如何在PHP中从数组中获取框图密钥数？

时间：2013-10-06 14:42:30

标签： php boxplot

假设我有一个包含以下值的数组：

$values = array(48,30,97,61,34,40,51,33,1);

我希望这些值能够绘制如下的方框图：

$box_plot_values = array(
    'lower_outlier'  => 1,
    'min'            => 8,
    'q1'             => 32,
    'median'         => 40,
    'q3'             => 56,
    'max'            => 80,
    'higher_outlier' => 97,
);

我将如何在PHP中执行此操作？

3 个答案:

答案 0 :(得分：5)

function box_plot_values($array)
{
    $return = array(
        'lower_outlier'  => 0,
        'min'            => 0,
        'q1'             => 0,
        'median'         => 0,
        'q3'             => 0,
        'max'            => 0,
        'higher_outlier' => 0,
    );

    $array_count = count($array);
    sort($array, SORT_NUMERIC);

    $return['min']            = $array[0];
    $return['lower_outlier']  = $return['min'];
    $return['max']            = $array[$array_count - 1];
    $return['higher_outlier'] = $return['max'];
    $middle_index             = floor($array_count / 2);
    $return['median']         = $array[$middle_index]; // Assume an odd # of items
    $lower_values             = array();
    $higher_values            = array();

    // If we have an even number of values, we need some special rules
    if ($array_count % 2 == 0)
    {
        // Handle the even case by averaging the middle 2 items
        $return['median'] = round(($return['median'] + $array[$middle_index - 1]) / 2);

        foreach ($array as $idx => $value)
        {
            if ($idx < ($middle_index - 1)) $lower_values[]  = $value; // We need to remove both of the values we used for the median from the lower values
            elseif ($idx > $middle_index)   $higher_values[] = $value;
        }
    }
    else
    {
        foreach ($array as $idx => $value)
        {
            if ($idx < $middle_index)     $lower_values[]  = $value;
            elseif ($idx > $middle_index) $higher_values[] = $value;
        }
    }

    $lower_values_count = count($lower_values);
    $lower_middle_index = floor($lower_values_count / 2);
    $return['q1']       = $lower_values[$lower_middle_index];
    if ($lower_values_count % 2 == 0)
        $return['q1'] = round(($return['q1'] + $lower_values[$lower_middle_index - 1]) / 2);

    $higher_values_count = count($higher_values);
    $higher_middle_index = floor($higher_values_count / 2);
    $return['q3']        = $higher_values[$higher_middle_index];
    if ($higher_values_count % 2 == 0)
        $return['q3'] = round(($return['q3'] + $higher_values[$higher_middle_index - 1]) / 2);

    // Check if min and max should be capped
    $iqr = $return['q3'] - $return['q1']; // Calculate the Inner Quartile Range (iqr)
    if ($return['q1'] > $iqr)                  $return['min'] = $return['q1'] - $iqr;
    if ($return['max'] - $return['q3'] > $iqr) $return['max'] = $return['q3'] + $iqr;

    return $return;
}

答案 1 :(得分：1)

Lilleman的代码很精彩。我非常感谢他处理中位数和q1 / q3的方法。如果我先回答这个问题，我会用更难但不必要的方式应对奇数甚至数量的价值观。我的意思是如果4种不同的模式使用4次（计数（值），4）。但他的方式只是整洁。我真的很佩服他的工作。

我想对max，min，higher_outliers和lower_outliers进行一些改进。因为q1 - 1.5 * IQR只是下限，我们应该找到大于此范围的最小值作为“min”＃。对于＆＃39; max＆＃39;这是相同的。此外，可能存在多个异常值。所以我想根据Lilleman的工作做一些改变。感谢。

function box_plot_values($array)
{
     $return = array(
    'lower_outlier'  => 0,
    'min'            => 0,
    'q1'             => 0,
    'median'         => 0,
    'q3'             => 0,
    'max'            => 0,
    'higher_outlier' => 0,
);

$array_count = count($array);
sort($array, SORT_NUMERIC);

