<rss xmlns:atom="http://www.w3.org/2005/Atom" version="2.0"><channel><title>推荐系统 - 标签 - ISLAND</title><link>https://youngxhui.top/tags/%E6%8E%A8%E8%8D%90%E7%B3%BB%E7%BB%9F/</link><description>推荐系统 - 标签 - ISLAND</description><generator>Hugo -- gohugo.io</generator><language>zh-CN</language><managingEditor>youngxhui@gmail.com (youngxhui)</managingEditor><webMaster>youngxhui@gmail.com (youngxhui)</webMaster><copyright>本作品采用知识共享署名-非商业性使用 4.0 国际许可协议进行许可。</copyright><lastBuildDate>Fri, 25 May 2018 09:21:51 +0000</lastBuildDate><atom:link href="https://youngxhui.top/tags/%E6%8E%A8%E8%8D%90%E7%B3%BB%E7%BB%9F/" rel="self" type="application/rss+xml"/><item><title>Apriori算法</title><link>https://youngxhui.top/2018/05/apriori%E7%AE%97%E6%B3%95/</link><pubDate>Fri, 25 May 2018 09:21:51 +0000</pubDate><author><name>youngxhui</name></author><guid>https://youngxhui.top/2018/05/apriori%E7%AE%97%E6%B3%95/</guid><description><![CDATA[<h1 id="前言" class="headerLink">
    <a href="#%e5%89%8d%e8%a8%80" class="header-mark"></a>前言</h1><p>下文所有内容基本总结于 《数据挖掘概念与技术 第三版》机械工业出版社</p>
<p>最近复习数据挖掘课程，就把复习的几个算法记录一下。</p>
<h1 id="什么是apriori算法" class="headerLink">
    <a href="#%e4%bb%80%e4%b9%88%e6%98%afapriori%e7%ae%97%e6%b3%95" class="header-mark"></a>什么是Apriori算法</h1><p><code>Apriori</code>算法是 Agrawal 和 R.Srikant 与1994 提出的，为布尔关联规则挖掘频繁项集的原创性算法。<code>Apriori</code>算法使用一宗称为逐层搜索的迭代方法，其中 <code>k</code> 项集用于探索 <code>(k+1)</code> 项。</p>
<p>基本为扫描数据库，累计每个项的计数，并且收集满足最小支持度的项，找出频繁1项集的集合。该集合记为L1。然后，使用L1找出频繁2项集的集合L2，如此一直找下去，知道不能再找到频繁k项集，找出每个Lk需要一次数据库的完整扫描。</p>
<h1 id="寻找频繁项集" class="headerLink">
    <a href="#%e5%af%bb%e6%89%be%e9%a2%91%e7%b9%81%e9%a1%b9%e9%9b%86" class="header-mark"></a>寻找频繁项集</h1><h2 id="先验性质" class="headerLink">
    <a href="#%e5%85%88%e9%aa%8c%e6%80%a7%e8%b4%a8" class="header-mark"></a>先验性质</h2><p>为了提高频繁项集逐层产生的效率，一种称为<strong>先验性质</strong>的重要性质用于压缩搜索空间。</p>
<blockquote>
  <p><strong>先验性质</strong>：频繁项集的所有非空子集也一定是频繁的。</p>

</blockquote><p>该表是一个事务数据库D。该数据库有9个事务，即|D|=9。假设最小支持度为2</p>
<table>
  <thead>
      <tr>
          <th><strong>TID</strong></th>
          <th><strong>商品ID的列表</strong></th>
          <th><strong>TID</strong></th>
          <th><strong>商品ID的列表</strong></th>
      </tr>
  </thead>
  <tbody>
      <tr>
          <td>T100</td>
          <td>I1,I2,I5</td>
          <td>T600</td>
          <td>I2,I3</td>
      </tr>
      <tr>
          <td>T200</td>
          <td>I2,I4</td>
          <td>T700</td>
          <td>I1,I3</td>
      </tr>
      <tr>
          <td>T300</td>
          <td>I2,I3</td>
          <td>T800</td>
          <td>I1,I2,I3,I5</td>
      </tr>
      <tr>
          <td>T400</td>
          <td>I1,I2,I4</td>
          <td>T900</td>
          <td>I1,I2,I3</td>
      </tr>
      <tr>
          <td>T500</td>
          <td>I1,I3</td>
          <td></td>
          <td></td>
      </tr>
  </tbody>
</table>
<ol>
<li>
<p>在算法第一次迭代的时候，每个项都是候选1项集的集合C1的成员。算法扫描，统计出每个项出现的次数。</p>
</li>
<li>
<p>可以确定频繁1项集的集合L1，它由满足最小支持段的候选1项集组成。</p>
</li>
<li>
<p>为发现频繁2项集的集合L2，算法使用L1 🔗 L1产生候选2项集的集合C2。注意剪枝，没有候选的C2中删除，因为这些候选的每个子集也是频繁的。</p>
</li>
<li>
<p>扫描D中的事务，累计C2中的每个候选项集的支持技术。</p>
</li>
<li>
<p>然后确定频繁2项集的集合L2，它由C2中满足最小支持度的候选2项集组成。</p>
</li>
<li>
<p>…………</p>
</li>
</ol>
<p>重复以上步骤，直到Cn = ∅，此算法终止，找出所有的频繁项集。</p>
<p><img class="tw-inline" loading="lazy" src='https://island-hexo.oss-cn-beijing.aliyuncs.com/apirori.png'     ></p>
<h1 id="由频繁项集产生关联规则" class="headerLink">
    <a href="#%e7%94%b1%e9%a2%91%e7%b9%81%e9%a1%b9%e9%9b%86%e4%ba%a7%e7%94%9f%e5%85%b3%e8%81%94%e8%a7%84%e5%88%99" class="header-mark"></a>由频繁项集产生关联规则</h1><p>一旦由数据库D中的事务找出频繁项集，就可以直接由它们产生强关联规则。对于置信度，可以用下面的公式计算。</p>
<blockquote>
  <p>confidence(A ＝&gt;B) = P（A | B）= support_count(A ∪ B) / support_count(A)</p>

</blockquote><p>继续使用上面的案例，该数据包含频繁项集 x = {I1,I2,I5},可有X产生哪些关联规则，并且计算{I1,I2},{I1,I5},{I1}置信度</p>
<p>产生的关联规则</p>
<p>{I1,I2},{I1,I5},{I2,I5},{I1},{I2},{I5}</p>
<p>{I1,I2} =&gt; I5, confidence = 2 / 4 = 50%</p>
<p>{I1,I5} =&gt; I2, confidence =  2 / 2 = 100%</p>
]]></description></item><item><title>协同过滤算法</title><link>https://youngxhui.top/2017/09/%E5%8D%8F%E5%90%8C%E8%BF%87%E6%BB%A4%E7%AE%97%E6%B3%95/</link><pubDate>Fri, 01 Sep 2017 17:52:48 +0000</pubDate><author><name>youngxhui</name></author><guid>https://youngxhui.top/2017/09/%E5%8D%8F%E5%90%8C%E8%BF%87%E6%BB%A4%E7%AE%97%E6%B3%95/</guid><description><![CDATA[<h1 id="前言" class="headerLink">
    <a href="#%e5%89%8d%e8%a8%80" class="header-mark"></a>前言</h1><p>在推荐系统中有很多算法，其中一种就是协同过滤算法。</p>
<h1 id="分类" class="headerLink">
    <a href="#%e5%88%86%e7%b1%bb" class="header-mark"></a>分类</h1><p>在协同过滤算法中，分为两类，基于用户的协同过滤(UserCF)和基于物品的协同过滤(ItemCF)。每个类型的适用情况不同。</p>
<p>在上述的两种类型中，基于用户的协同过滤算法是不太常用的，原因如下</p>
<ul>
<li>用户的数量是远远大于物品的数量，计算用户之间的相似度计算量会很大。</li>
<li>对于一个新用户很难找到相似度</li>
</ul>
<p>所以一般是不会使用基于用户的协同过滤，但是并不表示不用，一般用于新闻推送等。</p>
<h1 id="相似度计算" class="headerLink">
    <a href="#%e7%9b%b8%e4%bc%bc%e5%ba%a6%e8%ae%a1%e7%ae%97" class="header-mark"></a>相似度计算</h1><h2 id="离散图表示" class="headerLink">
    <a href="#%e7%a6%bb%e6%95%a3%e5%9b%be%e8%a1%a8%e7%a4%ba" class="header-mark"></a>离散图表示</h2><p>目前有五个用户对商品1和商品2的喜爱程度，分析用户之间的关系。</p>
<table>
  <thead>
      <tr>
          <th>用户</th>
          <th>商品1</th>
          <th>商品2</th>
      </tr>
  </thead>
  <tbody>
      <tr>
          <td>1</td>
          <td>3.3</td>
          <td>6.5</td>
      </tr>
      <tr>
          <td>2</td>
          <td>5.8</td>
          <td>2.6</td>
      </tr>
      <tr>
          <td>3</td>
          <td>3.6</td>
          <td>6.3</td>
      </tr>
      <tr>
          <td>4</td>
          <td>3.4</td>
          <td>5.8</td>
      </tr>
      <tr>
          <td>5</td>
          <td>5.2</td>
          <td>3.1</td>
      </tr>
  </tbody>
</table>
<p>寻找他们的相似度</p>
<p>通过离散图来进行表示</p>
<p><img class="tw-inline" loading="lazy" src='https://island-hexo.oss-cn-beijing.aliyuncs.com/shopping.png'     ></p>
<p>可以看出 <code>用户1</code> <code>用户3</code> <code>用户4</code> 之间是存在关系的。</p>
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  <pre style="counter-reset: codeblock;" class="tw-block tw-m-0 tw-p-0"><code 
    id="codeblock-id-9" 
    class="
      chroma 
      !tw-block 
      tw-p-0
      tw-m-0
      tw-transition-[max-height] 
      tw-duration-500 
      tw-ease-in-out 
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      "><span class="line"><span class="cl"><span class="c1"># coding = utf8</span>
</span></span><span class="line"><span class="cl"><span class="kn">import</span> <span class="nn">matplotlib.pyplot</span> <span class="k">as</span> <span class="nn">plt</span>
</span></span><span class="line"><span class="cl"><span class="kn">import</span> <span class="nn">numpy</span> <span class="k">as</span> <span class="nn">np</span>
</span></span><span class="line"><span class="cl">
</span></span><span class="line"><span class="cl"><span class="c1"># 散点图分析五个用户之间的相关程度</span>
</span></span><span class="line"><span class="cl"><span class="c1"># x轴为对于商品 1 的喜欢程度</span>
</span></span><span class="line"><span class="cl"><span class="c1"># y轴是对于商品 2 的喜欢程度</span>
</span></span><span class="line"><span class="cl"><span class="c1"># 数据如下</span>
</span></span><span class="line"><span class="cl"><span class="c1"># 商品1 [3.3, 5.8, 3.6, 3.4, 5.2]</span>
</span></span><span class="line"><span class="cl"><span class="c1"># 商品2 [6.5, 2.6, 6.3, 5.8, 3.1]</span>
</span></span><span class="line"><span class="cl">
</span></span><span class="line"><span class="cl"><span class="n">shop1</span> <span class="o">=</span> <span class="p">[</span><span class="mf">3.3</span><span class="p">,</span> <span class="mf">5.8</span><span class="p">,</span> <span class="mf">3.6</span><span class="p">,</span> <span class="mf">3.4</span><span class="p">,</span> <span class="mf">5.2</span><span class="p">]</span>
</span></span><span class="line"><span class="cl"><span class="n">shop2</span> <span class="o">=</span> <span class="p">[</span><span class="mf">6.5</span><span class="p">,</span> <span class="mf">2.6</span><span class="p">,</span> <span class="mf">6.3</span><span class="p">,</span> <span class="mf">5.8</span><span class="p">,</span> <span class="mf">3.1</span><span class="p">]</span>
</span></span><span class="line"><span class="cl"><span class="n">color</span> <span class="o">=</span> <span class="p">[</span><span class="s1">&#39;r&#39;</span><span class="p">,</span> <span class="s1">&#39;b&#39;</span><span class="p">,</span> <span class="s1">&#39;c&#39;</span><span class="p">,</span> <span class="s1">&#39;g&#39;</span><span class="p">,</span> <span class="s1">&#39;y&#39;</span><span class="p">,</span> <span class="s1">&#39;k&#39;</span><span class="p">,</span> <span class="s1">&#39;m&#39;</span><span class="p">,</span> <span class="s1">&#39;0xff0012&#39;</span><span class="p">]</span>
</span></span><span class="line"><span class="cl"><span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="mi">0</span><span class="p">,</span> <span class="nb">len</span><span class="p">(</span><span class="n">shop1</span><span class="p">)):</span>
</span></span><span class="line"><span class="cl">    <span class="n">plt</span><span class="o">.</span><span class="n">scatter</span><span class="p">(</span><span class="n">shop1</span><span class="p">[</span><span class="n">i</span><span class="p">],</span> <span class="n">shop2</span><span class="p">[</span><span class="n">i</span><span class="p">],</span> <span class="n">c</span><span class="o">=</span><span class="n">color</span><span class="p">[</span><span class="n">i</span><span class="p">])</span>
</span></span><span class="line"><span class="cl"><span class="n">plt</span><span class="o">.</span><span class="n">xlim</span><span class="p">(</span><span class="mi">0</span><span class="p">,</span> <span class="mi">10</span><span class="p">)</span>
