无论如何要简化这个老鼠环的老鼠窝?

时间:2011-08-30 19:55:33

标签: php oop loops methods foreach

这可行,但比地狱更难,基本上它是在子数组的两个独立部分中迭代,看看两个子数组的值除了1之外是否有最大公分母,如果存在,则将基值乘以1.5

提前抱歉草率的代码。

error_reporting(E_ALL);
ini_set('display_errors', '1');

class CSVParser
{

    public $output = NULL;
    public $digits = NULL;

    public function __construct($file)
    {


        if (!file_exists($file)) {
            throw new Exception("$file does not exist");
        }

        $this->contents = file_get_contents($file);
        $this->output = array();
        $this->digits = array();
        $this->factor = array();
    }

    public function parse($separatorChar1 = ',', $separatorChar2 = ';', $enclosureChar = '"', $newlineChar = "\n")
    {

        $lines = explode($newlineChar, $this->contents);
        foreach ($lines as $line) {
            if (strlen($line) == 0) continue;
            $group = array();
            list($part1, $part2) = explode($separatorChar2, $line);
            $group[] = array_map(array($this, "trim_value"), explode($separatorChar1, $part1), array("$enclosureChar \t"));
            $group[] = array_map(array($this, "trim_value"), explode($separatorChar1, $part2), array("$enclosureChar \t"));
            $this->output[] = $group;
        }
    }

    private function trim_value($value, $chars)
    {
        return preg_replace("#^( |" . $chars . ")+#", '', $value);
    }


    private function gcd($x,$y) 
    {
        do {
            $rest=$x%$y;
        $x=$y;
        $y=$rest;
        } while($rest!==0);
        return $x;
    }

    public function algorithm()
    {
        $alpha = array(
            'c' => str_split('bcdfghjklmnpqrstvwxz'),
            'v' => str_split('aeiouy')
        );
        $i=$k=0;
        foreach ($this->output as $item) {
            $cnt = 0;
            $this->digits[$i] = array();
            foreach ($item as $part) {
                $this->digits[$i][$cnt] = array();
                $new = array();
                foreach ($part as $str) { 
                    $v = count(array_intersect(str_split($str), $alpha['v']));
                    $c = count(array_intersect(str_split($str), $alpha['c']));
                    $t = strlen(str_replace(' ', '', $str));

                    $new = ($cnt == 0) 
                        ? array('v' => $v, 'c' => $c, 't' => $t, 'm' => ($t%2) ? $v * 1.5 : $c) 
                        : array('v' => $v, 'c' => $c, 't' => $t);

                    $this->digits[$i][$cnt][] = $new;
                }
                $cnt++;
            }
            $i++;
        }
        $h=$cuml=0; 
            foreach($this->digits as &$slice) { 
            foreach($slice[0] as &$sliceName){ 

                foreach($slice[1] as $sliceProduct) { 
                foreach($sliceProduct as $pKey=>$pVal) { 

                    foreach($sliceName as $nKey=>$nVal) { 
                        $tmp[$h] = ($this->gcd($pVal,$nVal) != 1) ? ++$cuml:'';
                    } 
                } 
                    $tmp[$h] = $sliceName['m']*$cuml*1.5; 
                    $h++; 
                    $cuml=0; 
                }$h=0; 

            $sliceName['f'] = $tmp; 
            $tmp=''; 
            } 

            } 
        foreach($this->digits as &$u){unset($u[1]);} 
    } 

}

$parser = new CSVParser("file.csv");
$parser->parse();   //print_r($parser->output);
$parser->algorithm();   print_r($parser->digits);

每个请求的示例CSV

Jeff Goes, Mika Enrar;Triple Threat, Dogs on  Bikes
Sonny Ray, Lars McGarvitch, Jason McKinley;Kasabian, Lords of Acid, Hard-Fi

输出

Array
(
    [0] => Array
    (
        [0] => Array
            (
                [0] => Array
                    (
                        [v] => 3
                        [c] => 3
                        [t] => 8
                        [m] => 3
                        [f] => Array
                            (
                                [0] => 40.5
                                [1] => 4.5 // Remainder.. So 'Jeff Goes' => 'Dogs on Bikes'
                            )

                    )

                [1] => Array
                    (
                        [v] => 3
                        [c] => 4
                        [t] => 9
                        [m] => 4.5
                        [f] => Array
                            (
                                [0] => 67.5 // High Score! So 'Mika Enrar' => 'Triple Threat'
                                [1] => 13.5
                            )

                    )

            )

    )

    [1] => Array
    (
        [0] => Array
            (
                [0] => Array
                    (
                        [v] => 4
                        [c] => 2
                        [t] => 8
                        [m] => 2
                        [f] => Array
                            (
                                [0] => 24
                                [1] => 12
                                [2] => 24 // Next Highest 'Sonny Ray' => 'Hard-Fi'
                            )

                    )

                [1] => Array
                    (
                        [v] => 3
                        [c] => 8
                        [t] => 14
                        [m] => 8
                        [f] => Array
                            (
                                [0] => 84 // High Score! (This is really a tie, but 'm' has the highest secondary value so...) 
                                [1] => 60 // 'Lars McGarvitch => 'Kasabian'
                                [2] => 84
                            )

                    )

                [2] => Array
                    (
                        [v] => 5
                        [c] => 5
                        [t] => 13
                        [m] => 7.5
                        [f] => Array
                            (
                                [0] => 0
                                [1] => 0 // The only one left 'Jason McKinley' => 'Lords of Acid'
                                [2] => 11.25
                            )

                    )

            )

    )

)

它的作用

到目前为止,这个类所做的是将csv拆分为一个数组,之前拆分内容;并进入两个子阵列后。计算两者的辅音和元音,找出每个C V或混合字母对的两个子部分之间是否存在最大公分母,并创建一个值以将带分配给产品。

真正需要做什么

生成的最高值应与创建该高值的波段相关联。所以我真正想做的是将一个名字与一个乐队联系起来,这取决于它最终产生的得分有多高。我大约一半=(

正如大家们所看到的,这段代码实际上是一团糟。我真正想要的是根据我正在生成的数字为乐队指定名称。

2 个答案:

答案 0 :(得分:4)

我必须同意其他所有人的意见......但我想补充一下:

而是更简单地搜索如何遍历 $this->digits ,您应该强烈考虑重新考虑数据的结构 $this->digits

此外,将所有内容整合到一个数组中并不总是有意义的。但是当它发生时,可以考虑结构,使其直观,并且可以轻松遍历。

如果没有关于这是做什么的更多信息,我们无法建议如何重组您的数据/类。一个开始就是给我们一个样本$this->digits数组的样子。此外,有关您的问题的一些更多信息将是好的(如何使用此方法)。

答案 1 :(得分:0)

如果它有效,你为什么要改变它?性能?重构?生意改变了?要求改变了?清洁代码撒玛利亚人?童子军规则?

当我遇到“意大利面条代码”时,除非我绝对必须改变它,否则我不管它。也就是说,我会写一些单元测试来验证“意大利面条代码”的输出,以便我知道我没有破坏任何东西或者让事情变得更糟。