php - 以小时为单位的平均时间

Question

我想获得平均小时数，但我遇到了问题，因为如果我使用 DateTime.add() 对象在超过 24 小时时添加时间，它会将时间转换为“日”。基本上我只想在“小时”内获得这些时间的平均值。请帮忙。

$time = array();
$time[] = '00:00:58';
$time[] = '00:45:08';
$time[] = '00:00:49';
$time[] = '00:11:26';
$time[] = '02:34:39';
$time[] = '02:18:24';
$time[] = '02:24:11';
$time[] = '01:23:19';
$time[] = '00:27:30';
$time[] = '02:14:50';
$time[] = '00:48:47';
$time[] = '02:19:52';
$time[] = '02:04:29';
$time[] = '01:08:46';
$time[] = '00:06:57';
$time[] = '00:35:31';
$time[] = '02:16:59';
$time[] = '02:23:33';
$time[] = '02:36:30';
$time[] = '02:11:47';
$time[] = '02:40:44';
$time[] = '02:03:42';
$time[] = '01:45:50';

Answer 1

尝试这个，

$total_time=0;

foreach($time as $t){
     $total_time+=strtotime($t)-1373241600;
}

所以 $total_time 将是一个整数，您可以将其除以 3600 和 count($time) 以获得平均小时数:)

$average_time = ($total_time/count($time))/3600;

Answer 2

如果您不想依赖时区设置，减去当天的时间戳或对 strtotime 的迟缓感到恼火，请改用简单的数学：

$timestamps = array_map(function($item){
    list($h, $m, $s) = explode(':', $item);
    return intval($h) * 3600 + intval($m) * 60 + intval($s);
}, $time);
$avg = ceil(array_sum($timestamps)/count($timestamps));
echo gmdate('H:i:s', $avg);

Answer 3

试试这个（注意闭包在 5.3+ 中可用）：

$timestamps = array_map(function($a){return strtotime($a);}, $time);
$avg = (array_sum($timestamps)/count($timestamps))/3600;