Question

页: 1 拥有一系列阵容

array(4) {
      [0]=>
      array(3) {
        ["a"]=>float(1000)
        ["b"]=>float(3)
        ["c"]=>float(500)
      }
      [1]=>
      array(3) {
        ["a"]=>float(1000)
        ["b"]=>float(852)
        ["c"]=>float(500)
      }
      [2]=>
      array(3) {
        ["a"]=>float(1000)
        ["b"]=>float(5)
        ["c"]=>float(500)
      }
      [3]=>
      array(1) {
        ["e"]=>float(1000)
      }
    }

其结果将总结同样关键因素的全部价值,即:

$result = 
  array(
      "a" =>3000,
      "b"=>900,
      "c"=>1500,
      "e"=>1000
  )

Anybody could help me todo this.

thanks.

Answer 1

Pseudo:

result <- new array                       # array holding result
foreach entry1 in array:                  # iterate outer array
    foreach entry2 in entry1:             # iterate each inner array
        if not exists result[entry2.key]: # if key is not already in result...
            result[entry2.key] = 0        # ... add key and set value to zero
        result[entry2.key] += value       # increment result for key with value from inner array

http://www.un.org/Depts/DGACM/index_chinese.htm

Answer 2

The trick for this is ofcourse to do some sort of iteration over your data, using those string-keys as identifiers. One way of approaching it would be using 2 nested foreaches ( one over the container, one over the individual keys and collecting the data in a central array:

$results = array();
foreach ($array as $elements)
{
        foreach ($elements as $key => $value)
        {   
                if (!isset($results[$key]))
                        $results[$key] = 0;

                $results[$key] += $value;
        }   
}

一种不同的方式是使植物检疫措施对你有利:

$results = array();

array_walk_recursive(
        $array,
        function($value, $key) use (&$results) {
                if (!isset($results[$key]))
                        $results[$key] = 0;

                $results[$key] += $value;
        }
);

Answer 3

This little function will do the job for you.

function SummarizeFosArray($array) {
    $results=array();
    foreach ($array as $a) {
      foreach ($a as $k=>$v) {
        $results[$k]+=$v;
      }
    }
    return $results;
}

Answer 4

页: 1

$array = array(
        array( a  => 1000,  b  =>3,  c => 500),
        array( a  => 1000,  b  =>852,  c => 500),
        array( a  => 1000,  b  =>5,  c => 500),
        array( e  => 1000)
        );

$result = array();  

foreach($array as $arr)
{
    foreach($arr as $a => $val){
        $result[$a] += $val;
    }
}

echo "<pre>";
print_r($result);
echo "</pre>";

阁下

Array
(
    [a] => 3000
    [b] => 860
    [c] => 1500
    [e] => 1000
)

Answer 5

rel=“nofollow”>array_walk_recursive( 职能。详细情况调查手册。

友情链接