php - 将时间:[Distance,速度]数据转换为距离:[Time,速度]数据

Question

我在想出一个算法时遇到了麻烦...

我有一系列 GPS 数据，以 1 秒为间隔记录时间、速度、距离。假设距离是米，速度是米/秒。可能有超过 2 小时的数据，或 7200 个点。这里的“时间”字段主要是作为引用。

因此，前 5 秒将是类似这样的值，其中 [1-5] 是秒。

$data = array(
  1 : array('distance'=>0, 'time'=>'2014-01-09 17:50:00', 'speed'=>0.0),
  2 : array('distance'=>2, 'time'=>'2014-01-09 17:50:01', 'speed'=>2.0),
  3 : array('distance'=>6, 'time'=>'2014-01-09 17:50:02', 'speed'=>4.0),
  4 : array('distance'=>10, 'time'=>'2014-01-09 17:50:03', 'speed'=>4.0),
  5 : array('distance'=>12, 'time'=>'2014-01-09 17:50:04', 'speed'=>2.0)
);

我想将其转换为以 1 米为间隔列出的数据，例如 [1-6] 是米。

$data = array(
  1 : array('seconds'=>1.5, 'time'=>'2014-01-09 17:50:01.500', 'speed'=>.666),
  2 : array('seconds'=>2, 'time'=>'2014-01-09 17:50:02', 'speed'=>2.0),
  3 : array('seconds'=>2.25, 'time'=>'2014-01-09 17:50:02.250', 'speed'=>4.0),
  4 : array('seconds'=>2.5, 'time'=>'2014-01-09 17:50:02.500', 'speed'=>4.0),
  5 : array('seconds'=>2.75, 'time'=>'2014-01-09 17:50:02.750', 'speed'=>4.0),
  6 : array('seconds'=>3, 'time'=>'2014-01-09 17:50:03', 'speed'=>4.0)
);

当然，这可以在没有时间字段的情况下完成。我在计算时遇到了问题，因为它绝对不是一对一的。如果我们从 7200 秒的数据开始，我们可能会根据所覆盖的距离(多于或少于 7200 米)最终得到或多或少的数据。

编辑(2014 年 1 月 10 日)

下面是这两个方法的实际实现。实际上，我很难决定我更喜欢哪种方法，迭代方法还是递归方法。我可能会去迭代

方法 1，迭代(@Ezequiel Muns，我做了非常小的修改):

function timeToDistance($data) {
  if(sizeof($data) == 0){ return; }
  $startTime = $data[0]['time'];

  $prev = null;
  $result = array();
  foreach ($data as $secs => $row) {
    $row['seconds'] = $secs; // to simplify passing in secs
    if ($prev == null) {
      // make sure we have a pair
      $prev = array( 'distance'=>0 );
    }
    foreach (distanceRowsBetween($startTime,$prev, $row) as $dist => $distRow) {
      $result[$dist] = $distRow;
    }
    $prev = $row;
  }
  return $result;
}

function distanceRowsBetween($startTime,$prevRow, $nextRow) {
  // Return the by-distance rows that are between $prevRow (exclusive)
  // and $nextRow (inclusive)
  $rows = array();
  $currDist = $prevRow['distance'];
  while (true) {
    // try to move to the next whole unit of distance
    $dDist = ceil($currDist) - $currDist;
    $dDist = $dDist == 0.0? 1.0 : $dDist; // dDist is 1 unit if currDist is whole
    $currDist += $dDist;
    if ($currDist > $nextRow['distance'])
      break;

    $currSpeed = $nextRow['speed'];
    $currSecs = strtotime($nextRow['time']) - strtotime($startTime);
    $currTime = $nextRow['time'];

    $rows[$currDist] =  array(
                          'speed' => $currSpeed,
                          'seconds' => $currSecs,
                          'time' => $currTime,
                         );
  }
  return $rows;
}

方法 2，递归(@Nathaniel Ford 伪代码，我的实际代码):

function data2dist($time_data = array()){
  $dist_data = array();
  if(sizeof($time_data) == 0){ return $dist_data; }

  $start_point = array_shift($time_data);
  $start_time = $start_point['time'];

  data2dist_sub($start_time, $time_data,$dist_data,$start_point);

  return $dist_data;
}

function data2dist_sub($start_time,&$time_data, &$dist_data, $start_point = array()){
  if(sizeof($time_data) == 0 && !isset($start_point)){
    return;
  }

  if(sizeof($dist_data) == 0){
    $prev_dist = 0;
  } else {
    $prev_dist = $dist_data[sizeof($dist_data)-1]['distance'];
  }
  // since distances are accumulating, get curr distance by subtracting last one
  $point_dist = $start_point['distance'] - $prev_dist;

  if($point_dist == 1){
    // exactly 1: perfect, add and continue
    $dist_data[] = $start_point;
    $start_point = array_shift($time_data);
  } else if($point_dist > 1){
    // larger than 1: effectively remove 1 from current point and send it forward
    $partial_point = $start_point;
    $partial_point['distance'] = 1 + $prev_dist;
    $dist_data[] = $partial_point;

  } else if($point_dist < 1){
    // less than 1, carry forward to the next item and continue (minor: this partial speed is absorbed into next item)
    $start_point = array_shift($time_data);
    if(!isset($start_point)){   return;   }

    $start_point['distance'] += $point_dist;
  }
  data2dist_sub($start_time,$time_data,$dist_data,$start_point);
}

Answer 1

您可以简化这一点，注意对于每一对连续的按时间行，您需要计算 0 个或更多按距离行，而这些仅取决于这两个按时间行。

因此，从一个函数开始进行这种更简单的计算，这是一个骨架，为了简单起见，将转换后的“秒”、“速度”和“时间”值的计算保留在外。

function distanceRowsBetween($prevRow, $nextRow) {
    // Return the by-distance rows that are between $prevRow (exclusive)
    // and $nextRow (inclusive)
    $rows = array();
    $currDist = $prevRow['distance'];
    while (true) {
        // try to move to the next whole unit of distance
        $dDist = ceil($currDist) - $currDist;
        $dDist = $dDist == 0.0? 1.0 : $dDist; // dDist is 1 unit if currDist is whole
        $currDist += $dDist;
        if ($currDist > $nextRow['distance'])
            break;

        // calculate $currSecs at distance $currDist
        // calculate $currSpeed 
        // calculate $currTime 

        $rows[$currDist] = array(
            'speed' => $currSpeed,
            'seconds' => $currSecs,
            'time' => $currTime,
        );
    }
    return $rows;
}

现在您已经有了这些，剩下的就是迭代输入中的每个连续对，并按距离累积结果行:

function timeToDistance($data) {
    $prev = null;
    $result = array();
    foreach ($data as $secs => $row) {
        $row['seconds'] = $secs; // to simplify passing in secs
        if ($prev == null) {
            $prev = $row; // make sure we have a pair
            continue;
        }
        foreach (distanceRowsBetween($prev, $row) as $dist => $distRow) {
            $result[$dist] = $distRow;
        }
        $prev = $row;
    }
    return $result;
}

请注意，在此函数中，我正在填充并传入行中的当前“秒”值，以减少传递到前一个函数的参数数量。