Question

我有这个数组：

$array = array(a, b, c, d, e, f, g);

我希望将它分成两个数组，具体取决于索引是偶数还是奇数，如下所示：

$odd = array(a, c, e, g);

$even = array(b, d, f);

提前致谢！

Answer 1

一种解决方案，使用匿名函数和array_walk：

$odd = array();
$even = array();
$both = array(&$even, &$odd);
array_walk($array, function($v, $k) use ($both) { $both[$k % 2][] = $v; });

这样就可以在数组中仅一次分隔项目，但它在“聪明”方面有点过分。它并不比经典，更冗长的

更好

$odd = array();
$even = array();
foreach ($array as $k => $v) {
    if ($k % 2 == 0) {
        $even[] = $v;
    }
    else {
        $odd[] = $v;
    }
}

Answer 2

使用array_filter（PHP＆gt; = 5.6）：

$odd = array_filter($array, function ($input) {return $input & 1;}, ARRAY_FILTER_USE_KEY);
$even = array_filter($array, function ($input) {return !($input & 1);}, ARRAY_FILTER_USE_KEY);

Answer 3

我不确定这是否是最优雅的方式，但它应该有魅力：

$odd=array();
$even=array();
$count=1;
foreach($array as $val)
{
    if($count%2==1)
    {
        $odd[]=$val;
    }
    else
    {
        $even[]=$val;
    }
    $count++;
}

Answer 4

作为一个近乎一线的人，我认为这将是我的最爱：

$even = $odd = array();
foreach( $array as $k => $v )  $k % 2  ?  $odd[] = $v  :  $even[] = $v;

还是一点点？速度：

$even = $odd = array();
foreach( $array as $k => $v )  ( $k & 1 ) === 0  ?  $even[] = $v  :  $odd[] = $v;

更冗长的变体：

$both = array( array(), array() );
// or, if $array has at least two elements:
$both = array();

foreach( $array as $k => $v )  $both[ $k % 2 ][] = $v;
list( $even, $odd ) = $both;

使用array_chunk：

$even = $odd = array();
foreach( array_chunk( $array, 2 ) as $chunk ){
  list( $even[], $odd[] ) = isset( $chunk[1]) ? $chunk : $chunk + array( null, null );
  // or, to force even and odd arrays to have the same count:
  list( $even[], $odd[] ) = $chunk + array( null, null );
}

如果$ array保证具有偶数个元素：

$even = $odd = array();
foreach( array_chunk( $array, 2 ) as $chunk )
  list( $even[], $odd[] ) = $chunk;

PHP 5.5.0+与array_column：

$chunks = array_chunk( $array, 2 );
$even = array_column( $chunks, 0 );
$odd  = array_column( $chunks, 1 );

旧版PHP版本的类似内容。键将是0,2,4，...和1,3,5，....如果您不喜欢这样，请同时应用array_values：

$even = array_intersect_key( $array, array_flip( range( 0, count( $array ), 2 )));
$odd  = array_intersect_key( $array, array_flip( range( 1, count( $array ), 2 )));

或

$even = array_intersect_key( $array, array_fill_keys( range( 0, count( $array ), 2 ), null ));
$odd  = array_intersect_key( $array, array_fill_keys( range( 1, count( $array ), 2 ), null ));

Answer 5

只需循环遍历它们并检查键是偶数还是奇数：

$odd = array();
$even = array();
foreach( $array as $key => $value ) {
    if( 0 === $key%2) { //Even
        $even[] = $value;
    }
    else {
        $odd[] = $value;
    }
}

Answer 6

一

$odd = $even = array();
for ($i = 0, $l = count($array ); $i < $l;) { // Notice how we increment $i each time we use it below, by two in total
    $even[] = $array[$i++];
    if($i < $l)
    {
       $odd[] = $array[$i++];
    }
}

两个

$odd = $even = array();
foreach (array_chunk($array , 2) as $chunk) {
    $even[] = $chunk[0];
    if(!empty( $chunk[1]))
    {
       $odd[] = $chunk[1];
    }
}

Answer 7

根据@ Jon的第二个变体，我在之后使用PHP Smarty v3模板引擎。这是用于显示带有一列或两列模板模型的新闻/博客。

在MySql查询之后，我将执行以下代码：

if(sizeof($results) > 0) {
    $data = array();
    foreach($results as $k => $res) {
        if($k % 2 == 0) {
            $res["class"] = "even";
            $data["all"][] = $data["even"][] = $res;
        }
        else {
            $res["class"] = "odd";
            $data["all"][] = $data["odd"][] = $res;
        }
    }
}

我使用Smarty语法获得了一个包含3个子数组（包括奇数/偶数类）的数组：

所有项目{foreach $data.all as $article}...{/foreach}
仅限奇数项目{foreach $data.odd as $article}...{/foreach}
偶数项目{foreach $data.even as $article}...{/foreach}

希望它可以帮助一些人...

Answer 8

$odd = [];
$even = [];
while (count($arr)) {
    $odd[] = array_shift($arr);
    $even[] = array_shift($arr);
}

Answer 9

    <?php

$array1 = array(0,1,2,3,4,5,6,7,8,9);
$oddarray = array();
$evenarray = array();

$count = 1;

echo "Original: ";
foreach ($array1 as $value)
 {
    echo "$value";
}

echo "<br> Even: ";

foreach ($array1 as $print) 
{
    if ($count%2==1) 
    {
        $evenarray = $print;
        echo "$print";
    }
    $count++;
}

echo "<br> Odd: ";

foreach ($array1 as $print2) {
    if ($count%2!=1) 
    {
        $oddarray[] = $print2;
        echo "$print2";
    }
    $count++;
}

?>

Output:

Original: 0123456789
Even: 02468
Odd: 13579

Answer 10

使用array_chunk和array_map的另一种功能解决方案。当源数组的大小为奇数时，最后一行从第二个数组中删除空项

list($odd, $even) = array_map(null, ...array_chunk($ar,2));
if(count($ar) % 2) array_pop($even);

通过索引偶数或奇数将数组拆分为两个数组

10 个答案: