Question

我在PHP中实现了线性回归的代码，其目的是为每个客户提供交货日期（无论每个客户在数据库中有多少交货日期），然后将这些交货日期（转换为天数）用作线性回归方程作为y参数，最终预测未来的交货日期/订单日期，并且基于该预测和与当前日期的比较，系统将采取诸如发送推送通知提醒之类的一些动作。我的实现代码如下：

$conn = mysql_connect($dbhost,$dbuser,$dbpass) or die (mysql_error());
    mysql_select_db($db, $conn) or die(mysql_error());



    //Outer Loop to save all eligable customer ID's

    $idarray=array();

$firstloop=mysql_query("Select id FROM ordering GROUP BY id",$conn);
$number_of_rows_initial=mysql_num_rows($firstloop);
while (($row = mysql_fetch_array($firstloop, MYSQL_ASSOC)) !== false){
  $idarray[] = $row; // add the row in to the results (data) array
 // print_r($idarray);
}


for($counter=0; $counter<$number_of_rows_initial; $counter++)   
        {
            $id=$idarray[$counter]['id'];



//First confirm if user is already registered via website

$result = mysql_query("SELECT TO_DAYS(date) FROM ordering WHERE id=$id",$conn);
$num_rows = mysql_num_rows($result);
// mysql_query($result,$conn) or die(mysql_error();

//echo $num_rows;


$data = array();
$days=array();
$xAxis=array();

$ldate=array();

while (($row = mysql_fetch_array($result, MYSQL_ASSOC)) !== false){
  $data[] = $row; // add the row in to the results (data) array
}

// print_r($data); // print result


//This part is working now!!!!

for ($x = $num_rows-1 ; $x > 0; $x--) {
    $days[$x] = $data[$x]['TO_DAYS(date)'] -$data[$x-1]['TO_DAYS(date)'];

    $xAxis[$x]=$x;
}

$days[0]=30;
print_r("<br />",$days);
//print_r($xAxis);




//my implementation of regression calculation

 // calculate sums
  $x_sum = (array_sum($xAxis));
  $y_sum = array_sum($days);
  echo "<br />","Customer ID is: ",$id,"<br />";
  echo "Sum of X Axis: ",$x_sum, "<br />";
  echo "Sum of Y Axis: ",$y_sum,"<br />" ;

  $xx_sum = 0;
  $xy_sum = 0;

  for($i = 0; $i < $num_rows; $i++) {

    $xy_sum+=($xAxis[$i]*$days[$i]);
    $xx_sum+=($xAxis[$i]*$xAxis[$i]);

  }

  // calculate slope
  $m = (($num_rows * $xy_sum) - ($x_sum * $y_sum)) / (($num_rows * $xx_sum) - ($x_sum * $x_sum));

  // calculate intercept
  $b = ($y_sum - ($m * $x_sum)) / $num_rows;

$predict=($m*($num_rows+1))+$b;

echo "The Slope of the Representative line is: ",$m,"<br />";
echo "The Y intercept is ",$b,"<br />";
echo "Predicted next date of delivery in days (from last delivery date) is: ", $predict,"<br />";


//Final Date Calculation

$lastdate=mysql_query("SELECT TO_DAYS(date) FROM ordering WHERE id=$id",$conn);
$num_rows_date = mysql_num_rows($lastdate);
while (($row_date = mysql_fetch_array($lastdate, MYSQL_ASSOC)) !== false){
  $ldate[] = $row_date; // add the row in to the results (data) array
}
echo "Total number of records for the given id are: ",$num_rows_date, "<br />";
$ldatefinal=$ldate[$num_rows_date-1]['TO_DAYS(date)'];

echo "Last date in the records for the given ID is ",$ldatefinal,"<br />";
echo "Last date + Prediction days = ",$ldatefinal+$predict,"<br />";


$currentdate=mysql_query("SELECT TO_DAYS(CURDATE())",$conn);

$dcrow = mysql_fetch_assoc($currentdate);

//$dhold=$dcrow['TO_DAYS(CURDATE())'];
//echo "Current Date is: ",$dcrow['TO_DAYS(CURDATE())'],"<br /" ;

//UPON EXECUTION NOTHING BELOW GETS ExEcuted ??

//echo "Last date + Predicted days = ";




if (( (($ldatefinal+$predict) - $dcrow['TO_DAYS(CURDATE())']) <3 )&& $predict>0){

echo "Time to Order<br />","</br />";
}else if((( (($ldatefinal+$predict) - $dcrow['TO_DAYS(CURDATE())']) >3 && $predict>0) )){
echo "You Got Gas!!<br />","</br />";
}else{
    echo "Linear Regression not Suitable for this Record ID","<br />";
}

}
mysql_close($conn);

除了预测带有负数的时候，每件事情看起来都很完美，我现在选择通过使用不适合此记录ID的“回声”线性回归来忽略这些，“
“;”但我相信必须有更好的方法来适应这种情况，我只是不确定如何在我的情况下，特别是使用日期格式。任何帮助将不胜感激。我必须说这个论坛，当然它的用户对我来说过去他们的时间和知识非常友好和慷慨，希望你们这次也能指出我正确的方向。 Hamood

如何在PHP中的线性回归实现中处理负预测值

0 个答案: