本文内容全部来自《集体智慧编程》一书，原书采用的是python，因为没有python编程环境，所以用PHP实现
<?php
//filename:test_collecting_preferences
//数据和代码来自《集体智慧编程》
//原文采用python实现，尝试用PHP进行转换
//@description 搜集用户偏好寻找相近用户
$datalist = array(
'Lisa Rose' => array(
'Lady in the Water' => 2.5,
'Snake on a Plane' => 3.5,
'Just My Luck' => 3.0,
'Superman Returns' => 3.5,
'You, Me and Dupree' => 2.5,
'The Night Listener'=> 3.0
),
'Gene Seymour' => array(
'Lady in the Water' => 3.0,
'Snake on a Plane' => 3.5,
'Just My Luck' => 1.5,
'Superman Returns' => 5.0,
'You, Me and Dupree' => 3.5,
'The Night Listener'=> 3.0
),
'Michael Phillips' => array(
'Lady in the Water' => 2.5,
'Snake on a Plane' => 3.0,
'Superman Returns' => 3.5,
'The Night Listener'=> 4.0
),
'Claudia Puig' => array(
'Snake on a Plane' => 3.5,
'Just My Luck' =>3.0,
'Superman Returns' => 4.0,
'You, Me and Dupree' => 2.5,
'The Night Listener'=>4.5
),
'Mick LaSalle' => array(
'Lady in the Water' => 3.0,
'Snake on a Plane' => 4.0,
'Just My Luck' => 2.0,
'Superman Returns' => 3.0,
'You, Me and Dupree' => 2.0,
'The Night Listener'=> 3.0
),
'Jack Matthews' => array(
'Lady in the Water' => 3.0,
'Snake on a Plane' => 4.0,
'Superman Returns' => 5.0,
'You, Me and Dupree' => 3.5,
'The Night Listener'=> 3.0
),
'Toby' => array(
'Snake on a Plane' => 4.5,
'Superman Returns' => 4.0,
'You, Me and Dupree' => 1.0,
),
);
//欧几里德距离
//它以经过人们的一致评价的物品为坐标轴，然后将参与评价的人绘制到图上，并考查他们彼此间的距离远近。
//偏好越相似的人，距离越近。不过我们还需要一个函数来对偏好越相近的情况给出越大的值，
//为此我们可以将函数值加1（这样可以避免遇到被零整除的错误），并取其倒数
//公式是 1 / (1 + sqrt ( pow( data[a][1] - data[b][1] .... ) ))
function sim_distance ( $datalist , $person1 , $person2)
{
$si = array();
foreach ( $datalist[$person1] as $moviename => $grade ){
if( array_key_exists( $moviename, $datalist[$person2] )){
$si[$moviename] = 1;
}
}
if( emptyempty( $si )){
return 0;
}
$powers = 0;
foreach ( $si as $moviename=>$val ){
$powers += pow( ($datalist[$person1][$moviename] - $datalist[$person2][$moviename] ), 2 );//两者影评分数相减的平方值
}
return 1 / (1+ sqrt($powers));
}
//测试 'Lisa Rose' 和 'Gene Seymour' 的相似度评价
//原书上求出来是 0.29429805508554946 , PHP 的结果是 0.29429805508555，默认精度没有python高
echo( sim_distance( $datalist , 'Lisa Rose' , 'Gene Seymour') );
echo( '<br/>' );
//皮尔逊相关系数
//该相关系统是判断两组数据与某一直线拟合程序的一种度量。对应的公司比欧几里德距离评价的计算公式要复杂
//但是它在数据不是很规范时（如影评者对影片的评价总是相对于平均水平偏离很大），会倾向于给出更好的结果
//皮尔逊相关度评价法首先会找出两位评论者都曾评过的物品
//计算两者的评分总和与平方和，并求得评分的乘积之和，最后，利用这个结果计算出相关系数
function sim_person ( $datalist ,$person1 , $person2)
{
$si = array();
foreach ( $datalist[$person1] as $moviename => $grade ){
if( array_key_exists( $moviename, $datalist[$person2] )){
$si[$moviename] = 1;
}
}
if( emptyempty( $si )){
return 1;
}
$n = count( $si );
$sum1 = $sum1Sq = $sum2 = $sum2Sq = $pSum = 0;
foreach ( $si as $moviename => $val ){
$sum1 += $datalist[$person1][$moviename]; //个人影评分数累加
$sum1Sq += pow( $datalist[$person1][$moviename], 2 );//个人影评分数平方的累加
$sum2 += $datalist[$person2][$moviename];
$sum2Sq += pow( $datalist[$person2][$moviename], 2 );
$pSum += ( $datalist[$person1][$moviename] * $datalist[$person2][$moviename]);//两人影评之乘积
}
$num = $pSum - ( $sum1 * $sum2 / $n); // 正常情况下，我怎么都觉得这是1吧？
$den = sqrt( ( $sum1Sq - pow( $sum1, 2 ) / $n) * ( $sum2Sq - pow( $sum2, 2 ) / $n) );
if ( $den == 0 ){
return 0;
}
return ($num / $den );
}
//继续测试 'Lisa Rose' 和 'Gene Seymour' 的相似度评价
//原书上求出来是 0.396059017191 , PHP 的结果是 0.39605901719067，这回。。。位数超过了python
echo( sim_person( $datalist , 'Lisa Rose' , 'Gene Seymour') );
?>
有点长，随便看看吧
