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<a href="/2023/04/14/why-Item-based-recommender-often-works-better-than-user-based-in-practe/" class="post-title-link" itemprop="url">why Item based recommender often works better than user based in practe</a>
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<time title="创建时间：2023-04-14 21:18:51 / 修改时间：21:47:22" itemprop="dateCreated datePublished" datetime="2023-04-14T21:18:51+08:00">2023-04-14</time>
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<h1 id="BigData-Question-Item-based-vs-User-based-in-filtering"><a href="#BigData-Question-Item-based-vs-User-based-in-filtering" class="headerlink" title="BigData Question:Item-based vs User-based in filtering"></a>BigData Question:Item-based vs User-based in filtering</h1><p>Q:In practice, it has been observed that item-item often works better than user-user</p>
<p>A:</p>
<ul>
<li>数据稀疏性：用户-用户的矩阵通常是稀疏的，用户可能只评价了几个项目。在这种情况下，用户-用户协同过滤可能不是很有效，因为没有足够的数据来找到相似的用户。另一方面，项目-项目协同过滤仍然有效，因为它依赖于项目之间的相似性，而这些相似性通常比用户-用户相似性更密集。</li>
<li>冷启动：对于用户-用户协方法来说，新用户几乎没有关于他们的信息可以与其他用户进行比较。但是item-item方法来说，它受到较小的冷启动影响</li>
<li>稳定性：在很多情况下，物品比用户更稳定。例如，电影的特征可能会随着时间的推移保持不变，而用户的偏好可能会发生变化。这意味着项目之间的相似性可能比用户之间的相似性更稳定，从而使项目-项目协同过滤更加有效。</li>
<li>可扩展性：对于user-user方法来说，系统中的用户越多，寻找最近的 K 个邻居的成本就越大。。另一方面，item-item filtering只需要计算item之间的相似度，通常计算量较小。</li>
</ul>
<p>总体而言，项目-项目过滤在实践中通常比用户-用户过滤效果更好，因为它具有可扩展性、处理稀疏性的能力以及更好的稳定性。</p>
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<a href="/2023/04/14/CMake%E8%AF%AD%E6%B3%95%E4%BB%8B%E7%BB%8D/" class="post-title-link" itemprop="url">CMake语法介绍</a>
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<span class="post-meta-item-text">发表于</span>
<time title="创建时间：2023-04-14 16:22:36 / 修改时间：19:25:29" itemprop="dateCreated datePublished" datetime="2023-04-14T16:22:36+08:00">2023-04-14</time>
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<p><font face="黑体" size="4">一、CMake介绍</font><br><font face="宋体" size=4></p>
<p>&nbsp;<br>CMake是一种用于项目管理的工具，适用于多种编程语言组成的项目。<br></font></p>
<p><font face="黑体" size="4">二、CMake关键字</font></p>
<font face="宋体" size=4>
<ul>
<li><p>PROJECT 关键字，指定工程名字和支持语言，格式为PROJECT(name,option),name为项目名，option:CXX 支持C++;C 支持C;JAVA 支持java.默认为空，支持所有语言</p>
</li>
<li><p>SET关键字<br>用来显式指定变量,比如SET($SRC_LIST,main.cpp) SRC_LIST变量包含了main.cpp</p>
</li>
<li><p>MESSAGE关键字 用来输出关键信息的，有3种状态的信息：</p>
<ul>
<li>1.SEND_ERROR,产生错误，生成过程跳过</li>
<li>2.STATUS,前缀为- 的信息</li>
<li>3.FATAL_ERROR,立即终止所有的cmake进程</li>
</ul>
</li>
<li><p>ADD_EXECUTABLE关键字 生成可执行文件。格式为ADD_EXECUTABLE(name ${SRC_LIST})</p>
</li>
<li><p>ADD_SUBDIRECTORY(name1,name2) name1是子目录文件夹名，name2是中间文件放置的文件夹名</p>
</li>
</ul>
<p><font face="黑体" size="4">二、语法规则</font></p>
<font face="宋体" size=4>
<ul>
<li>变量使用${}取值，但在IF语句中直接取值</li>
<li>指令(参数1，2，3…)可以使用空格或分号隔开</li>
<li>指令大小写无关，参数和变量大小写相关</li>
</ul>
<p>注意：如果源文件名中有空格，需要使用””</p>
<p><font face="黑体" size="4">四、内部构建和外部构建</font></p>
<font face="宋体" size=4>
<ul>
<li>内部构建就是在源文件目录内直接运行，产生的临时文件很多，不好清理</li>
<li>外部构建就是新建一个build文件夹，在里面运行cmake ..,之后在build目录下，运行make来构建工程</li>
