从 csv 文件计算 PHP 中的平均持续时间。平均时间似乎是 0 毫秒
Calculate Average Time Duration in PHP from a csv file. Average time seems 0 milisseconds
我有一个包含这些值的 csv 文件:
Iteration, Duration
0,00:00:02.694414
81,00:00:02.790214
82,00:00:02.933225
83,00:00:03.077099
84,00:00:03.220184
85,00:00:03.363437
在它的第二列,我有一个 H:i:s.ss
格式的持续时间,我需要计算平均持续时间:sum(Duration)/csv_size
为了做到这一点,我使用了这个函数:
/**
* convert a DateTimeIntervalIn milliseconds
* @return int Time in miliseconds
*/
function intervalInMiliseconds(DateInterval $interval) : float
{
return ($interval->days*86400 + $interval->h*3600
+ $interval->i*60 + $interval->s)*1000 + ($interval->f*1000);
}
/**
* Undocumented function
*
* @return int
*/
function calculateAverageTime(string $csvFile)
{
if ( !file_exists($csvFile) ) {
throw new Exception('File not found.');
}
$file = fopen($csvFile,"r");
$count=0;
$duration= new DateTime('00:00');
$durationCalc= clone $duration;
while(($data = fgetcsv($file, 1000, ",")) !== FALSE){
$duration->add( DateInterval::createFromDateString($data[1]));
$count++;
}
if($count == 0)
{
throw new Exception("No Data In the CSV");
}
$duration=$duration->diff($durationCalc);
$averageSeconds=intervalInMiliseconds($duration)/$count;
fclose($file);
return $averageSeconds;
}
$time = calculateAverageTime(__DIR__.'/timestamps.csv');
echo $time;
但我的问题是如何以毫秒为单位格式化 DateInterval
以便能够计算以毫秒为单位的平均时间我得到 0,这个结果对我来说似乎不合适。对于上面给出的数据,我希望是几分之一毫秒,但不是 0。
所以你知道如何解决这个问题吗?
编辑 1:
快速 var_dump 显示
var_dump($datetime);
显示无法将我的时间戳解析为 DateInterval:
class DateInterval#1 (16) {
public $y =>
int(0)
public $m =>
int(0)
public $d =>
int(0)
public $h =>
int(0)
public $i =>
int(0)
public $s =>
int(0)
public $f =>
double(0)
public $weekday =>
int(0)
public $weekday_behavior =>
int(0)
public $first_last_day_of =>
int(0)
public $invert =>
int(0)
public $days =>
bool(false)
public $special_type =>
int(0)
public $special_amount =>
int(0)
public $have_weekday_relative =>
int(0)
public $have_special_relative =>
int(0)
}
那么如何将时间戳从 csv 转换为 DateInterval?
用 milliseconds/microseconds 和 DateTime 求和时间可能会导致不准确。
我会完全按照@04FS 评论中的建议去做。
原理:
$durations = [
"00:00:02.694414",
"00:00:02.790214",
"00:00:02.933225",
"00:00:03.077099",
"00:00:03.220184",
"00:00:03.363437"
];
$sum = 0.0;
foreach($durations as $duration){
list($h,$m,$s) = explode(':',$duration);
$sum += $h * 3600 + $m * 60 + $s;
}
$avg = $sum/count($durations);
echo $avg; //3.0130955
函数array_column()可用于从二维数组 (csv) 中提取持续时间。
我有一个包含这些值的 csv 文件:
Iteration, Duration
0,00:00:02.694414
81,00:00:02.790214
82,00:00:02.933225
83,00:00:03.077099
84,00:00:03.220184
85,00:00:03.363437
在它的第二列,我有一个 H:i:s.ss
格式的持续时间,我需要计算平均持续时间:sum(Duration)/csv_size
为了做到这一点,我使用了这个函数:
/**
* convert a DateTimeIntervalIn milliseconds
* @return int Time in miliseconds
*/
function intervalInMiliseconds(DateInterval $interval) : float
{
return ($interval->days*86400 + $interval->h*3600
+ $interval->i*60 + $interval->s)*1000 + ($interval->f*1000);
}
/**
* Undocumented function
*
* @return int
*/
function calculateAverageTime(string $csvFile)
{
if ( !file_exists($csvFile) ) {
throw new Exception('File not found.');
}
$file = fopen($csvFile,"r");
$count=0;
$duration= new DateTime('00:00');
$durationCalc= clone $duration;
while(($data = fgetcsv($file, 1000, ",")) !== FALSE){
$duration->add( DateInterval::createFromDateString($data[1]));
$count++;
}
if($count == 0)
{
throw new Exception("No Data In the CSV");
}
$duration=$duration->diff($durationCalc);
$averageSeconds=intervalInMiliseconds($duration)/$count;
fclose($file);
return $averageSeconds;
}
$time = calculateAverageTime(__DIR__.'/timestamps.csv');
echo $time;
但我的问题是如何以毫秒为单位格式化 DateInterval
以便能够计算以毫秒为单位的平均时间我得到 0,这个结果对我来说似乎不合适。对于上面给出的数据,我希望是几分之一毫秒,但不是 0。
所以你知道如何解决这个问题吗?
编辑 1: 快速 var_dump 显示
var_dump($datetime);
显示无法将我的时间戳解析为 DateInterval:
class DateInterval#1 (16) {
public $y =>
int(0)
public $m =>
int(0)
public $d =>
int(0)
public $h =>
int(0)
public $i =>
int(0)
public $s =>
int(0)
public $f =>
double(0)
public $weekday =>
int(0)
public $weekday_behavior =>
int(0)
public $first_last_day_of =>
int(0)
public $invert =>
int(0)
public $days =>
bool(false)
public $special_type =>
int(0)
public $special_amount =>
int(0)
public $have_weekday_relative =>
int(0)
public $have_special_relative =>
int(0)
}
那么如何将时间戳从 csv 转换为 DateInterval?
用 milliseconds/microseconds 和 DateTime 求和时间可能会导致不准确。 我会完全按照@04FS 评论中的建议去做。
原理:
$durations = [
"00:00:02.694414",
"00:00:02.790214",
"00:00:02.933225",
"00:00:03.077099",
"00:00:03.220184",
"00:00:03.363437"
];
$sum = 0.0;
foreach($durations as $duration){
list($h,$m,$s) = explode(':',$duration);
$sum += $h * 3600 + $m * 60 + $s;
}
$avg = $sum/count($durations);
echo $avg; //3.0130955
函数array_column()可用于从二维数组 (csv) 中提取持续时间。