freq1[word] * freq2[word]

10-05 21:00
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