求合并数组的高效方法

[ohmygirl]

//原始的数组如下：
  $test=array(
             'titan'=>array(0=>array('date'=>'2011-08-11','DAU'=>'222','DNU'=>'111','DNUF'=>'444'),
			                1=>array('date'=>'2011-08-12','DAU'=>'333','DNU'=>'222','DNUF'=>'333'),
							2=>array('date'=>'2011-08-13','DAU'=>'444','DNU'=>'333','DNUF'=>'555')),
             'apollo'=>array(0=>array('date'=>'2011-08-11','DAU'=>'298','DNU'=>'290','DNUF'=>'900'),
			                 1=>array('date'=>'2011-08-12','DAU'=>'340','DNU'=>'490','DNUF'=>'599'),
							 2=>array('date'=>'2011-08-13','DAU'=>'234','DNU'=>'345','DNUF'=>'532')),
             'bossvegas'=>array(0=>array('date'=>'2011-08-11','DAU'=>'343','DNU'=>'532','DNUF'=>'342'),
			                1=>array('date'=>'2011-08-12','DAU'=>'323','DNU'=>'234','DNUF'=>'521'),
							2=>array('date'=>'2011-08-13','DAU'=>'342','DNU'=>'342','DNUF'=>'523')), 
             'fishbowl'=>array(0=>array('date'=>'2011-08-11','DAU'=>'342','DNU'=>'523','DNUF'=>'522'),
			                1=>array('date'=>'2011-08-12','DAU'=>'333','DNU'=>'324','DNUF'=>'555'),
							2=>array('date'=>'2011-08-13','DAU'=>'943','DNU'=>'533','DNUF'=>'666')),   
			 );
	

//需要合并成如下的形式：		 
	$resultArray=array(
	              0=>array(0=>array('date'=>'2011-08-11','DAU-titan'=>'222','DAU-apollo'=>'298','DAU-bossvegas'=>'','DAU-fishbowl'=>''), 
				           1=>array('date'=>'2011-08-12','DAU-titan'=>'','DAU-apollo'=>'','DAU-bossvegas'=>'','DNU-fishbowl'=>''),
				           2=>array('date'=>'2011-08-13','DAU-titan'=>'','DAU-apollo'=>'','DAU-bossvegas'=>'','DAU-fishbowl'=>'')),
			      
				  1=>array(0=>array('date'=>'2011-08-11','DNU-titan'=>'111','DNU-apollo'=>'290','DNU-bossvegas'=>'','DNU-fishbowl'=>''), 
				           1=>array('date'=>'2011-08-12','DNU-titan'=>'','DNU-apollo'=>'','DNU-bossvegas'=>'','DNU-fishbowl'=>''),
				           2=>array('date'=>'2011-08-13','DNU-titan'=>'','DNU-apollo'=>'','DNU-bossvegas'=>'','DNU-fishbowl'=>'')),
			      
				  2=>array(0=>array('date'=>'2011-08-11','DNUF-titan'=>'444','DNUF-apollo'=>'900','DAUF-bossvegas'=>'','DNUF-fishbowl'=>''), 
				           1=>array('date'=>'2011-08-12','DNUF-titan'=>'','DNUF-apollo'=>'','DAUF-bossvegas'=>'','DNUF-fishbowl'=>''),
				           2=>array('date'=>'2011-08-13','DNUF-titan'=>'','DNUF-apollo'=>'','DAUF-bossvegas'=>'','DAUF-fishbowl'=>''))	
	 );

求合并数组的高效方法。

[乌镇程序员]

$result = array();
$i = 0;
while($ar = current($test)) {
   $key = key($test);
   $result[$i] = array();
   foreach($ar as $v) {
      //print_r($v);
      $ar_keys = array_keys($v);
      $ar_values = array_values($v);
      for($j = 1; $j < count($ar_keys); $j ++) $ar_keys[$j] = $key.$ar_keys[$j];
      array_push($result[$i], array_combine($ar_keys, $ar_values));
   }
   $i ++;
   next($test);
}

print_r($result);

[乌镇程序员]

没看仔细。。这是个三维数组，第三维数组元素的键名拼接的规则是怎么样的啊，莫非原始数组$test只有固定的4个元素？

[乌镇程序员]

1楼代码作废，特此声明。。。合并规则你要说清楚。

[ohmygirl]

[Quote=引用 2 楼 t5500 的回复:]

没看仔细。。这是个三维数组，第三维数组元素的键名拼接的规则是怎么样的啊，莫非原始数组$test只有固定的4个元素？
[/Quote]

恩，是个三维数组，原始的数组中有7个元素，我只写了7个，对应的把原数组的第三维除了date外的键值拼为DAU-fishbowl这种形式。

[ohmygirl]

我只写了4个

[乌镇程序员]

看不懂合并规则。。合并后第三维元素的值是怎么处理的，原始数据中的数字值是有含义的吧，跟第一维的键名是怎么关联起来的？

[ohmygirl]

具体的合并规则是这样的：把时间相同的各条记录合并，相同的部分合并到一条记录中：
如第一维有四个：titan,apollo,bossvegas,fishbowl
那么生成的结果数组中第三维数组应该包含五个（date,DAU-titan,DAU-fishbowl,DAU-bossvegas,DAU-apollo）

生成的结果数组的元素个数与原始数组第三维数组元素个数相等（也就是按照date合并的）。

[ohmygirl]

解决：

  $gamelist=array(0=>array('date','DAU-apollo','DAU-bossvegas','DAU-fishbowl','DAU-titan'),
	                1=>array('date','DNU-apollo','DNU-bossvegas','DNU-fishbowl','DNU-titan'),
					2=>array('date','DNUF-apollo','DNUF-bossvegas','DNUF-fishbowl','DNUF-titan'));
	
	
	for($i=0;$i<=2;$i++)
	  for($j=0;$j<=2;$j++)
	     foreach($gamelist[$i] as $k)
		   { 
		     if($k=='date')
		  	    $result[$i][$j]['date'] = $test['titan'][$j]['date'];
	         else
               {
			    $str = substr($k,4);   
			    $result[$i][$j][$k] = $test[$str][$j][$k]; 
			   }		   
		 }
	   
	
	
	echo '<pre>';	  
    var_export($result);
    echo '</pre>';