Question

我试图计算并删除重复的数组。要成为一个副本，整个数组将使用键和值与另一个匹配。

Array
(
    [0] => Array
        (
            [name] => Superman
            [time] => 60 
        )

    [1] => Array
        (
            [name] => Superman
            [time] => 60
        )

    [2] => Array
        (
            [name] => Superman
            [time] => 50
        )

    [3] => Array
        (
            [name] => Superman
            [time] => 40
        )

    [4] => Array
        (
            [name] => Superman
            [time] => 50
        )

    [5] => Array
        (
            [name] => Superman
            [time] => 60
        )

)

分为：

Array
(
    [0] => Array
        (
            [name] => Superman
            [time] => 60
            [count] => 3
        )

    [1] => Array
        (
            [name] => Superman
            [time] => 50
            [count] => 2
        )

    [2] => Array
        (
            [name] => Superman
            [time] => 40
        )

)

我遇到了this answer that can remove the duplicates，但我很难看到我如何算他们。

$input = array_map("unserialize", array_unique(array_map("serialize", $input)));

Answer 1

在这种情况下，您也可以使用array_count_values。例如：

$values = array(
    array('name' => 'Superman', 'time' => 60),
    array('name' => 'Superman', 'time' => 60),
    array('name' => 'Superman', 'time' => 50),
    array('name' => 'Superman', 'time' => 40),
    array('name' => 'Superman', 'time' => 50),
    array('name' => 'Superman', 'time' => 60),
);

// map out into string then unique them
$uniques = array_map("unserialize", array_unique(array_map("serialize", $values)));
$count = array_count_values(array_map("serialize", $values)); // map out the arrays then get the counts

// then to merge the count
foreach($uniques as &$batch) {
    foreach($count as $array => $v) {
        if(unserialize($array) == $batch) { // if this particular key count is equal to this unique array, then push the count
            $batch['count'] = $v;
        }
    }
}

echo '<pre>';
print_r($uniques);

Sample Output

Answer 2

快速而又脏，但您可以获得所要求的数据结构：

$data = array(
    array("name" => "Superman", "time" => 60),
    array("name" => "Superman", "time" => 60),
    array("name" => "Superman", "time" => 50),
    array("name" => "Superman", "time" => 40),
    array("name" => "Superman", "time" => 50),
    array("name" => "Superman", "time" => 60),
);

// count the occurrences
$occurrences = array();
for ($i = 0, $l = count($data); $i < $l; $i++) {
    $serialized = serialize($data[$i]);
    if (!isset($occurrences[$serialized])) {
        $occurrences[$serialized] = 1;
    }
    else {
        $occurrences[$serialized] = $occurrences[$serialized] + 1;
    }
}

// get the wanted structure
$uniques = array();
foreach ($occurrences as $serialized => $count) {
    $unserialized = unserialize($serialized);
    if ($count > 1) {
        $unserialized['count'] = $count;
    }
    $uniques[] = $unserialized;
}

print_r($uniques);

Answer 3

不像我想的那样紧凑，但它完成了工作：

function unserialize_unique_count($input, $k = 'count') {
    $a = [];
    foreach ($input as $d) {
        $s = serialize($d);
        $a[$s] = (isset($a[$s]) ? ($a[$s] + 1) : 1);
    }
    foreach ($a as $s => $c) {
        $a[$s] = unserialize($s) + [ $k => $c ];
    }
    return array_values($a);
}

$grouped_with_count = unserialize_unique_count($input);

工作原理：第一个循环序列化并计数。第二个独特的合并。为O（n）。

使用方法：将多维数组作为参数＃1传递。使用额外的键＆＃34;计数＆＃34;持有点数。如果你想让count键不是＆＃34; count＆＃34;，那就把第二个参数传递给函数。

Answer 4

获取数组及其值
在paypal_address上创建唯一键
使用唯一键创建临时数组
使用临时数组中的相应唯一键存储所有值

我将时间用作唯一键。

 $tmpArr = Array();
$cnt = sizeof($arr);
for($i=0;$i<$cnt;$i++){
        $time = $arr[$i]['time'];
        if(!is_array($tmpArr[$time])){
                $tmpArr[$time] = Array();
                $tmpArr[$time]['count'] = 0;
        }
        $tmpArr[$time]['time'] = $arr[$i]['time'];
        $tmpArr[$time]['name'] = $arr[$i]['name'];
        $tmpArr[$time]['count'] = $tmpArr[$time]['count'] + 1;

}
print_r($tmpArr);

注意：根据您的要求对代码进行更改

计算并删除重复的数组

4 个答案: