BSAS顺序聚类算法matlab实现代码（注释很全）_算法_聚类_其他

<html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta charset="utf-8"> <meta name="generator" content="pdf2htmlEX"> <meta http-equiv="X-UA-Compatible" content="IE=edge,chrome=1"> <link rel="stylesheet" href="https://static.pudn.com/base/css/base.min.css"> <link rel="stylesheet" href="https://static.pudn.com/base/css/fancy.min.css"> <link rel="stylesheet" href="https://static.pudn.com/prod/directory_preview_static/62ba849323dfab40401dd628/raw.css"> <script src="https://static.pudn.com/base/js/compatibility.min.js"></script> <script src="https://static.pudn.com/base/js/pdf2htmlEX.min.js"></script> <script> try{ pdf2htmlEX.defaultViewer = new pdf2htmlEX.Viewer({}); }catch(e){} </script> <title></title> </head> <body> <div id="sidebar" style="display: none"> <div id="outline"> </div> </div> <div id="pf1" class="pf w0 h0" data-page-no="1"><div class="pc pc1 w0 h0"><img class="bi x0 y0 w1 h1" alt="" src="https://static.pudn.com/prod/directory_preview_static/62ba849323dfab40401dd628/bg1.jpg"><div class="c x0 y1 w2 h2"><div class="t m0 x1 h3 y2 ff1 fs0 fc0 sc0 ls0 ws0"><span class="_ _0"> </span><span class="ff2">算法实例</span></div><div class="t m0 x2 h3 y3 ff3 fs0 fc0 sc0 ls0 ws0">10<span class="_ _1"> </span><span class="ff2">个模式样本点：<span class="ff4">{x</span></span></div><div class="t m0 x3 h4 y4 ff4 fs1 fc0 sc0 ls0 ws0">1</div><div class="t m0 x4 h5 y3 ff4 fs0 fc0 sc0 ls0 ws0">(0 0), x</div><div class="t m0 x5 h4 y4 ff4 fs1 fc0 sc0 ls0 ws0">2</div><div class="t m0 x6 h5 y3 ff4 fs0 fc0 sc0 ls0 ws0">(3 8), x</div><div class="t m0 x7 h4 y4 ff4 fs1 fc0 sc0 ls0 ws0">3</div><div class="t m0 x8 h5 y3 ff4 fs0 fc0 sc0 ls0 ws0">(2 2), x</div><div class="t m0 x9 h4 y4 ff4 fs1 fc0 sc0 ls0 ws0">4</div><div class="t m0 xa h5 y3 ff4 fs0 fc0 sc0 ls0 ws0">(1 1), x</div><div class="t m0 xb h4 y4 ff4 fs1 fc0 sc0 ls0 ws0">5</div><div class="t m0 xc h5 y3 ff4 fs0 fc0 sc0 ls0 ws0">(5 </div><div class="t m0 x1 h5 y5 ff4 fs0 fc0 sc0 ls0 ws0">3), x</div><div class="t m0 xd h4 y6 ff4 fs1 fc0 sc0 ls0 ws0">6</div><div class="t m0 xe h5 y5 ff4 fs0 fc0 sc0 ls0 ws0">(4 8), x</div><div class="t m0 xf h4 y6 ff4 fs1 fc0 sc0 ls0 ws0">7</div><div class="t m0 x10 h5 y5 ff4 fs0 fc0 sc0 ls0 ws0">(6 3), x</div><div class="t m0 x11 h4 y6 ff4 fs1 fc0 sc0 ls0 ws0">8</div><div class="t m0 x12 h5 y5 ff4 fs0 fc0 sc0 ls0 ws0">(5 4), x</div><div class="t m0 x13 h4 y6 ff4 fs1 fc0 sc0 ls0 ws0">9</div><div class="t m0 x14 h5 y5 ff4 fs0 fc0 sc0 ls0 ws0">(6 4), x</div><div class="t m0 x15 h4 y6 ff4 fs1 fc0 sc0 ls0 ws0">10</div><div class="t m0 x16 h5 y5 ff4 fs0 fc0 sc0 ls0 ws0">(7 5)}</div><div class="t m0 x17 h3 y7 ff2 fs0 fc0 sc0 ls0 ws0">第一步<span class="fc1 sc0">：选任意一个模式样本作为第一个聚类中心，</span>如<span class="_ _1"> </span><span class="ff3">z</span></div><div class="t m0 x18 h6 y8 ff3 fs1 fc0 sc0 ls0 ws0">1 </div><div class="t m0 x19 h7 y7 ff3 fs0 fc0 sc0 ls0 ws0">= x</div><div class="t m0 x1a h6 y8 ff3 fs1 fc0 sc0 ls0 ws0">1</div><div class="t m0 x17 h3 y9 ff2 fs0 fc0 sc0 ls0 ws0">第二步<span class="fc1 sc0">：选距离</span><span class="_ _1"> </span><span class="ff3"><span class="fc1 sc0">z</span></span></div><div class="t m0 x1b h6 ya ff3 fs1 fc0 sc0 ls0 ws0"><span class="fc1 sc0">1</span></div><div class="t m0 x1c h3 y9 ff2 fs0 fc0 sc0 ls0 ws0"><span class="fc1 sc0">最远的样本作为第二个聚类中心。</span></div><div class="t m0 x1d h3 yb ff2 fs0 fc0 sc0 ls0 ws0"><span class="fc1 sc0">经计算，</span><span class="ff3"><span class="fc1 sc0">|| </span><span class="fc1 sc0">x</span></span></div><div class="t m0 x1e h6 yc ff3 fs1 fc0 sc0 ls0 ws0"><span class="fc1 sc0">6</span><span class="fc1 sc0"> </span></div><div class="t m0 x3 h7 yb ff3 fs0 fc0 sc0 ls0 ws0"><span class="fc1 sc0">-</span><span class="fc1 sc0"> </span><span class="fc1 sc0">z</span></div><div class="t m0 x1f h6 yc ff3 fs1 fc0 sc0 ls0 ws0"><span class="fc1 sc0">1</span><span class="fc1 sc0"> </span></div><div class="t m0 x20 h3 yb ff3 fs0 fc0 sc0 ls0 ws0"><span class="fc1 sc0">||</span><span class="ff2"><span class="fc1 sc0">最大，所以</span><span class="_ _1"> </span></span><span class="fc1 sc0">z</span></div><div class="t m0 x21 h6 yc ff3 fs1 fc0 sc0 ls0 ws0"><span class="fc1 sc0">2</span><span class="fc1 sc0"> </span></div><div class="t m0 x7 h7 yb ff3 fs0 fc0 sc0 ls0 ws0"><span class="fc1 sc0">= </span><span class="fc1 sc0">x</span></div><div class="t m0 x22 h6 yc ff3 fs1 fc0 sc0 ls0 ws0"><span class="fc1 sc0">6</span></div><div class="t m0 x17 h3 yd ff2 fs0 fc0 sc0 ls0 ws0">第三<span class="_ _2"></span>步：<span class="_ _2"></span>逐个<span class="_ _2"></span>计算<span class="_ _2"></span>各模<span class="_ _2"></span>式样<span class="_ _2"></span>本<span class="_ _3"></span><span class="ff3">{x</span></div><div class="t m0 x23 h6 ye ff3 fs1 fc0 sc0 ls0 ws0">i</div><div class="t m0 x24 h3 yd ff3 fs0 fc0 sc0 ls0 ws0">, i <span class="_ _2"></span>= 1,2,<span class="_ _2"></span>…,N}<span class="ff2">与<span class="_ _2"></span></span>{z</div><div class="t m0 x25 h6 ye ff3 fs1 fc0 sc0 ls0 ws0">1</div><div class="t m0 x26 h7 yd ff3 fs0 fc0 sc0 ls0 ws0">, z</div><div class="t m0 x27 h6 ye ff3 fs1 fc0 sc0 ls0 ws0">2</div><div class="t m0 x28 h3 yd ff3 fs0 fc0 sc0 ls0 ws0">}<span class="ff2">之<span class="_ _2"></span>间的<span class="_ _2"></span>距</span></div><div class="t m0 x1d h3 yf ff2 fs0 fc0 sc0 ls0 ws0">离，即</div><div class="t m0 x29 h7 y10 ff3 fs0 fc0 sc0 ls0 ws0">D</div><div class="t m0 x2a h6 y11 ff3 fs1 fc0 sc0 ls0 ws0">i1</div><div class="t m0 x2b h7 y10 ff3 fs0 fc0 sc0 ls0 ws0"> = || x</div><div class="t m0 x2c h6 y11 ff3 fs1 fc0 sc0 ls0 ws0">i </div><div class="t m0 x2d h7 y10 ff3 fs0 fc0 sc0 ls0 ws0">- z</div><div class="t m0 x2e h6 y11 ff3 fs1 fc0 sc0 ls0 ws0">1 </div><div class="t m0 x2f h7 y10 ff3 fs0 fc0 sc0 ls0 ws0">||</div><div class="t m0 x29 h7 y12 ff3 fs0 fc0 sc0 ls0 ws0">D</div><div class="t m0 x2a h6 y13 ff3 fs1 fc0 sc0 ls0 ws0">i2</div><div class="t m0 x2b h7 y12 ff3 fs0 fc0 sc0 ls0 ws0"> = || x</div><div class="t m0 x2c h6 y13 ff3 fs1 fc0 sc0 ls0 ws0">i </div><div class="t m0 x2d h7 y12 ff3 fs0 fc0 sc0 ls0 ws0">– z</div><div class="t m0 x30 h6 y13 ff3 fs1 fc0 sc0 ls0 ws0">2 </div><div class="t m0 x31 h7 y12 ff3 fs0 fc0 sc0 ls0 ws0">||</div><div class="t m0 x1d h3 y14 ff2 fs0 fc0 sc0 ls0 ws0">并选出其中的最小距离<span class="_ _1"> </span><span class="ff3">min(D</span></div><div class="t m0 x32 h6 y15 ff3 fs1 fc0 sc0 ls0 ws0">i1</div><div