具有错误字符容差的最长公共子串

Posted 2023-02-22

【中文标题】具有错误字符容差的最长公共子串

【英文标题】：Longest Common Substring with wrong character tolerance

【发布时间】：2012-09-26 09:06:23

【问题描述】：

我在这里找到了一个脚本，在查找最低公共子字符串时效果很好。

但是，我需要它来容忍一些不正确/缺失的字符。我希望能够输入所需的相似度百分比，或者指定允许的缺失/错误字符数。

比如我要查找这个字符串：

大黄色校车

这个字符串的内部：

那天下午他们乘坐的是大黄校车

这是我目前使用的代码：

function longest_common_substring($words) 
    $words = array_map('strtolower', array_map('trim', $words));
    $sort_by_strlen = create_function('$a, $b', 'if (strlen($a) == strlen($b))  return strcmp($a, $b);  return (strlen($a) < strlen($b)) ? -1 : 1;');
    usort($words, $sort_by_strlen);

    // We have to assume that each string has something in common with the first
    // string (post sort), we just need to figure out what the longest common
    // string is. If any string DOES NOT have something in common with the first
    // string, return false.
    $longest_common_substring = array();
    $shortest_string = str_split(array_shift($words));

    while (sizeof($shortest_string)) 
        array_unshift($longest_common_substring, '');
        foreach ($shortest_string as $ci => $char) 
            foreach ($words as $wi => $word) 
                if (!strstr($word, $longest_common_substring[0] . $char)) 
                    // No match
                    break 2;
                
            
            // we found the current char in each word, so add it to the first longest_common_substring element,
            // then start checking again using the next char as well
            $longest_common_substring[0].= $char;
        
        // We've finished looping through the entire shortest_string.
        // Remove the first char and start all over. Do this until there are no more
        // chars to search on.
        array_shift($shortest_string);
    

    // If we made it here then we've run through everything
    usort($longest_common_substring, $sort_by_strlen);

    return array_pop($longest_common_substring);

非常感谢任何帮助。

更新

php levenshtein 函数限制为 255 个字符，而我正在搜索的一些 haystacks 是 1000+ 个字符。

【问题讨论】：

我想说您应该使用自定义字符串比较函数，该函数将使用一个符号容差。算法可能是这样的：一次将两个符号与最长的公共子字符串进行比较，一旦找到其中一个，就可以逐个符号进行比较。如果不匹配，检查容差阈值，如果失败，继续搜索 LCS 的可能开始。如果成功，添加容差并检查下一个符号，将它们相互比较，并首先比较未处理的 LCS 符号。如果成功，继续检查，就像刚刚发现遗漏或错误一样。

Wagner-Fischer 可能会给您一个很好的起点。您也许可以查看矩阵上生成的对角线并在此基础上解决问题。 en.wikipedia.org/wiki/Wagner%E2%80%93Fischer_algorithm我也会考虑的。

【参考方案1】：

将其写为第二个答案，因为它根本不是基于我以前的（坏的）答案。

此代码基于http://en.wikipedia.org/wiki/Wagner%E2%80%93Fischer_algorithm 和http://en.wikipedia.org/wiki/Approximate_string_matching#Problem_formulation_and_algorithms

给定 $needle，它返回 $haystack 的一个（可能是几个）最小 levenshtein 子字符串。现在，levenshtein 距离只是编辑距离的一种度量，它实际上可能并不适合您的需求。在这个度量上，'hte' 更接近于 'he' 而不是 'the'。我放入的一些示例显示了这种技术的局限性。我相信这比我之前给出的答案要可靠得多，但请告诉我它是如何为您工作的。

// utility function - returns the key of the array minimum
function array_min_key($arr)

    $min_key = null;
    $min = PHP_INT_MAX;
    foreach($arr as $k => $v) 
        if ($v < $min) 
            $min = $v;
            $min_key = $k;
        
    
    return $min_key;


// Calculate the edit distance between two strings
function edit_distance($string1, $string2)

    $m = strlen($string1);
    $n = strlen($string2);
    $d = array();

    // the distance from '' to substr(string,$i)
    for($i=0;$i<=$m;$i++) $d[$i][0] = $i;
    for($i=0;$i<=$n;$i++) $d[0][$i] = $i;

