使用Levenstein距离重新排列单词

Question

摘要

我试图在php中找到名称匹配百分比，但在此之前，我需要根据第一个字符串重新排列字符串中的单词。

源代码是关于什么的

我有两根绳子。首先，我将两个字符串都添加到数组中，若在字符串中找到空格，则将其添加到数组。$arraydataBaseName和$arraybankData来自我的第一个数组，即$arraydataBaseName。我正在搜索$arraybankData的所有值并获取密钥。我得到了正确的密钥排列，但无法将其特定位置的值排列到新数组中。

$dataBaseName = "Jardine Lloyd Thompson";
$bankdata = "Thompson Thompson Jardine"; 
$replacedataBaseName = preg_replace("#[s]+#", " ", $dataBaseName);
$replacebankData = preg_replace("#[s]+#", " ", $bankdata); 
$arraydataBaseName = explode(" ",$replacedataBaseName);
$arraybankData = explode(" ",$replacebankData); 
echo "<br/>";
print_r($arraydataBaseName);
$a="";
$i="";
$arraysize =  count($arraydataBaseName);
$push=array();
for($i=0;$i< $arraysize;$i++)
{     
if(array_search($arraybankData[$i],$arraydataBaseName)>0)
{
${"$a$i"} =  array_search($arraybankData[$i],$arraydataBaseName); 
//echo ${"$a$i"};
array_push($push,${"$a$i"});
}    
}
print_r($push); 

案例1：

输入

DatabaseName=Jardine Lloyd-Thompson

银行名称=汤普森怡和银行

输出

预期输出=Jardine Lloyd-Thompson

案例2：##

输入

DatabaseName=Jardine Lloyd-Thompson

银行名称=Thoapson Jordine Llayd

如果在上面的DatabaseName中没有找到单词，那么预期的搜索将基于leventish算法单词，该单词的距离较小，将被视为关键

输出

预期输出=Jordine Llayd Thoapson

问题描述

问题更新

当用户输入的$bankdata包含更多无法匹配的单词时，我需要将这些单词附加到末尾。

Answer 1

这是一个简单的版本，随后逐字逐句地找到最佳匹配。

declare (strict_types=1);
$dataBaseName = 'Jardine Lloyd Thompson';
$bankdataRows =
[
'Thompson Jardine Lloyd',
'Blaaa  Llayd Thoapson   f***ing user input   Jordine   aso. ',
];
// assume the "database" is already stored trimmed since it is server-side controlled
$dbWords = preg_split("#[s]+#", $dataBaseName);
foreach ($bankdataRows as $bankdata)
{
// here we trim the data received from client-side.
$bankWords = preg_split("#[s]+#", trim($bankdata));
$result    = [];
if(!empty($bankWords))
foreach ($dbWords as $dbWord)
{
$idx   = null;
$least = PHP_INT_MAX;
foreach ($bankWords as $k => $bankWord)
if (($lv = levenshtein($bankWord, $dbWord)) < $least)
{
$least = $lv;
$idx   = $k;
}
$result[] = $bankWords[$idx];
unset($bankWords[$idx]);
}
$result = array_merge($result, $bankWords);
var_dump($result);
}

结果

array(3) {
[0] =>
string(7) "Jardine"
[1] =>
string(5) "Lloyd"
[2] =>
string(8) "Thompson"
}
array(8) {
[0] =>
string(7) "Jordine"
[1] =>
string(5) "Llayd"
[2] =>
string(8) "Thoapson"
[3] =>
string(5) "Blaaa"
[4] =>
string(7) "f***ing"
[5] =>
string(4) "user"
[6] =>
string(5) "input"
[7] =>
string(4) "aso."
}

查看现场小提琴

您可能需要扩展这种方法，首先计算每个可能组合的Levenstein距离，然后选择最佳的整个匹配。

Answer 2

我已经分解了案例1和案例2中的代码。
但很明显，如果var_export为false，则使用相同的变量执行案例2的代码。

//Case 1:
$DatabaseName = "Jardine Lloyd Thompson";
$BankName = "Thompson Jardine Lloyd";
//Split and sort them
$data = explode(" ", $DatabaseName);
$bank = explode(" ", $BankName);
sort($data);
sort($bank);
Var_export(($data == $bank)); //true
//Case 2
$DatabaseName = "Jardine Lloyd Thompson";
$BankName = "Thoapson Jordine Llayd";
//Split and sort
$data = explode(" ", $DatabaseName);
$bank = explode(" ", $BankName);
sort($data);
sort($bank);
// Loop and accumulate the levenshtein return
$lev = 0;
foreach($data as $key => $name){
$lev += levenshtein($name, $bank[$key]);
}
echo PHP_EOL . $lev; // 3 letters "off"

https://3v4l.org/eP5PE

同一代码中情况1和2的示例。

$DatabaseName = "Jardine Lloyd Thompson";
$BankName = "Thoapson Jordine Llayd";
$data = explode(" ", $DatabaseName);
$bank = explode(" ", $BankName);
sort($data);
sort($bank);
if($data == $bank){
echo "true";
exit;
// No need to do levenshtein
}
$lev = 0;
foreach($data as $key => $name){
$lev += levenshtein($name, $bank[$key]);
}
echo PHP_EOL . $lev;

https://3v4l.org/RJSiB