PHP for for循環for循環。試圖對多個數組列進行求和以得到組合

Question

我想循環遍歷一個二維數組並自動獲取列的組合的總和。

假設我有一個名爲$ a的4列的排列：0,1,2,3，

$a=array(); 
$a[0][0]=1; 
$a[0][1]=3; 
$a[0][2]=5; 

$a[1][0]=10; 
$a[1][1]=2; 
$a[1][2]=5; 
$a[1][3]=7; 

$a[2][0]=9; 
$a[2][1]=8; 
$a[2][2]=9; 
$a[2][3]=8; 

$a[3][0]=9; 
$a[3][1]=8; 
$a[3][2]=9; 
$a[3][3]=8; 
$a[3][4]=1; 

，我試圖總結了所有喜歡和列（0組合， 0; 1,0; 2; 0.3; 0）等使用此代碼

for($i=0;$i<count($a[0]);$i++){ 
for($l=0;$l<count($a[1]);$l++){ 
for($s=0;$s<count($a[2]);$s++){ 
for($m=0;$m<count($a[3]);$m++){ 
echo $sum[]= $a[0][$i]+$a[1][$l]+$a[2][$s]+$a[3][$m]; 
echo $sum; 
    echo "<br>"; 
    } 
    } 
} 
} 

?> 

和代碼的工作，問題是，我手動做這些for循環，必須有某種方式在我可以通過某種方式簡化這個插入列數的計數？

我想是這樣

$numberofcolumns=4; 

for($n=0;$n<$numberofcolumns;$n++){ 
for($i=0;$i<count($a[$n]);$i++){ 
for($m=0;$m<count($a[$n+1]);$m++){ 
echo $sums[]= $a[$n][$i]+$a[$n+1][$m]; 
} 
} 
} 

，但不起作用，必須有某種方式來簡化for循環，這樣我就不必手動鍵入for循環每列

有人有線索嗎？

Answer 1

您可以使用遞歸，或者只是直接嵌套循環這一點，但是當使用組合或置換，可能​​性的總數可能會爆炸併成爲一個龐大的數字，這會消耗大量的內存以至於無法運行代碼。使用迭代器是交換cpu效率以提高內存效率的好方法。這是我寫的一個迭代器。

class CartesianProductIterator implements Iterator { 
    protected $iterators; 

    function __construct(array $iters) { 
     $this->iterators = $iters; 
    } 

    function rewind() { 
     foreach ($this->iterators as $it) { 
      $it->rewind(); 
     } 
    } 

    function current() { 
     $values = array(); 
     foreach ($this->iterators as $it) { 
      $values[] = $it->current(); 
     } 
     return $values; 
    } 

    function key() { 
     return null; 
    } 

    function next() { 
     /*  
     loop them in reverse, but exclude first 
     why? example, odometer: 55199 
     you always check the rightmost digit first to see if incrementing it would roll it over and need to be "rewound" to 0, 
     which causes the digit to the left to increase as well, which may also cause it to roll over as well, and so on... 
     looping in reverse operates from right column to the left. 
     we dont rewind the first column because if the leftmost column is on its last element and needs to roll over 
     then this iterator has reached its end, and so rewind() needs to be explicitly called 
     */ 
     for ($i = count($this->iterators) - 1; $i > 0; --$i) { 
      $it = $this->iterators[$i]; 
      $it->next(); 
      if ($it->valid()) { 
       // were done advancing because we found a column that didnt roll over 
       return; 
      } else { 
       $it->rewind(); 
      } 
     } 

     //if execution reached here, then all of the columns have rolled over, so we must attempt to roll over the left most column 
     $this->iterators[0]->next(); 
    } 

    function valid() { 
     return $this->iterators[0]->valid(); 
    } 
} 

然後用它作爲

$iterators = array(); 
foreach ($a as $columnNumber => $values) { 
    $iterators[] = new ArrayIterator($values); 
} 
foreach (new CartesianProductIterator($iterators) as $combo) { 
    // combo has 1 value from each of the ArrayIterators we instantiated 
    printf("summing %s = %d\n", join('+', $combo), array_sum($combo)); 
} 

繼承人演示http://codepad.org/UasdgvWf

Answer 2

您可以使用RecursiveIteratorIterator

嘗試

$a = array(); 
$a [0] [0] = 1; 
$a [0] [1] = 3; 
$a [0] [2] = 5; 

$a [1] [0] = 10; 
$a [1] [1] = 2; 
$a [1] [2] = 5; 
$a [1] [3] = 7; 

$a [2] [0] = 9; 
$a [2] [1] = 8; 
$a [2] [2] = 9; 
$a [2] [3] = 8; 

$a [3] [0] = 9; 
$a [3] [1] = 8; 
$a [3] [2] = 9; 
$a [3] [3] = 8; 
$a [3] [4] = 1; 

$sum = 0; 
$array = new RecursiveIteratorIterator (new RecursiveArrayIterator ($a)); 
foreach ($array as $key => $value) { 
     $sum += $value; 
} 
echo $sum; 

輸出

102 

使用$array = new RecursiveIteratorIterator (new RecursiveArrayIterator ($a[1]));得到各部分的總和......

Answer 3

如果我正確認識你，這個功能應該可以做到這一點：

<?php 
function recursive_sum($arr) { 

    $sum = 0; 

    foreach($arr as $value) { 
     if(is_array($value)) { 
      $sum += recursive_sum($value); 
     } 
     else { 
      $sum += $value; 
     } 
    } 

    return $sum; 
} 
?> 

只需撥打recursive_sum($a)把所有的值的總和數組中，像這樣：

<?php 
    echo recursive_sum($a); 
?>  

Answer 4

<? //PHP 5.4+ 
$a=[]; 
$a[0][0]=1; 
$a[0][1]=3; 
$a[0][2]=5; 

$a[1][0]=10; 
$a[1][1]=2; 
$a[1][2]=5; 
$a[1][3]=7; 

$a[2][0]=9; 
$a[2][1]=8; 
$a[2][2]=9; 
$a[2][3]=8; 

$a[3][0]=9; 
$a[3][1]=8; 
$a[3][2]=9; 
$a[3][3]=8; 
$a[3][4]=1; 

//This is downright evil, but it works. 
eval(\array_reduce(
    \array_reverse(\array_keys($a)), 
    static function($eval, $key){ 
    return "foreach(\$a[$key]as\$i$key)$eval+\$i$key"; 
    }, 
    '{echo$sum[]=0') . ';echo"$sum<br/>";}'); 
?>