当值相同时，将 2 个数组合并到第三个新数组中

Question

我有一个名为 job counts 的数组，如下 $job=>

Array
        (
            [0] => Array([count] => 3[yearmonth] => 2019-7)
        
            [1] => Array([count] => 3[yearmonth] => 2019-9)
        
            [2] => Array([count] => 5[yearmonth] => 2019-10))           
    )

第二个日期数组为 $date=>

Array
    ([0] => Array([yearmonth] => 2019-6)
    
     [1] => Array([yearmonth] => 2019-7)
    
     [2] => Array([yearmonth] => 2019-8)
    
     [3] => Array([yearmonth] => 2019-9)
    
     [4] => Array([yearmonth] => 2019-10)
)

作业数组不是连续数组。所以在这里我想要的是如果一个条目在 $date 中可用但在 $job 中不存在，那么创建第三个数组，其中 count = 0 和 yearmonth 值。 类似下面的东西

Array
    ([0] => Array([count] => 0[yearmonth] => 2019-6)
    
     [1] => Array([count] => 3[yearmonth] => 2019-7)
    
     [2] => Array([count] => 0[yearmonth] => 2019-8)
    
     [3] => Array([count] => 3[yearmonth] => 2019-9)
    
     [4] => Array([count] => 5[yearmonth] => 2019-10)
)

Answer 1

要在迭代 $dates 数组时有效地在 $jobs 数组中搜索日期，请创建一个“查找”数组。

查找应具有表示关系数据的键 (yearmonth)。

使用 $dates 数组中遇到的每个日期，检查 yearmonth 值是否表示为查找中的键 - 如果是，则使用 count 值，如果不是，则设置为 0。

& 变量前的意思是“通过引用修改——这样原始数组就发生了变异，而不需要声明一个新的输出数组。

?? 是“空合并运算符”，如果变量为 null/未声明，则允许使用备用值。

代码：(Demo)

$jobs = [
    ['count' => 3, 'yearmonth' => '2019-7'],
    ['count' => 3, 'yearmonth' => '2019-9'],
    ['count' => 5, 'yearmonth' => '2019-10'],         
];
$dates = [
    ['yearmonth' => '2019-6'],
    ['yearmonth' => '2019-7'],
    ['yearmonth' => '2019-8'],
    ['yearmonth' => '2019-9'],
    ['yearmonth' => '2019-10'],
];

$lookup = array_column($jobs, 'count', 'yearmonth');

foreach ($dates as &$date) {
    $date['count'] = $lookup[$date['yearmonth']] ?? 0;
}

var_export($dates);

输出：

array (
  0 => 
  array (
    'yearmonth' => '2019-6',
    'count' => 0,
  ),
  1 => 
  array (
    'yearmonth' => '2019-7',
    'count' => 3,
  ),
  2 => 
  array (
    'yearmonth' => '2019-8',
    'count' => 0,
  ),
  3 => 
  array (
    'yearmonth' => '2019-9',
    'count' => 3,
  ),
  4 => 
  array (
    'yearmonth' => '2019-10',
    'count' => 5,
  ),
)

---

或者，如果您要求声明一个新的输出数组并且关联子数组必须与您的问题中的顺序相同，那么这是我对相同技术的调整：

代码：(Demo)

$lookup = array_column($jobs, null, 'yearmonth');

$result = [];
foreach ($dates as $date) {
    $result[] = $lookup[$date['yearmonth']] ?? ['count' => 0] + $date;
}

var_export($result);

查找数组现在包含完整的子数组数据。 array_column() 使用null 作为第二个参数以防止隔离单行，并使用yearmonth 作为第三个参数分配第一级键。

foreach 循环不再“通过引用修改”。空合并运算符仍然用于避免调用isset()，是简洁编写回退数据的理想选择。我在包含count 元素的硬编码数组和$date 数组之间使用“联合运算符”——避免调用array_merge() 或手动编写['yearmonth' => $date['yearmonth']。这是一种合适的合并技术，因为键是唯一的，并且与每个子数组相关。

Answer 2

永远不要低估能够在 php 数组中以非常灵活的方式/方式使用键的能力。

<?php
//Copied from mickmackusa
$jobs = [
    ['count' => 3, 'yearmonth' => '2019-7'],
    ['count' => 3, 'yearmonth' => '2019-9'],
    ['count' => 5, 'yearmonth' => '2019-10'],         
];
$dates = [
    ['yearmonth' => '2019-6'],
    ['yearmonth' => '2019-7'],
    ['yearmonth' => '2019-8'],
    ['yearmonth' => '2019-9'],
    ['yearmonth' => '2019-10'],
];
//End copy

/*
    Make the keys of $jobs be the same as values
    of yearmonth: (This makes it easy to compare later on)

    Array
    (
        [0] => Array
            (
                [count] => 3
                [yearmonth] => 2019-7
            )

        [1] => Array
            (
                [count] => 3
                [yearmonth] => 2019-9
            )

        [2] => Array
            (
                [count] => 5
                [yearmonth] => 2019-10
            )

        [2019-7] => Array
            (
                [count] => 3
                [yearmonth] => 2019-7
            )

        [2019-9] => Array
            (
                [count] => 3
                [yearmonth] => 2019-9
            )

        [2019-10] => Array
            (
                [count] => 5
                [yearmonth] => 2019-10
            )

    )

*/
foreach($jobs as $key=>$item) {
    $jobs[$item['yearmonth']] = $item;
}

//Create a third array $third_arr based on your $jobs and $dates array
$third_arr = [];
foreach($dates as $item) {
    $key_value = $item['yearmonth'];
    if (isset($jobs[$key_value]['yearmonth'])) {
        //Available in dates and present in $jobs
        //Just copy values from the $jobs item which relates to this yearmonth
        $third_arr[] = $jobs[$key_value];
    }
    else {
        //Available in dates but not present in $job
        $third_arr[] = ['count'=>0, 'yearmonth'=>$key_value];
    }
}

第三个数组$third_arr的输出：

Array
(
    [0] => Array
        (
            [count] => 0
            [yearmonth] => 2019-6
        )

    [1] => Array
        (
            [count] => 3
            [yearmonth] => 2019-7
        )

    [2] => Array
        (
            [count] => 0
            [yearmonth] => 2019-8
        )

    [3] => Array
        (
            [count] => 3
            [yearmonth] => 2019-9
        )

    [4] => Array
        (
            [count] => 5
            [yearmonth] => 2019-10
        )

)

以上代码的压缩版本如下所示：

foreach($jobs as $key=>$item) {
    $jobs[$item['yearmonth']] = $item;
}
$third_arr = [];
foreach($dates as $item) {
    $kvalue = $item['yearmonth'];
    isset($jobs[$kvalue]['yearmonth']) ? 
    $third_arr[] = $jobs[$kvalue] : $third_arr[] = ['count'=>0, 'yearmonth'=>$kvalue];
}

Answer 3

因此，如果您将count => 0 添加到$date 数组并在yearmonth 上建立索引，您可以在yearmonth 上索引$job 并将其合并到$date：

$result = array_merge(array_column(array_map(function($v) { $v['count'] = 0; return $v; },
                                   $date), null, 'yearmonth'),
                      array_column($job, null, 'yearmonth'));