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<p><font face="黑体" size="4">五、安装</font></p>
<font face="宋体" size=4>
INSTALL:
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<h1 id="序列比对DP问题特征分析"><a href="#序列比对DP问题特征分析" class="headerlink" title="序列比对DP问题特征分析"></a><center><font face="黑体" size="6">序列比对DP问题特征分析</font></center></h1><h2 id="一些例题"><a href="#一些例题" class="headerlink" title="一些例题"></a>一些例题</h2><p><a target="_blank" rel="noopener" href="https://leetcode.cn/problems/uncrossed-lines/">1035. 不相交的线</a></p>
<p><a target="_blank" rel="noopener" href="https://leetcode.cn/problems/longest-increasing-subsequence/">300. 最长递增子序列</a></p>
<p><a target="_blank" rel="noopener" href="https://leetcode.cn/problems/delete-operation-for-two-strings/">583. 两个字符串的删除操作</a></p>
<p><a target="_blank" rel="noopener" href="https://leetcode.cn/problems/minimum-ascii-delete-sum-for-two-strings/">712. 两个字符串的最小ASCII删除和</a> </p>
<p><a target="_blank" rel="noopener" href="https://leetcode.cn/problems/longest-common-subsequence/">1143. 最长公共子序列</a></p>
<h2 id="进阶"><a href="#进阶" class="headerlink" title="进阶"></a>进阶</h2><p><a target="_blank" rel="noopener" href="https://leetcode.cn/problems/edit-distance/">72 编辑距离 hard</a></p>
<p><font face="黑体" size="4"><strong>摘要</strong>：</font><font face="宋体" size="4"><br>本文是对动态规划问题的一个类型——序列比对问题的特征分析。<br></font></p>
<p><font face="黑体" size="4">关键词：</font><font face="宋体" size="4">动态规划、序列比对</font></p>
<h2 id="一、问题引入"><a href="#一、问题引入" class="headerlink" title="一、问题引入"></a><font face="黑体" size="5">一、问题引入</font></h2><h3 id="1-序列比对"><a href="#1-序列比对" class="headerlink" title="1.序列比对"></a><font face="黑体" size="4">1.序列比对</font></h3><font face="宋体" size="4">
序列比对是生物信息领域的经典问题，简单来说，就是求两个序列之间的相似性。例如以下两个碱基序列:
<p>序列X:ATTG 序列Y:ATGC,如何比对这两个序列的相似性？</p>
<p>这里可以引入点阵图的方式来看一下这种问题</p>
<center>
<table>
<thead>
<tr>
<th></th>
<th>-</th>
<th>A</th>
<th>T</th>
<th>T</th>
<th>G</th>
</tr>
</thead>
<tbody><tr>
<td>-</td>
<td>0</td>
<td>0</td>
<td>0</td>
<td>0</td>
<td>0</td>
</tr>
<tr>
<td>A</td>
<td>0</td>
<td>1</td>
<td>0</td>
<td>0</td>
<td>0</td>
</tr>
<tr>
<td>T</td>
<td>0</td>
<td>0</td>
<td>1</td>
<td>1</td>
<td>0</td>
</tr>
<tr>
<td>G</td>
<td>0</td>
<td>0</td>
<td>0</td>
<td>0</td>
<td>1</td>
</tr>
<tr>
<td>C</td>
<td>0</td>
<td>0</td>
<td>0</td>
<td>0</td>
<td>0</td>
</tr>
</tbody></table>
</center>
<h2 id="二、特征分析"><a href="#二、特征分析" class="headerlink" title="二、特征分析"></a><font face="黑体" size="5">二、特征分析</font></h2><font face="宋体" size="4">
<p>这类问题基本可以抽象为1.两个序列 2.求公共子序列相关问题 3.不能改变相对顺序</p>
<h2 id="三、解决方法"><a href="#三、解决方法" class="headerlink" title="三、解决方法"></a><font face="黑体" size="5">三、解决方法</font></h2><p>按照动态规划的几个基本步骤来说：最重要的就是找出递推公式，这一类问题一般可以用列表法找出递推公式.</p>
<p>通常递推公式dp[i][j]和dp[i-1][j-1],{dp[i-1][j],dp[i][j-1]}有关</p>
<p>比如 <a target="_blank" rel="noopener" href="https://leetcode.cn/problems/uncrossed-lines/">1035. 不相交的线</a></p>
<p><a target="_blank" rel="noopener" href="https://leetcode.cn/problems/longest-increasing-subsequence/">300. 最长递增子序列</a></p>
<p>这两个就是典型的序列比对问题，我们可以提取出递推公式dp[i][j]=s2[i-1]==s1[j-1]?dp[i-1][j-1]+1:max(dp[i-1][j],dp[i][j-1])</p>
