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php - 如何获得将第二个数组的元素作为参数对数组的所有元素应用操作的结果数组?

转载 作者:行者123 更新时间:2023-12-03 20:47:05 25 4
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好的 - 所以这个应该很简单。我有两个数组。数组 1 的结构如下:

[
[
'value' => 3337
'end_time' => '2012-10-07T07:00:00+0000'
],
[
'value' => 2811
'end_time' => '2012-10-09T07:00:00+0000'
],
[
'value' => 1318
'end_time' => '2012-10-14T07:00:00+0000'
]

和数组 2:

[  
[
'_id' => '2012-10-07'
'value' => 50
],
[
'_id' => '2012-10-09'
'value' => 15
],
[
'_id' => '2012-10-10'
'value' => 2
]
]

SO....我想根据 array1[_id] = substr(array2['end_time'], 0, 10) 将数组 1 的元素除以数组 2 的元素。

数组 2 没有数组 1 中所有 ID 的结束时间值。

生成的数组应该类似于这样:

[
'2012-10-01' => 0.11
'2012-10-02' => 0 (if this date is not available in the second array).
'2012-10-03' => 0.12312
]

这样做:

$kermit = array();
foreach($dauresult['data']['0']['values'] as $subdau) {
foreach($revenue['results'] as $subrev) {
$date = substr($subdau['end_time'], 0, 10);
if($date == $subrev['_id']) {
$kermit[$date] = $subrev['value']/$subdau['value'];
}
}
}

我可以得到一个如下所示的数组:

Array
(
[2012-09-30] => 0.0044950554390171
[2012-10-01] => 0.019565990750623
[2012-10-05] => 0.015487397509869
[2012-10-07] => 0.020177562550444
[2012-10-09] => 0.0075150300601202
[2012-10-10] => 0.00095831336847149
[2012-10-11] => 0.0010183299389002
[2012-10-12] => 0.0010126582278481
[2012-10-13] => 0.029866666666667
[2012-10-14] => 0.029779630732579
[2012-10-15] => 0.011926058437686
[2012-10-18] => 0.0018844221105528
[2012-10-19] => 0.0005941770647653
[2012-10-21] => 0.0023781212841855
[2012-10-27] => 0.0011820330969267
[2012-10-28] => 0.0011467889908257
)

但我不能做的是让缺失的日期(10-20、10-16 等)显示为 0。愚蠢的问题,但这让我有点抓狂。有什么建议么?谢谢!-蕨类

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