</span></span><span class="line"><span class="cl"><span class="n">plt</span><span class="o">.</span><span class="n">ylim</span><span class="p">(</span><span class="mi">0</span><span class="p">,</span> <span class="mi">10</span><span class="p">)</span>
</span></span><span class="line"><span class="cl"><span class="n">plt</span><span class="o">.</span><span class="n">ylabel</span><span class="p">(</span><span class="s2">&#34;shopping 2&#34;</span><span class="p">)</span>
</span></span><span class="line"><span class="cl"><span class="n">plt</span><span class="o">.</span><span class="n">xlabel</span><span class="p">(</span><span class="s2">&#34;shopping 1&#34;</span><span class="p">)</span>
</span></span><span class="line"><span class="cl"><span class="n">plt</span><span class="o">.</span><span class="n">legend</span><span class="p">(</span><span class="s1">&#39;12345&#39;</span><span class="p">)</span>
</span></span><span class="line"><span class="cl"><span class="n">plt</span><span class="o">.</span><span class="n">show</span><span class="p">()</span></span></span></code></pre>
</div>
<h2 id="欧几里得距离" class="headerLink">
    <a href="#%e6%ac%a7%e5%87%a0%e9%87%8c%e5%be%97%e8%b7%9d%e7%a6%bb" class="header-mark"></a>欧几里得距离</h2><p>欧几里得距离是最常用的距离公式,用于计算两点之间的距离。公式如下：</p>
<p><img class="tw-inline" loading="lazy" src='https://island-hexo.oss-cn-beijing.aliyuncs.com/render.gif'     ></p>
<p>通过计算两点之间的距离来表示他们之间的相关程度。</p>
<p>我们通过 <code>python</code> 计算 欧几里得距离。</p>
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          <p class="tw-select-none !tw-my-1">python</p>
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  <pre style="counter-reset: codeblock;" class="tw-block tw-m-0 tw-p-0"><code 
    id="codeblock-id-10" 
    class="
      chroma 
      !tw-block 
      tw-p-0
      tw-m-0
      tw-transition-[max-height] 
      tw-duration-500 
      tw-ease-in-out 
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      group-[.is-wrap]:tw-text-wrap
      tw-overflow-y-hidden
      tw-overflow-x-auto
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      "><span class="line"><span class="cl"><span class="c1"># coding = utf8</span>
</span></span><span class="line"><span class="cl">
</span></span><span class="line"><span class="cl"><span class="c1">#        1    2     3   4   5</span>
</span></span><span class="line"><span class="cl"><span class="n">shop1</span> <span class="o">=</span> <span class="p">[</span><span class="mf">3.3</span><span class="p">,</span> <span class="mf">5.8</span><span class="p">,</span> <span class="mf">3.6</span><span class="p">,</span> <span class="mf">3.4</span><span class="p">,</span> <span class="mf">5.2</span><span class="p">]</span>
</span></span><span class="line"><span class="cl"><span class="n">shop2</span> <span class="o">=</span> <span class="p">[</span><span class="mf">6.5</span><span class="p">,</span> <span class="mf">2.6</span><span class="p">,</span> <span class="mf">6.3</span><span class="p">,</span> <span class="mf">5.8</span><span class="p">,</span> <span class="mf">3.1</span><span class="p">]</span>
</span></span><span class="line"><span class="cl"><span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="mi">0</span><span class="p">,</span> <span class="nb">len</span><span class="p">(</span><span class="n">shop1</span><span class="p">)</span> <span class="o">-</span> <span class="mi">1</span><span class="p">):</span>
</span></span><span class="line"><span class="cl">    <span class="k">for</span> <span class="n">j</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="n">i</span> <span class="o">+</span> <span class="mi">1</span><span class="p">,</span> <span class="nb">len</span><span class="p">(</span><span class="n">shop2</span><span class="p">)):</span>
</span></span><span class="line"><span class="cl">        <span class="n">distance</span> <span class="o">=</span> <span class="p">(</span><span class="n">shop1</span><span class="p">[</span><span class="n">i</span><span class="p">]</span> <span class="o">-</span> <span class="n">shop1</span><span class="p">[</span><span class="n">j</span><span class="p">])</span> <span class="o">**</span> <span class="mi">2</span> <span class="o">+</span> <span class="p">(</span><span class="n">shop2</span><span class="p">[</span><span class="n">i</span><span class="p">]</span> <span class="o">-</span> <span class="n">shop2</span><span class="p">[</span><span class="n">j</span><span class="p">])</span> <span class="o">**</span> <span class="mi">2</span>
</span></span><span class="line"><span class="cl">        <span class="n">correlation</span> <span class="o">=</span> <span class="mi">1</span> <span class="o">/</span> <span class="p">(</span><span class="mi">1</span> <span class="o">+</span> <span class="n">distance</span><span class="p">)</span> <span class="c1"># 将范围缩小至0 ~ 1 之间</span>
</span></span><span class="line"><span class="cl">        <span class="k">if</span> <span class="n">correlation</span> <span class="o">&gt;</span> <span class="mf">0.60</span><span class="p">:</span>  <span class="c1"># 根据需求更换相关系数</span>
</span></span><span class="line"><span class="cl">            <span class="nb">print</span><span class="p">(</span><span class="s2">&#34;第 </span><span class="si">{}</span><span class="s2"> and </span><span class="si">{}</span><span class="s2">  distance is </span><span class="si">{}</span><span class="s2">&#34;</span><span class="o">.</span><span class="n">format</span><span class="p">(</span><span class="n">i</span> <span class="o">+</span> <span class="mi">1</span><span class="p">,</span> <span class="n">j</span> <span class="o">+</span> <span class="mi">1</span><span class="p">,</span> <span class="n">correlation</span><span class="p">))</span></span></span></code></pre>
</div>
<p>计算结果</p>
<hr>
<blockquote>
  <p>第 1 和 2  相关系数是 0.04452359750667854</p>

</blockquote><blockquote>
  <p>第 1 和 3  相关系数是 0.8849557522123891</p>

</blockquote><blockquote>
  <p>第 1 和 4  相关系数是 0.6666666666666665</p>

</blockquote><blockquote>
  <p>第 1 和 5  相关系数是 0.061842918985776124</p>

</blockquote><blockquote>
  <p>第 2 和 3  相关系数是 0.051203277009728626</p>

</blockquote><blockquote>
  <p>第 2 和 4  相关系数是 0.058823529411764705</p>

</blockquote><blockquote>
  <p>第 2 和 5  相关系数是 0.6211180124223604</p>

</blockquote><blockquote>
  <p>第 3 和 4  相关系数是 0.7751937984496123</p>

</blockquote><blockquote>
  <p>第 3 和 5  相关系数是 0.07246376811594203</p>

</blockquote><blockquote>
  <p>第 4 和 5  相关系数是 0.08673026886383348</p>

</blockquote><hr>
<p>得出结论，基本和离散图相似。</p>
<p>这样就可以根据需求来判断用户之间的相似度。</p>
<h2 id="皮尔逊相关系数" class="headerLink">
    <a href="#%e7%9a%ae%e5%b0%94%e9%80%8a%e7%9b%b8%e5%85%b3%e7%b3%bb%e6%95%b0" class="header-mark"></a>皮尔逊相关系数</h2><p>皮尔逊相关系数同样也是计算距离的一种方式</p>
<p>通过相关系数可以计算出他们的相关程度。</p>
<h3 id="适用范围" class="headerLink">
    <a href="#%e9%80%82%e7%94%a8%e8%8c%83%e5%9b%b4" class="header-mark"></a>适用范围</h3><p>皮尔逊相关系数是有一定的适用条件。</p>
<blockquote>
  <p>当两个变量的标准差都不为零时，相关系数才有定义，皮尔逊相关系数适用于：
(1)、两个变量之间是线性关系，都是连续数据。
(2)、两个变量的总体是正态分布，或接近正态的单峰分布。
(3)、两个变量的观测值是成对的，每对观测值之间相互独立。</p>

