0

我有 +-100 点/值的小矩阵、测试结果及其彼此之间的距离（0-10、10 是最接近的）：http: //vis.arcs.cz。

我想在 2D 中可视化它以快速找到一组接近的值。所以我需要处理这个矩阵并获得二维点的坐标。

方法可能是多维缩放，但我无法找到算法、库或扩展，也无法在 PHP 代码中使用数学公式。

我过去两天在谷歌上搜索，最接近的结果是 http://www.php.net/manual/en/lapack.leastsquaresbysvd.php - ？（不能作为新手分享更多链接）

我将感谢任何适用于 php 项目的解决方案（用 C++ 编译的 MathLab 代码......）。

4

1 回答 1

0 
index.php:

<?php
require("data.php");
require("Mds.php");
$mds = new Mds();
$mds->prepareData($data);
//cislo udava pocet iteraci algoritmu,
//idealne by melo mit hodnotu pocetUzlu na druhou, prakticky muze byt o dost mensi, 5 az 10 krat pocet uzlu.
$mds->scale(100);
//mira splneni podminek - vraci cislo < mensi nez 1. Cim blizsi k 1, tim lepe vyhovuji vzdalenosti
//z vypocteneho rozlozeni zadanym vzdalenostem.
//v nemetrickych datech ovsem dochazi rychle k poklesu az do zapornych hodnot, protoze ve 2D neexistuje
//pro nektere body zadne spravne misto.
echo $mds->getRSquared();
?>
<!DOCTYPE HTML>
<html>
<head>
<script type="text/javascript" src="http://canvasjs.com/assets/script/canvasjs.min.js"></script>
<script type="text/javascript">
window.onload = function () {
var chart = new CanvasJS.Chart("chartContainer",
{
title:{
text: "MDS",
fontFamily: "arial black",
fontColor: "black"
},
axisX: {
title:"",
titleFontFamily: "arial"

},
axisY:{
title: "",
titleFontFamily: "arial",
titleFontSize: 12
},

data: [
{
type: "scatter",
toolTipContent: "<span style='"'color: {color};'"'><strong>{name}</strong></span> x: {y}, y: {y}",
dataPoints:

<?php
echo $mds->printCoords('{x: %X, y: %Y, name: "%LABEL"}');
?>
}]
});

chart.render();
}
</script>

</head>
<body>
<div id="chartContainer" style="height: 600px; width: 600px;">
</div>

<a href="mds.zip">mds.zip</a>
</body>
</html>


data.php:



$data = array( 
array ("Jablko", "Voda", 2),
array("Jablko", "Ohen", 7),
array("Jablko", "Cervena", 3),
array("Jablko", "Zelena", 2),

array("Voda", "Ohen", 8),
array("Voda", "Cervena", 8),
array("Voda", "Zelena", 2),

array("Ohen", "Cervena", 1),
array("Ohen", "Zelena", 5),

array("Cervena", "Zelena", 3)
);


Mds.php:

<?php

class Mds {
//node - pole tvaru [[x, y, label], ...]
public $nodes = array();

//tvaru [source][target]=distance
public $givenDistances = array();
public $currentDistances = array();


public function prepareData($data) {
$nodesMap = array();
$xxxx = 10000;
$xxx= 1000;

// $xxxx = 5000;
// $xxx = 600;

foreach($data as $link) {
$source = $link[0];
$target = $link[1];
if(!isset($nodesMap[$source])) {
$nodesMap[$source] = true;
$this->nodes[]=array((float) rand(1,$xxxx) / $xxx, (float) rand(1,$xxxx) / $xxx, $source);
}
if(!isset($nodesMap[$target])) {
$nodesMap[$target] = true;
$this->nodes[]= array((float) rand(1,$xxxx) / $xxx, (float) rand(1,$xxxx) / $xxx, $target);
}
}
//vytvori matici pro ulozeni vzdalenosti
foreach($this->nodes as $node) {
$this->givenDistances[$node[2]] = array();
$this->currentDistances[$node[2]] = array();
}
foreach($data as $link) {
$source = $link[0];
$target = $link[1];
$distance = $link[2];
$this->givenDistances[$source][$target] = $distance;
$this->givenDistances[$target][$source] = $distance;
}
}