class="t m0 x16 h7 y14 ff3 fs0 fc0 sc0 ls0 ws0">, D</div><div class="t m0 x33 h6 y15 ff3 fs1 fc0 sc0 ls0 ws0">i2</div><div class="t m0 x34 h3 y14 ff3 fs0 fc0 sc0 ls0 ws0">)<span class="ff2">，</span>i = 1,2,…,N</div><div class="t m0 x17 h3 y16 ff2 fs0 fc0 sc0 ls0 ws0">第四步：在<span class="_ _2"></span>所有模式样本的最<span class="_ _2"></span>小值中选出最<span class="_ _2"></span>大距离，若该<span class="_ _2"></span>最大值</div><div class="t m0 x1d h3 y17 ff2 fs0 fc0 sc0 ls0 ws0">达<span class="_ _3"></span>到<span class="_ _2"></span><span class="ff3">||z</span></div><div class="t m0 x35 h6 y18 ff3 fs1 fc0 sc0 ls0 ws0">1<span class="_ _3"></span> </div><div class="t m0 x36 h7 y17 ff3 fs0 fc0 sc0 ls0 ws0">- <span class="_ _2"></span>z</div><div class="t m0 x37 h6 y18 ff3 fs1 fc0 sc0 ls0 ws0">2<span class="_ _3"></span> </div><div class="t m0 x12 h3 y17 ff3 fs0 fc0 sc0 ls0 ws0">||<span class="_ _3"></span><span class="ff2">的<span class="_ _2"></span>一<span class="_ _3"></span>定<span class="_ _2"></span>比<span class="_ _3"></span>例<span class="_ _3"></span>以<span class="_ _2"></span>上<span class="_ _3"></span>，<span class="_ _3"></span>则<span class="_ _2"></span>相<span class="_ _3"></span>应<span class="_ _2"></span>的<span class="_ _3"></span>样<span class="_ _3"></span>本<span class="_ _2"></span>点<span class="_ _3"></span>取<span class="_ _3"></span>为<span class="_ _2"></span>第</span></div><div class="t m0 x1d h3 y19 ff2 fs0 fc0 sc0 ls0 ws0">三个聚类中心<span class="_ _1"> </span><span class="ff3">z</span></div><div class="t m0 x1f h6 y1a ff3 fs1 fc0 sc0 ls0 ws0">3</div><div class="t m0 x2c h3 y19 ff2 fs0 fc0 sc0 ls0 ws0">，即</div><div class="t m0 x38 h3 y1b ff2 fs0 fc0 sc0 ls0 ws0">若<span class="_ _1"> </span><span class="ff3">max{min(D</span></div><div class="t m0 x39 h6 y1c ff3 fs1 fc0 sc0 ls0 ws0">i1</div><div class="t m0 x2c h7 y1b ff3 fs0 fc0 sc0 ls0 ws0">, D</div><div class="t m0 x3a h6 y1c ff3 fs1 fc0 sc0 ls0 ws0">i2</div><div class="t m0 x3b h5 y1b ff3 fs0 fc0 sc0 ls0 ws0">), i = 1,2,…,N} ><span class="ff4">θ</span>||z</div><div class="t m0 x3c h6 y1c ff3 fs1 fc0 sc0 ls0 ws0">1 </div><div class="t m0 x3d h7 y1b ff3 fs0 fc0 sc0 ls0 ws0">- z</div><div class="t m0 x3e h6 y1c ff3 fs1 fc0 sc0 ls0 ws0">2 </div><div class="t m0 x3f h3 y1b ff3 fs0 fc0 sc0 ls0 ws0">||<span class="ff2">，则<span class="_ _1"> </span></span>z</div><div class="t m0 x40 h6 y1c ff3 fs1 fc0 sc0 ls0 ws0">3 </div><div class="t m0 x41 h7 y1b ff3 fs0 fc0 sc0 ls0 ws0">= x</div><div class="t m0 x42 h6 y1c ff3 fs1 fc0 sc0 ls0 ws0">i</div><div class="t m0 x38 h3 y1d ff2 fs0 fc0 sc0 ls0 ws0">否<span class="_ _2"></span>则<span class="_ _2"></span>，<span class="_ _2"></span>若<span class="_ _3"></span>找不<span class="_ _2"></span>到<span class="_ _3"></span>适合<span class="_ _2"></span>要<span class="_ _3"></span>求的<span class="_ _3"></span>样本<span class="_ _3"></span>作为<span class="_ _3"></span>新的<span class="_ _2"></span>聚<span class="_ _2"></span>类<span class="_ _3"></span>中心<span class="_ _3"></span>，</div><div class="t m0 x38 h3 y1e ff2 fs0 fc0 sc0 ls0 ws0">则找聚类中心的过程结束。</div></div></div><div class="pi" data-data='{"ctm":[1.611850,0.000000,0.000000,1.611850,0.000000,0.000000]}'></div></div> </body> </html>