</blockquote><p>同样这个算法也是有缺陷的,但数据越少的时候,波动性越大。当数据很少的时候不建议使用。</p>
<p>皮尔逊相关系数公式有四个</p>
<p>公式一：</p>
<p><img class="tw-inline" loading="lazy" src='https://island-hexo.oss-cn-beijing.aliyuncs.com/person1.gif'     ></p>
<p>公式二：</p>
<p><img class="tw-inline" loading="lazy" src='https://island-hexo.oss-cn-beijing.aliyuncs.com/person2.gif'     ></p>
<p>公式三：</p>
<p><img class="tw-inline" loading="lazy" src='https://island-hexo.oss-cn-beijing.aliyuncs.com/person3.gif'     ></p>
<p>公式四：</p>
<p><img class="tw-inline" loading="lazy" src='https://island-hexo.oss-cn-beijing.aliyuncs.com/person4.gif'     ></p>
<p>相关系数</p>
<ul>
<li>0.8-1.0     极强相关</li>
<li>0.6-0.8     强相关</li>
<li>0.4-0.6     中等程度相关</li>
<li>0.2-0.4     弱相关</li>
<li>0.0-0.2     极弱相关或无相关</li>
</ul>
<p>我们通过第四个公式进行计算</p>
<p><img class="tw-inline" loading="lazy" src='https://island-hexo.oss-cn-beijing.aliyuncs.com/person4.gif'     ></p>
<h3 id="简单案例" class="headerLink">
    <a href="#%e7%ae%80%e5%8d%95%e6%a1%88%e4%be%8b" class="header-mark"></a>简单案例</h3><p>通过 <code>python</code> 计算出相关系数</p>
<div class="code-block highlight is-closed show-line-numbers  tw-group tw-my-2">
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          <p class="tw-select-none !tw-my-1">python</p>
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  <pre style="counter-reset: codeblock;" class="tw-block tw-m-0 tw-p-0"><code 
    id="codeblock-id-11" 
    class="
      chroma 
      !tw-block 
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      "><span class="line"><span class="cl"><span class="c1"># coding=utf8</span>
</span></span><span class="line"><span class="cl">
</span></span><span class="line"><span class="cl">
</span></span><span class="line"><span class="cl"><span class="c1"># 皮尔孙相关系数计算</span>
</span></span><span class="line"><span class="cl"><span class="k">def</span> <span class="nf">PearsonCorrelationSimilarity</span><span class="p">(</span><span class="n">vec1</span><span class="p">,</span> <span class="n">vec2</span><span class="p">):</span>
</span></span><span class="line"><span class="cl">    <span class="n">value</span> <span class="o">=</span> <span class="nb">range</span><span class="p">(</span><span class="nb">len</span><span class="p">(</span><span class="n">vec1</span><span class="p">))</span>
</span></span><span class="line"><span class="cl">
</span></span><span class="line"><span class="cl">    <span class="n">sum_vec1</span> <span class="o">=</span> <span class="nb">sum</span><span class="p">([</span><span class="n">vec1</span><span class="p">[</span><span class="n">i</span><span class="p">]</span> <span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="n">value</span><span class="p">])</span>
</span></span><span class="line"><span class="cl">    <span class="n">sum_vec2</span> <span class="o">=</span> <span class="nb">sum</span><span class="p">([</span><span class="n">vec2</span><span class="p">[</span><span class="n">i</span><span class="p">]</span> <span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="n">value</span><span class="p">])</span>
</span></span><span class="line"><span class="cl">
</span></span><span class="line"><span class="cl">    <span class="n">square_sum_vec1</span> <span class="o">=</span> <span class="nb">sum</span><span class="p">([</span><span class="nb">pow</span><span class="p">(</span><span class="n">vec1</span><span class="p">[</span><span class="n">i</span><span class="p">],</span> <span class="mi">2</span><span class="p">)</span> <span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="n">value</span><span class="p">])</span>
</span></span><span class="line"><span class="cl">    <span class="n">square_sum_vec2</span> <span class="o">=</span> <span class="nb">sum</span><span class="p">([</span><span class="nb">pow</span><span class="p">(</span><span class="n">vec2</span><span class="p">[</span><span class="n">i</span><span class="p">],</span> <span class="mi">2</span><span class="p">)</span> <span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="n">value</span><span class="p">])</span>
</span></span><span class="line"><span class="cl">
</span></span><span class="line"><span class="cl">    <span class="n">product</span> <span class="o">=</span> <span class="nb">sum</span><span class="p">([</span><span class="n">vec1</span><span class="p">[</span><span class="n">i</span><span class="p">]</span> <span class="o">*</span> <span class="n">vec2</span><span class="p">[</span><span class="n">i</span><span class="p">]</span> <span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="n">value</span><span class="p">])</span>
</span></span><span class="line"><span class="cl">
</span></span><span class="line"><span class="cl">    <span class="n">numerator</span> <span class="o">=</span> <span class="n">product</span> <span class="o">-</span> <span class="p">(</span><span class="n">sum_vec1</span> <span class="o">*</span> <span class="n">sum_vec2</span> <span class="o">/</span> <span class="nb">len</span><span class="p">(</span><span class="n">vec1</span><span class="p">))</span>
</span></span><span class="line"><span class="cl">    <span class="n">dominator</span> <span class="o">=</span> <span class="p">((</span><span class="n">square_sum_vec1</span> <span class="o">-</span> <span class="nb">pow</span><span class="p">(</span><span class="n">sum_vec1</span><span class="p">,</span> <span class="mi">2</span><span class="p">)</span> <span class="o">/</span> <span class="nb">len</span><span class="p">(</span><span class="n">vec1</span><span class="p">))</span> <span class="o">*</span> <span class="p">(</span>
