查找和替换阵副本 [英] Find and replace duplicates in Array

问题描述

PHP数组问题
我需要与应用将填补阵列用一些随机值，但如果阵列是重复我的应用程序无法正常工作。所以，我需要编写脚本code，将查找重复，并与一些其它值替换它们。
好了，比如我有一个数组：

PHP Array Question I need to make app with will fill array with some random values, but if in array are duplicates my app not working correctly. So i need to write script code which will find duplicates and replace them with some other values. Okay so for example i have an array:

<?PHP
$charset=array(123,78111,0000,123,900,134,00000,900);

function arrayDupFindAndReplace($array){

// if in array are duplicated values then -> Replace duplicates with some other numbers which ones im able to specify.
return $ArrayWithReplacedValues;
}
?>

这样的结果应是同一阵列替换为重复值。

So result shall be same array with replaced duplicated values.

感谢您的帮助。

推荐答案

您可以只跟踪你已经迄今所看到和替换，当您去的话。

You can just keep track of the words that you've seen so far and replace as you go.

// words we've seen so far
$words_so_far = array();
// for each word, check if we've encountered it so far
//    - if not, add it to our list
//    - if yes, replace it
foreach($charset as $k => $word){
    if(in_array($word, $words_so_far)){
        $charset[$k] = $your_replacement_here;
    }
    else {
        $words_so_far[] = $word;
    }
}

有关有点优化的解决方案（对于那里有没有那么多的重复情况下），使用array_count_values​​（）的（参考这里）来计算的次数就出现了。

For a somewhat-optimized solution (for cases where there are not that many duplicates), use array_count_values() (reference here) to count the number of times it shows up.

// counts the number of words
$word_count = array_count_values($charset);
// words we've seen so far
$words_so_far = array();
// for each word, check if we've encountered it so far
//    - if not, add it to our list
//    - if yes, replace it
foreach($charset as $k => $word){
    if($word_count[$word] > 1 && in_array($word, $words_so_far)){
        $charset[$k] = $your_replacement_here;
    }
    elseif($word_count[$word] > 1){
        $words_so_far[] = $word;
    }
}

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