如何在一个新的多维数组中分离具有相同索引的键值

Question

我有下面的多维数组，需要帮助在特定的编队中转换多维

Array
(
[0] => Array
(
[0] => 20
[1] => 15
[2] => 0
)
[1] => Array
(
[0] => 28
[1] => 45
[2] => 0
)
[2] => Array
(
[0] => 20
[1] => 5
[2] => 50
)
[3] => Array
(
[0] => 48
[1] => 20
[2] => 0
)
)

并且想要像这样转换：

Array
(
[0] => Array
(
[0] => 20
[1] => 28
[2] => 20
[3] => 48
)
[1] => Array
(
[0] => 15
[1] => 45
[2] => 5
[3] => 20
)
[2] => Array
(
[0] => 0
[1] => 0
[2] => 50
[3] => 0
)
)

在上面的输出中，我在一个新的数组中分离了具有相同索引的键值。

有可能吗？

Answer 1

您可以使用array_map函数

$result = array_map(null, ...$array);

演示

Answer 2

这里有一个解决方案：

<?php
$array =
[
0 =>
[
0 => 20,
1 => 15,
2 => 0
],
1 =>
[
0 => 28,
1 => 45,
2 => 0
],
2 =>
[
0 => 20,
1 => 5,
2 => 50
],
3 =>
[
0 => 48,
1 => 20,
2 => 0
]
];
$result = [];
echo '<pre>';
foreach ($array as $key => $value) {
for ($i = 0; $i<count($value); $i++) {
$result[$i][] = $value[$i];
}
}

var_dump($result);

结果：

array(3) {
[0]=>
array(4) {
[0]=>
int(20)
[1]=>
int(28)
[2]=>
int(20)
[3]=>
int(48)
}
[1]=>
array(4) {
[0]=>
int(15)
[1]=>
int(45)
[2]=>
int(5)
[3]=>
int(20)
}
[2]=>
array(4) {
[0]=>
int(0)
[1]=>
int(0)
[2]=>
int(50)
[3]=>
int(0)
}
}

Answer 3

这段代码运行良好，正在做您想要做的事情，请尝试在编辑器上复制粘贴并运行它。

<?php 
$originalArray = Array(
Array
(
20,
15,
0
),
Array
(
28,
45,
0
),
Array
(
20,
5,
50
),
Array
(
48,
20,
0
)
); //echo"<pre>";print_R($originalArray); 00 10 20 30 40 : 01,11,21,31 :  02,12,22,32,    
$newArray =array();
//echo count($originalArray);
//echo count($originalArray[1]);
$count = array_map('count', $originalArray);
$min = array_keys($count , max($count))[0];
$largest_arr = $originalArray[$min];
for($i=0;$i<count($largest_arr);$i++){
array_push($newArray,$originalArray[$i]);
for($j=0;$j<count($originalArray);$j++){
echo $j."--".$i."---".count($originalArray)."----".$originalArray[$j][$i]."<br>";
array_push($newArray[$i],$originalArray[$j][$i]);
}
}
echo "<pre>";print_R($newArray);
?>

Answer 4

我从PHP中找到了解决方案-如何在后翻转2D阵列的行和列

$out = array();
foreach ($array as  $rowkey => $row) {
foreach($row as $colkey => $col){
$out[$colkey][$rowkey]=$col;
}
}

这非常适合我。