</span></span><span class="line"><span class="cl">        <span class="n">square_sum_vec2</span> <span class="o">-</span> <span class="nb">pow</span><span class="p">(</span><span class="n">sum_vec2</span><span class="p">,</span> <span class="mi">2</span><span class="p">)</span> <span class="o">/</span> <span class="nb">len</span><span class="p">(</span><span class="n">vec2</span><span class="p">)))</span> <span class="o">**</span> <span class="mf">0.5</span>
</span></span><span class="line"><span class="cl">
</span></span><span class="line"><span class="cl">    <span class="k">if</span> <span class="n">dominator</span> <span class="o">==</span> <span class="mi">0</span><span class="p">:</span>
</span></span><span class="line"><span class="cl">        <span class="k">return</span> <span class="mi">0</span>
</span></span><span class="line"><span class="cl">    <span class="n">result</span> <span class="o">=</span> <span class="n">numerator</span> <span class="o">/</span> <span class="p">(</span><span class="n">dominator</span> <span class="o">*</span> <span class="mf">1.0</span><span class="p">)</span>
</span></span><span class="line"><span class="cl">
</span></span><span class="line"><span class="cl">    <span class="k">return</span> <span class="n">result</span>
</span></span><span class="line"><span class="cl">
</span></span><span class="line"><span class="cl">
</span></span><span class="line"><span class="cl"><span class="k">if</span> <span class="vm">__name__</span> <span class="o">==</span> <span class="s1">&#39;__main__&#39;</span><span class="p">:</span>
</span></span><span class="line"><span class="cl">    <span class="c1"># 五个用户对五个商品的评价</span>
</span></span><span class="line"><span class="cl">    <span class="n">user1</span> <span class="o">=</span> <span class="p">[</span><span class="mf">3.3</span><span class="p">,</span> <span class="mf">5.8</span><span class="p">,</span> <span class="mf">3.6</span><span class="p">,</span> <span class="mf">3.4</span><span class="p">,</span> <span class="mf">5.2</span><span class="p">]</span>
</span></span><span class="line"><span class="cl">    <span class="n">user2</span> <span class="o">=</span> <span class="p">[</span><span class="mf">6.5</span><span class="p">,</span> <span class="mf">2.6</span><span class="p">,</span> <span class="mf">6.3</span><span class="p">,</span> <span class="mf">5.8</span><span class="p">,</span> <span class="mf">3.1</span><span class="p">]</span>
</span></span><span class="line"><span class="cl">    <span class="n">user3</span> <span class="o">=</span> <span class="p">[</span><span class="mf">5.5</span><span class="p">,</span> <span class="mf">3.2</span><span class="p">,</span> <span class="mf">6.5</span><span class="p">,</span> <span class="mf">4.7</span><span class="p">,</span> <span class="mf">4.4</span><span class="p">]</span>
</span></span><span class="line"><span class="cl">    <span class="n">user4</span> <span class="o">=</span> <span class="p">[</span><span class="mf">4.4</span><span class="p">,</span> <span class="mf">6.2</span><span class="p">,</span> <span class="mf">2.3</span><span class="p">,</span> <span class="mf">5.1</span><span class="p">,</span> <span class="mf">3.3</span><span class="p">]</span>
</span></span><span class="line"><span class="cl">    <span class="n">user5</span> <span class="o">=</span> <span class="p">[</span><span class="mf">2.1</span><span class="p">,</span> <span class="mf">5.2</span><span class="p">,</span> <span class="mf">4.2</span><span class="p">,</span> <span class="mf">2.2</span><span class="p">,</span> <span class="mf">4.1</span><span class="p">]</span>
</span></span><span class="line"><span class="cl">
</span></span><span class="line"><span class="cl">    <span class="n">userlist</span> <span class="o">=</span> <span class="p">[</span><span class="n">user1</span><span class="p">,</span> <span class="n">user2</span><span class="p">,</span> <span class="n">user3</span><span class="p">,</span> <span class="n">user4</span><span class="p">,</span> <span class="n">user5</span><span class="p">]</span>
</span></span><span class="line"><span class="cl">    <span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="mi">0</span><span class="p">,</span> <span class="nb">len</span><span class="p">(</span><span class="n">userlist</span><span class="p">)</span> <span class="o">-</span> <span class="mi">1</span><span class="p">):</span>
</span></span><span class="line"><span class="cl">        <span class="k">for</span> <span class="n">j</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="n">i</span> <span class="o">+</span> <span class="mi">1</span><span class="p">,</span> <span class="nb">len</span><span class="p">(</span><span class="n">userlist</span><span class="p">)):</span>
</span></span><span class="line"><span class="cl">            <span class="n">result</span> <span class="o">=</span> <span class="n">PearsonCorrelationSimilarity</span><span class="p">(</span><span class="n">userlist</span><span class="p">[</span><span class="n">i</span><span class="p">],</span> <span class="n">userlist</span><span class="p">[</span><span class="n">j</span><span class="p">])</span>
</span></span><span class="line"><span class="cl">            <span class="nb">print</span> <span class="s2">&#34;user</span><span class="si">%d</span><span class="s2"> 和 user</span><span class="si">%d</span><span class="s2"> 的相关系数是</span><span class="si">%f</span><span class="s2">&#34;</span> <span class="o">%</span> <span class="p">(</span><span class="n">i</span> <span class="o">+</span> <span class="mi">1</span><span class="p">,</span> <span class="n">j</span> <span class="o">+</span> <span class="mi">1</span><span class="p">,</span> <span class="n">result</span><span class="p">)</span></span></span></code></pre>
</div>
<blockquote>
  <p>user1 和 user2 的相关系数是0.999774</p>