protected function countCurrentDistances() {
$mean = 0;
$i = 0;

foreach($this->nodes as $nodeA) {
foreach($this->nodes as $nodeB) {
$dist = sqrt(($nodeB[0]-$nodeA[0])*($nodeB[0]-$nodeA[0]) + ($nodeB[1]-$nodeA[1])*($nodeB[1]-$nodeA[1]));
$this->currentDistances[$nodeA[2]][$nodeB[2]] = $dist;

if($nodeA[2]!==$nodeB[2]) {
$mean += $this->givenDistances[$nodeA[2]][$nodeB[2]];
}
}
}
// echo "<pre>";
// var_dump($this->currentDistances);
// echo "</pre>";
$check = array();

$nodesCount = count($this->nodes);
$this->mean = $mean/($nodesCount*$nodesCount);
}

public function getRSquared() {
$this->countCurrentDistances();

$sum = 0;
$SStot = 0;
$SSres = 0;
foreach($this->nodes as $nodeA) {
foreach($this->nodes as $nodeB) {
$aLab = $nodeA[2];
$bLab = $nodeB[2];
if($aLab===$bLab) {
continue;
}

$given = $this->givenDistances[$aLab][$bLab];
$computed = $this->currentDistances[$aLab][$bLab];

$SSres+=(($given-$computed)*($given-$computed));
$SStot+= ($given-$this->mean)*($given-$this->mean);
}
}

return 1 - ($SSres/$SStot);
}

protected function iterate($inverseCoefStrenth=1) {
for($i=0; $i<count($this->nodes); $i++) {
$nodeA = $this->nodes[$i];
$move = array(0,0);
$moves = 0;

for($j=0; $j<count($this->nodes); $j++) {
if($j===$i) {
continue;
}

$nodeB = $this->nodes[$j];
$dist = $this->givenDistances[$nodeA[2]][$nodeB[2]];

$AB = array($nodeB[0]-$nodeA[0], $nodeB[1]-$nodeA[1]);
$lAB = sqrt(($AB[0]*$AB[0])+($AB[1]*$AB[1]));

$coef = ($lAB - $dist)/$lAB;

$AA1 = array($AB[0]*$coef, $AB[1]*$coef);

$moves++;
$move[0]+=$AA1[0];
$move[1]+=$AA1[1];
}

if($moves>0) {
$resultingMoveX = $move[0]/($moves*$inverseCoefStrenth);
$resultingMoveY = $move[1]/($moves*$inverseCoefStrenth);
$this->nodes[$i][0]+=$resultingMoveX;
$this->nodes[$i][1]+=$resultingMoveY;
}
}
}


public function scale($iterations, $coefIncrease=0) {
$basicCoef=1;
for($i=0; $i<$iterations; $i++) {
$this->iterate($basicCoef);
//echo $this->getRSquared();
//echo " | ";

$basicCoef+=$coefIncrease;
}
}

public function coordsAsString() {
$coords = array();
foreach($this->nodes as $node) {
$coords[]="[{$node[0]}, {$node[1]}]";
}

$res = join(", ",$coords);
return "[$res]";
}

public function printCoords($pattern='[%X,%Y,"%LABEL"]') {
$coords = array();
foreach($this->nodes as $node) {
$x = $node[0];
$y = $node[1];
$label = $node[2];
$res = str_replace('%X', $x, $pattern);
$res = str_replace('%Y', $y, $res);
$res = str_replace('%LABEL', $label, $res);
$coords[]=$res;
}

$res = join(", ",$coords);
return "[$res]";
}

public function pointsCoords() {
$coords = array();
foreach($this->nodes as $node) {
$res['x'] = round($node[0]*100);
$res['y'] = round($node[1]*100);
$res['label'] = $node[2];
$coords[] = $res;
}
return $coords;
}

}
于 2014-06-27T13:00:14.227 回答