如何将数组的相同键与其值相加答案

【问题标题】：How to sum of same keys of arrays with their values如何将数组的相同键与其值相加
【发布时间】：2016-06-06 12:02:19
【问题描述】：

我必须将多维数组的数组键总结为：

)

在这里，我想得到数组键的总和，例如，键 [qty_of_leads_by_levels] 的值应该用相同的键和输出数组求和：

 [qty_of_leads_by_levels] => Array
    (
        [0] => 17814.49                   //sum
        [1] => 8965.52
        [2] => 6971.02
        [3] => 5320.75
        [4] => 2843.33
        [5] => 1422.7
        [6] => 356.72
        [7] => 90.22
        [8] => 69.28
    )      

 [avg_conv_cof_arry] => Array
            (
                [0] =>                 //sum
                [1] =>
                [2] => 
                [3] => 
                [4] => 
                [5] => 
                [6] => 
                [7] => 
                [8] =>
            )

      [avg_deal_size] =>                //sum

      [total_bgt] =>                    //sum

【问题讨论】：

标签： php multidimensional-array

【解决方案1】：

首先将您的关联数组分配给 $inputArray。

$sumArr = array();
    foreach($inputArray as $key1=>$value1){
    foreach($value1 as $key2=>$value2){
    if(is_array($value2)){
    foreach($value2 as $key3=>$value3){
        if(isset($sumArr[$key3])) {
                 $sumArr[$key3] = $sumArr[$key3] + $value3;
             } else {
                 $sumArr[$key3] = $value3;
             }

    }
    }
    }
    }

   print_r($sumArr);

【讨论】：

【解决方案2】：

你可以通过循环数组来做到：

$qty_of_leads_by_levels = array();
foreach ($your_arr as $key_internal=>$arr_internal){
    foreach ($arr_internalas $key_internal_2=>$value_internal_2){
        foreach ($value_internal_2 as $key=>$value){
             if(isset($qty_of_leads_by_levels[$key])) {
                 $qty_of_leads_by_levels[$key] = $qty_of_leads_by_levels[$key] + $value
             } else {
                 $qty_of_leads_by_levels[$key] = $value;
             }
        }
   }
}

print_r($qty_of_leads_by_levels);

【讨论】：

【解决方案3】：

您可以通过使用array_walk_recursive 使用传递引用&$final 来做到这一点对于每个item 递归。

$final = array();

array_walk_recursive( $array, function ( $item, $key ) use ( &$final ) {
    $final[ $key ] = isset( $final[ $key ] ) ? $item + $final[ $key ] : $item;
} );
unset( $final['avg_deal_size'], $final['total_bgt'] );

echo '<pre>';print_r($final);echo '</pre>';

见最后的Demo输出here

输出：

<pre>Array
(
    [0] => 17814.49
    [1] => 8965.52
    [2] => 6971.02
    [3] => 5320.75
    [4] => 2843.33
    [5] => 1422.7
    [6] => 356.72
    [7] => 90.22
    [8] => 69.28
)
</pre>

【讨论】：

【解决方案4】：

我尝试使用RecursiveArrayIterator 来维护所有密钥。看看下面的解决方案：

$array = array
(
    array
    (
        'qty_of_leads_by_levels' => array
        (
            0 => 6054,
            1 => 454,

        ),

        'avg_conv_cof_arry' => array
        (
            0 => 0.08,
            1 => 0.1

        ),

        'avg_deal_size' => 1,
        'total_bgt' => 1
    ),

    array
    (
        'qty_of_leads_by_levels' => array
        (
            0 => 11393,
            1 => 8144,

        ),

        'avg_conv_cof_arry' => array
        (
            0 => 0.41,
            1 => 0.42

        ),

        'avg_deal_size' => 2,
        'total_bgt' => 2
    ),

    array
    (
        'qty_of_leads_by_levels' => array
        (
            0 => 366,
            1 => 366

        ),

        'avg_conv_cof_arry' => array
        (
            0 => 1,
            1 => 1

        ),

        'avg_deal_size' => 3,
        'total_bgt' => 3
    )
);

$new_array = array();
$iterator = new RecursiveIteratorIterator(new RecursiveArrayIterator($array));

foreach ($iterator as $key => $value) {
    $keys = array();
    $keys[] = $key;
    for ($i = $iterator->getDepth() - 1; $i >= 0; $i--) {
        $keys[] = $iterator->getSubIterator($i)->key();
    }

    //get key path
    $key_paths = array_reverse($keys);

    if((count($key_paths) > 2)) {
        if(!isset($new_array[$key_paths[1]][$key])){
            $new_array[$key_paths[1]][$key] = 0;
        }
        $new_array[$key_paths[1]][$key] += $value;
    }
    else {
        if(!isset($new_array[$key_paths[1]])){
            $new_array[$key_paths[1]] = 0;
        }
        $new_array[$key_paths[1]] += $value;
    }
}

print_r($new_array);

输出

Array
(
    [qty_of_leads_by_levels] => Array
        (
            [0] => 17813
            [1] => 8964
        )

    [avg_conv_cof_arry] => Array
        (
            [0] => 1.49
            [1] => 1.52
        )

    [avg_deal_size] => 6
    [total_bgt] => 6
)

【讨论】：

【解决方案5】：

<?php
$sumArray = [];
foreach($array as $r) 
    for($i = 0; $i < count($r); $i++)
        $sumArray[$i] += $r['qty_of_leads_by_levels'][$i];
?>

【讨论】：

【解决方案6】：

以下适用于 PHP >4

$arr = array_column($input, 'qty_of_leads_by_levels');

$sumArray = array();
array_walk($arr, function ($val, $key) use ($sumArray){
    foreach($val as $k => $v) {
        $sumArray[$k] = isset($sumArray[$k]) ? $sumArray[$k]+$v : $v;
    }
});

print_r($sumArray);

输出：

Array
(
    [0] => 17814.49
    [1] => 8965.52
    [2] => 6971.02
    [3] => 5320.75
    [4] => 2843.33
    [5] => 1422.7
    [6] => 356.72
    [7] => 90.22
    [8] => 69.28
)

【讨论】：

【解决方案7】：

我通过以下方式进行了尝试：

$qty_of_leads_by_levels = array();
$avg_conv_cof_arry = array();
$avg_deal_size = 0;
$total_bgt = 0;


for($i=0;$i<count($arr1);$i++){

    foreach($arr1[$i]['qty_of_leads_by_levels'] as $k=>$v){
        $qty_of_leads_by_levels[$k] += $v;  
    }

    foreach($arr1[$i]['avg_conv_cof_arry'] as $k1=>$v1){
        $avg_conv_cof_arry[$k1] = $avg_conv_cof_arry[$k1] + $v1;
    }

    $avg_deal_size += $arr1[$i]['avg_deal_size'];
    $total_bgt += $arr1[$i]['total_bgt'];
}

【讨论】：