</blockquote><blockquote>
  <p>user1 和 user3 的相关系数是-0.847758</p>

</blockquote><blockquote>
  <p>user1 和 user4 的相关系数是-0.841816</p>

</blockquote><blockquote>
  <p>user1 和 user5 的相关系数是-0.915237</p>

</blockquote><blockquote>
  <p>user2 和 user3 的相关系数是-0.841741</p>

</blockquote><blockquote>
  <p>user2 和 user4 的相关系数是-0.835320</p>

</blockquote><blockquote>
  <p>user2 和 user5 的相关系数是-0.909975</p>

</blockquote><blockquote>
  <p>user3 和 user4 的相关系数是0.998987</p>

</blockquote><blockquote>
  <p>user3 和 user5 的相关系数是0.976272</p>

</blockquote><blockquote>
  <p>user4 和 user5 的相关系数是0.969782</p>

</blockquote><p>通过计算出来的结果可以看出来等到的结果</p>
<h2 id="余弦距离" class="headerLink">
    <a href="#%e4%bd%99%e5%bc%a6%e8%b7%9d%e7%a6%bb" class="header-mark"></a>余弦距离</h2><p>余弦相似度用向量空间中两个向量夹角的余弦值作为衡量两个个体间差异的大小。余弦值越接近1，就表明夹角越接近0度，也就是两个向量越相似，这就叫&quot;余弦相似性&quot;。</p>
<h3 id="公式" class="headerLink">
    <a href="#%e5%85%ac%e5%bc%8f" class="header-mark"></a>公式</h3><p>一般余弦定理这个公式运用与文本分析。</p>
<p>分别采用下列的文本进行</p>
<h3 id="示例" class="headerLink">
    <a href="#%e7%a4%ba%e4%be%8b" class="header-mark"></a>示例</h3><p>A：这里距离北京有多远，火车多长时间</p>
<p>B:北京离这里有多远，坐火车多长时间</p>
<h4 id="对ab进行分词" class="headerLink">
    <a href="#%e5%af%b9ab%e8%bf%9b%e8%a1%8c%e5%88%86%e8%af%8d" class="header-mark"></a>对A，B进行分词</h4><p>A:这里/距离/北京/有/多远，火车/多长/时间</p>
<p>B:北京/离/这里/有/多远，坐/火车/多长/时间</p>
<h4 id="列出所有的词" class="headerLink">
    <a href="#%e5%88%97%e5%87%ba%e6%89%80%e6%9c%89%e7%9a%84%e8%af%8d" class="header-mark"></a>列出所有的词</h4><p>这里 距离 北京 有 多远 火车 多长 时间 离 坐</p>
<h4 id="进行词频统计" class="headerLink">
    <a href="#%e8%bf%9b%e8%a1%8c%e8%af%8d%e9%a2%91%e7%bb%9f%e8%ae%a1" class="header-mark"></a>进行词频统计</h4><p>A: 这里(1)距离(1)北京(1)有(1)多远(1)火车(1)多长(1)时间(1)离(0)坐(0)</p>
<p>B: 这里(1)距离(0)北京(1)有(1)多远(1)火车(1)多长(1)时间(1)离(1)坐(1)</p>
<h4 id="计算句子的词频向量" class="headerLink">
    <a href="#%e8%ae%a1%e7%ae%97%e5%8f%a5%e5%ad%90%e7%9a%84%e8%af%8d%e9%a2%91%e5%90%91%e9%87%8f" class="header-mark"></a>计算句子的词频向量</h4><p>A (1,1,1,1,1,1,1,1,0,0)</p>
<p>B (1,0,1,1,1,1,1,1,1,1)</p>
<h4 id="通过余弦定理计算相似度" class="headerLink">
    <a href="#%e9%80%9a%e8%bf%87%e4%bd%99%e5%bc%a6%e5%ae%9a%e7%90%86%e8%ae%a1%e7%ae%97%e7%9b%b8%e4%bc%bc%e5%ba%a6" class="header-mark"></a>通过余弦定理计算相似度</h4><div class="code-block highlight is-open show-line-numbers  tw-group tw-my-2">
  <div class="
    
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  <pre style="counter-reset: codeblock;" class="tw-block tw-m-0 tw-p-0"><code 
    id="codeblock-id-12" 
    class="
      chroma 
      !tw-block 
      tw-p-0
      tw-m-0
      tw-transition-[max-height] 
      tw-duration-500 
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      "><span class="line"><span class="cl"><span class="c1"># coding=utf8</span>
</span></span><span class="line"><span class="cl">
</span></span><span class="line"><span class="cl"><span class="kn">import</span> <span class="nn">math</span>
</span></span><span class="line"><span class="cl">
</span></span><span class="line"><span class="cl"><span class="c1"># 余弦定理</span>
</span></span><span class="line"><span class="cl"><span class="k">def</span> <span class="nf">cos</span><span class="p">(</span><span class="n">v1</span><span class="p">,</span> <span class="n">v2</span><span class="p">):</span>
</span></span><span class="line"><span class="cl">    <span class="n">l</span> <span class="o">=</span> <span class="mf">0.0</span>
</span></span><span class="line"><span class="cl">    <span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="mi">0</span><span class="p">,</span> <span class="nb">len</span><span class="p">(</span><span class="n">v1</span><span class="p">)):</span>
</span></span><span class="line"><span class="cl">        <span class="n">l</span> <span class="o">+=</span> <span class="n">v1</span><span class="p">[</span><span class="n">i</span><span class="p">]</span> <span class="o">*</span> <span class="n">v2</span><span class="p">[</span><span class="n">i</span><span class="p">]</span>
</span></span><span class="line"><span class="cl">    <span class="n">v</span> <span class="o">=</span> <span class="mf">0.0</span>
</span></span><span class="line"><span class="cl">    <span class="n">w</span> <span class="o">=</span> <span class="mf">0.0</span>
</span></span><span class="line"><span class="cl">    <span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="mi">0</span><span class="p">,</span> <span class="nb">len</span><span class="p">(</span><span class="n">v1</span><span class="p">)):</span>
</span></span><span class="line"><span class="cl">        <span class="n">v</span> <span class="o">+=</span> <span class="n">math</span><span class="o">.</span><span class="n">pow</span><span class="p">(</span><span class="n">v1</span><span class="p">[</span><span class="n">i</span><span class="p">],</span> <span class="mi">2</span><span class="p">)</span>
</span></span><span class="line"><span class="cl">        <span class="n">w</span> <span class="o">+=</span> <span class="n">math</span><span class="o">.</span><span class="n">pow</span><span class="p">(</span><span class="n">v2</span><span class="p">[</span><span class="n">i</span><span class="p">],</span> <span class="mi">2</span><span class="p">)</span>
</span></span><span class="line"><span class="cl">    <span class="n">cos</span> <span class="o">=</span> <span class="n">l</span> <span class="o">/</span> <span class="p">(</span><span class="n">math</span><span class="o">.</span><span class="n">sqrt</span><span class="p">(</span><span class="n">v</span><span class="p">)</span> <span class="o">*</span> <span class="n">math</span><span class="o">.</span><span class="n">sqrt</span><span class="p">(</span><span class="n">w</span><span class="p">))</span>
</span></span><span class="line"><span class="cl">    <span class="k">return</span> <span class="n">cos</span>
</span></span><span class="line"><span class="cl">
</span></span><span class="line"><span class="cl"><span class="k">if</span> <span class="vm">__name__</span> <span class="o">==</span> <span class="s1">&#39;__main__&#39;</span><span class="p">:</span>
</span></span><span class="line"><span class="cl">    <span class="n">v1</span> <span class="o">=</span> <span class="p">[</span><span class="mi">1</span><span class="p">,</span> <span class="mi">1</span><span class="p">,</span> <span class="mi">1</span><span class="p">,</span> <span class="mi">1</span><span class="p">,</span> <span class="mi">1</span><span class="p">,</span> <span class="mi">1</span><span class="p">,</span> <span class="mi">1</span><span class="p">,</span> <span class="mi">1</span><span class="p">,</span> <span class="mi">0</span><span class="p">,</span> <span class="mi">0</span><span class="p">]</span>
</span></span><span class="line"><span class="cl">    <span class="n">v2</span> <span class="o">=</span> <span class="p">[</span><span class="mi">1</span><span class="p">,</span> <span class="mi">0</span><span class="p">,</span> <span class="mi">1</span><span class="p">,</span> <span class="mi">1</span><span class="p">,</span> <span class="mi">1</span><span class="p">,</span> <span class="mi">1</span><span class="p">,</span> <span class="mi">1</span><span class="p">,</span> <span class="mi">1</span><span class="p">,</span> <span class="mi">1</span><span class="p">,</span> <span class="mi">1</span><span class="p">]</span>
</span></span><span class="line"><span class="cl">    <span class="n">l</span> <span class="o">=</span> <span class="n">cos</span><span class="p">(</span><span class="n">v1</span><span class="p">,</span> <span class="n">v2</span><span class="p">)</span>
</span></span><span class="line"><span class="cl">    <span class="nb">print</span> <span class="n">l</span></span></span></code></pre>
</div>
<p>计算出的相似度为：</p>
<blockquote>
  <p>0.824957911384</p>

</blockquote><p>说明这两句话基本相等</p>
<p>这是一些基本的相似度计算方法。</p>
]]></description></item></channel></rss>