Question

我的言辞清单中有一些是言词,例如,

palanca
plato
platopalanca

I need to remove "plato" and "palanca" and let only "platopalanca". Used array_unique to remove duplicates, but those composed words are tricky...

Should I sort the list by word length and compare one by one? A regular expression is the answer?

更新: 言词清单更大,有好有坏,而不仅仅是相关词。

更新2:我可以安全地将阵列化为扼杀。

更新3:我试图避免这样做,好像这样说。必须有更有效的方式这样做。

Well, I think that a buble-sort like approach is the only possible one :-( I don t like it, but it s what i have... Any better approach?

function sortByLengthDesc($a,$b){
return strlen($a)-strlen($b);
}

usort($words, sortByLengthDesc );
$count = count($words);
for($i=0;$i<=$count;$i++) {
    for($j=$i+1;$j<$count;$j++) {
        if(strstr($words[$j], $words[$i]) ){
            $delete[]=$i;
        }
    }
}
foreach($delete as $i) {
    unset($words[$i]);
}

update 5: Sorry all. I m A moron. Jonathan Swift make me realize I was asking the wrong question. Given x words which START the same, I need to remove the shortests ones.

"hot, dog, stand, hotdogstand" should become "dog, stand, hotdogstand"
"car, pet, carpet" should become "pet, carpet"
"palanca, plato, platopalanca" should become "palanca, platopalanca"
"platoother, other" should be untouchedm they both start different

Answer 1

我认为,你需要再界定这个问题,以便我们能够作出坚定的回答。这里有一些病理学清单。应删除哪些物品?

hot, dog, hotdogstand.
hot, dog, stand, hotdogstand
hot, dogs, stand, hotdogstand

www.un.org/Depts/DGACM/index_spanish.htm SOME CODE

这部法典应当比你拥有的更有效:

$words = array( hatstand , hat , stand , hot , dog , cat , hotdogstand , catbasket );

$count = count($words);

for ($i=0; $i<=$count; $i++) {
    if (isset($words[$i])) {
        $len_i = strlen($words[$i]);
        for ($j=$i+1; $j<$count; $j++) {
            if (isset($words[$j])) {
                $len_j = strlen($words[$j]);

                if ($len_i<=$len_j) {
                    if (substr($words[$j],0,$len_i)==$words[$i]) {
                        unset($words[$i]);  
                    }
                } else {
                    if (substr($words[$i],0,$len_j)==$words[$j]) {
                        unset($words[$j]);
                    }
                }
            }
        }
    }
}

foreach ($words as $word) {
    echo "$word<br>";
}

你们可以选择,在休息前的阵列中储存字长。

Answer 2

您可以逐字看一看,阵列中的任何字从一开始,还是从一开始。如果是的话,应当删除这个词(unset(unset)()。

Answer 3

您可将言辞纳入一个阵列,按字母顺序排列阵列,然后按目前的指数核对下一个字,从而形成文字。如果是的话,你可以删除现行指数中的字句和后面几部分。

与此类似:

$array = array( palanca ,  plato ,  platopalanca );
// ok, the example array is already sorted alphabetically, but anyway...
sort($array);

// another array for words to be removed
$removearray = array();

// loop through the array, the last index won t have to be checked
for ($i = 0; $i < count($array) - 1; $i++) {

  $current = $array[$i];

  // use another loop in case there are more than one combined words
  // if the words are case sensitive, use strpos() instead to compare
  while ($i < count($array) && stripos($array[$i + 1], $current) === 0) {
    // the next word starts with the current one, so remove current
    $removearray[] = $current;
    // get the other word to remove
    $removearray[] = substr($next, strlen($current));
    $i++;
  }

}

// now just get rid of the words to be removed
// for example by joining the arrays and getting the unique words
$result = array_unique(array_merge($array, $removearray));

Answer 4

Regex可以工作。如果限制的开始和结束适用,你可以在监管范围内加以界定。

^ defines the start $ defines the end

类似情况

foreach($array as $value)
{
    //$term is the value that you want to remove
    if(preg_match( /^  . $term .  $/ , $value))
    {
        //Here you can be confident that $term is $value, and then either remove it from
        //$array, or you can add all not-matched values to a new result array
    }
}

避免你的问题

但是,如果你只是检查两种价值观是平等的,=将同样工作(而且可能比以前更快)。

如果数额和价值清单是巨大的,那就是一个最有效的战略,但这是一个简单的解决办法。

如果履约是一个问题,分类(注解提供的sort功能),清单可能更有用,然后将清单放在一边。我在把该守则摆在我前面之前将实际检验这一想法。

友情链接