    // fill-in the edit distance matrix
    for($j=1; $j<=$n; $j++)
    
        for($i=1; $i<=$m; $i++)
        
            // Using, for example, the levenshtein distance as edit distance
            list($p_i,$p_j,$cost) = levenshtein_weighting($i,$j,$d,$string1,$string2);
            $d[$i][$j] = $d[$p_i][$p_j]+$cost;
        
    

    return $d[$m][$n];


// Helper function for edit_distance()
function levenshtein_weighting($i,$j,$d,$string1,$string2)

    // if the two letters are equal, cost is 0
    if($string1[$i-1] === $string2[$j-1]) 
        return array($i-1,$j-1,0);
    

    // cost we assign each operation
    $cost['delete'] = 1;
    $cost['insert'] = 1;
    $cost['substitute'] = 1;

    // cost of operation + cost to get to the substring we perform it on
    $total_cost['delete'] = $d[$i-1][$j] + $cost['delete'];
    $total_cost['insert'] = $d[$i][$j-1] + $cost['insert'];
    $total_cost['substitute'] = $d[$i-1][$j-1] + $cost['substitute'];

    // return the parent array keys of $d and the operation's cost
    $min_key = array_min_key($total_cost);
    if ($min_key == 'delete') 
        return array($i-1,$j,$cost['delete']);
     elseif($min_key == 'insert') 
        return array($i,$j-1,$cost['insert']);
     else 
        return array($i-1,$j-1,$cost['substitute']);
    


// attempt to find the substring of $haystack most closely matching $needle
function shortest_edit_substring($needle, $haystack)

    // initialize edit distance matrix
    $m = strlen($needle);
    $n = strlen($haystack);
    $d = array();
    for($i=0;$i<=$m;$i++) 
        $d[$i][0] = $i;
        $backtrace[$i][0] = null;
    
    // instead of strlen, we initialize the top row to all 0's
    for($i=0;$i<=$n;$i++) 
        $d[0][$i] = 0;
        $backtrace[0][$i] = null;
    

    // same as the edit_distance calculation, but keep track of how we got there
    for($j=1; $j<=$n; $j++)
    
        for($i=1; $i<=$m; $i++)
        
            list($p_i,$p_j,$cost) = levenshtein_weighting($i,$j,$d,$needle,$haystack);
            $d[$i][$j] = $d[$p_i][$p_j]+$cost;
            $backtrace[$i][$j] = array($p_i,$p_j);
        
    

    // now find the minimum at the bottom row
    $min_key = array_min_key($d[$m]);
    $current = array($m,$min_key);
    $parent = $backtrace[$m][$min_key];

    // trace up path to the top row
    while(! is_null($parent)) 
        $current = $parent;
        $parent = $backtrace[$current[0]][$current[1]];
    

    // and take a substring based on those results
    $start = $current[1];
    $end = $min_key;
    return substr($haystack,$start,$end-$start);


// some testing
$data = array( array('foo',' foo'), array('fat','far'), array('dat burn','rugburn'));
$data[] = array('big yellow school bus','they rode the bigyellow schook bus that afternoon');
$data[] = array('bus','they rode the bigyellow schook bus that afternoon');
$data[] = array('big','they rode the bigyellow schook bus that afternoon');
$data[] = array('nook','they rode the bigyellow schook bus that afternoon');
$data[] = array('they','console, controller and games are all in very good condition, only played occasionally. includes power cable, controller charge cable and audio cable. smoke free house. pes 2011 super street fighter');
$data[] = array('controker','console, controller and games are all in very good condition, only played occasionally. includes power cable, controller charge cable and audio cable. smoke free house. pes 2011 super street fighter');

foreach($data as $dat) 
    $substring = shortest_edit_substring($dat[0],$dat[1]);
    $dist = edit_distance($dat[0],$substring);
    printf("Found |%s| in |%s|, matching |%s| with edit distance %d\n",$substring,$dat[1],$dat[0],$dist);

【讨论】：

这很完美，非常感谢！如果有什么方法可以给你买一瓶啤酒（或一盒），请告诉我！