预测数据趋势

Question

早上好，斯塔克。我正在尝试找出一种预测数据趋势的方法。我想知道是否有更好的方法来做到这一点。是否有任何内置函数或库可供我研究？

$PopulationOfTexas = array(
    1999 => 20.56, // in millions
    2000 => 21.56,
    2001 => 22.56,
    2002 => 23.56
);

//generate an array sohwing the difference in each year compared to the previous year
$differneces = array();
$lastyear = null;
foreach($PopulationOfTexas as $k=>$v){
    if(empty($lastyear)){$lastyear = $k; continue;}
    $differneces[$k] = $k - $lastyear;
    $lastyear = $k;

    //use this later
    $lastitem = array("year"=>$k, "data"=>$v);
}

//get the average difference per year
$count = 0;
$total = 0;
foreach($differneces as $k=>$v){
    $count++;
    $total += $v;
}

$average = number_format(($total/$count), 2);

//make a prediction
$predictions = array();
for($i=0;$i<5;$i++){
    $year = isset($year) ? $year+1 : $lastitem["year"]+1;
    $prediction = isset($prediction) ? $prediction+floatval($average) : $lastitem["data"]+floatval($average);
    $predictions[$year] = $prediction;
}

print_r($predictions);

Answer 1

该算法完全错误，因为它计算的是数组键（年份值、1999、2000 等）而不是数组值（人口）的平均增长，因此结果始终为 1。

这被您的样本人口数据总是增加一这一事实掩盖了，如果您添加了更多变化，您可能会发现错误。修复：

foreach($PopulationOfTexas as $k=>$v){
    if(empty($lastyear)){$lastyear = $v; continue;}
    $differneces[$k] = $v - $lastyear;
    $lastyear = $v;

    //use this later
    $lastitem = array("year"=>$k, "data"=>$v);
}

更笼统地说，该算法非常简单，因为它会预测平坦的增加/减少。

预测数据趋势

Predicting Trends in Data

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