$return['min']            = $array[0];
$return['lower_outlier']  = array();
$return['max']            = $array[$array_count - 1];
$return['higher_outlier'] = array();
$middle_index             = floor($array_count / 2);
$return['median']         = $array[$middle_index]; // Assume an odd # of items
$lower_values             = array();
$higher_values            = array();

// If we have an even number of values, we need some special rules
if ($array_count % 2 == 0)
{
    // Handle the even case by averaging the middle 2 items
    $return['median'] = round(($return['median'] + $array[$middle_index - 1]) / 2);

    foreach ($array as $idx => $value)
    {
        if ($idx < ($middle_index - 1)) $lower_values[]  = $value; // We need to remove both of the values we used for the median from the lower values
        elseif ($idx > $middle_index)   $higher_values[] = $value;
    }
}
else
{
    foreach ($array as $idx => $value)
    {
        if ($idx < $middle_index)     $lower_values[]  = $value;
        elseif ($idx > $middle_index) $higher_values[] = $value;
    }
}

$lower_values_count = count($lower_values);
$lower_middle_index = floor($lower_values_count / 2);
$return['q1']       = $lower_values[$lower_middle_index];
if ($lower_values_count % 2 == 0)
    $return['q1'] = round(($return['q1'] + $lower_values[$lower_middle_index - 1]) / 2);

$higher_values_count = count($higher_values);
$higher_middle_index = floor($higher_values_count / 2);
$return['q3']        = $higher_values[$higher_middle_index];
if ($higher_values_count % 2 == 0)
    $return['q3'] = round(($return['q3'] + $higher_values[$higher_middle_index - 1]) / 2);

// Check if min and max should be capped
$iqr = $return['q3'] - $return['q1']; // Calculate the Inner Quartile Range (iqr)

$return['min'] = $return['q1'] - 1.5*$iqr; // This ( q1 - 1.5*IQR ) is actually the lower bound,
                                           // We must compare every value in the lower half to this.
                                           // Those less than the bound are outliers, whereas
                                           // The least one that greater than this bound is the 'min'
                                           // for the boxplot.
foreach( $lower_values as  $idx => $value )
{
    if( $value < $return['min'] )  // when values are less than the bound
    {
        $return['lower_outlier'][$idx] = $value ; // keep the index here seems unnecessary
                                                  // but those who are interested in which values are outliers 
                                                  // can take advantage of this and asort to identify the outliers
    }else
    {
        $return['min'] = $value; // when values that greater than the bound
        break;  // we should break the loop to keep the 'min' as the least that greater than the bound
    }
}

$return['max'] = $return['q3'] + 1.5*$iqr; // This ( q3 + 1.5*IQR ) is the same as previous.
foreach( array_reverse($higher_values) as  $idx => $value )
{
    if( $value > $return['max'] )
    {
        $return['higher_outlier'][$idx] = $value ;
    }else
    {
        $return['max'] = $value;
        break;
    }
}
    return $return;
}

我希望这对那些对此问题感兴趣的人有所帮助。如果有更好的方法可以知道哪些值是异常值，请Pls给我添加评论。谢谢！

答案 2 :(得分：0)

我有一个不同的解决方案来计算较低和较高的胡须。与ShaoE的解决方案一样，它找到的最小值大于或等于下限（Q1 - 1.5 * IQR），反之亦然。

我使用array_filter迭代数组，将值传递给回调函数并返回一个数组，其中只包含回调赋予true的值（参见php.net's array_filter manual）。在这种情况下，返回大于下限的值，并将其用作min的输入，// get lower whisker $whiskerMin = min(array_filter($array, function($value) use($quartile1, $iqr) { return $value >= $quartile1 - 1.5 * $iqr; } )); // get higher whisker vice versa $whiskerMax = max(array_filter($array, function($value) use($quartile3, $iqr) { return $value <= $quartile3 + 1.5 * $iqr; } )); 本身返回的值最小。

protected function _getStatusText

请注意，它忽略了异常值，我只测试了正值。

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