<p>注意点:</p>
<ul>
<li>dp初始化为dp[m+1][n+1] (m,n分别为两个序列的长度)，这相当于做了一个填充，或者理解为插入了初始”惩罚”</li>
<li>递推公式中需要根据目标是惩罚还是相似度来决定是max还是min,后面是否需要加reward还是penality</li>
</ul>
<p>算法伪代码</p>
<figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br></pre></td><td class="code"><pre><span class="line"></span><br><span class="line"><span class="built_in">seq_align</span>(string s1,string s2){</span><br><span class="line"> m=s1.<span class="built_in">size</span>(),n=s2.<span class="built_in">size</span>();</span><br><span class="line"> dp[m+<span class="number">1</span>][n+<span class="number">1</span>];</span><br><span class="line"> <span class="comment">//initial,这里的m和n都是可以取到的</span></span><br><span class="line"> <span class="keyword">for</span> i -> <span class="number">0</span>-m:</span><br><span class="line"> dp[i][<span class="number">0</span>]=i*delta;<span class="comment">//这里的惩罚需要根据题意设置，比如有的题目不需要设置惩罚，delta=0</span></span><br><span class="line"> <span class="keyword">for</span> j -> <span class="number">0</span>-n:</span><br><span class="line"> dp[<span class="number">0</span>][j]=j*delta</span><br><span class="line"> <span class="keyword">for</span> i -> <span class="number">1</span>-m:</span><br><span class="line"> <span class="keyword">for</span> j -> <span class="number">1</span>-n:</span><br><span class="line"> dp[i][j]=s1[i<span class="number">-1</span>]==s2[j<span class="number">-1</span>]?dp[i<span class="number">-1</span>][j<span class="number">-1</span>]:<span class="built_in">min</span>(dp[i<span class="number">-1</span>][j],dp[i][j<span class="number">-1</span>])+delta;</span><br><span class="line"> <span class="keyword">return</span> dp[m][n] </span><br><span class="line">}</span><br></pre></td></tr></table></figure>
<h2 id="四、节省空间"><a href="#四、节省空间" class="headerlink" title="四、节省空间"></a><font face="黑体" size="5">四、节省空间</font></h2><font face="宋体" size="4">
<p>动态规划的提升一般集中在节省空间，对于这一类问题，我们可以观察到我们每次更新dp[i][…]都依赖前一个状态dp[i-1][…],因此我们可以用一个dp[][2]的数组存储状态，每次更新dp[][1]，之后将dp[][1]赋值给dp[][0]。</p>
<figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br></pre></td><td class="code"><pre><span class="line"><span class="built_in">seq_align</span>(string s1,string s2){</span><br><span class="line"> m=s1.<span class="built_in">size</span>(),n=s2.<span class="built_in">size</span>();</span><br><span class="line"> dp[n+<span class="number">1</span>][<span class="number">2</span>];</span><br><span class="line"> <span class="comment">//initial,这里的m和n都是可以取到的</span></span><br><span class="line"> <span class="keyword">for</span> i -> <span class="number">0</span>-n:</span><br><span class="line"> dp[i][<span class="number">0</span>]=i*delta;<span class="comment">//这里的惩罚需要根据题意设置，比如有的题目不需要设置惩罚，delta=0</span></span><br><span class="line"> <span class="keyword">for</span> i -> <span class="number">1</span>-m:</span><br><span class="line"> dp[<span class="number">0</span>][<span class="number">1</span>]=i*delta;</span><br><span class="line"> <span class="keyword">for</span> j -> <span class="number">1</span>-n:</span><br><span class="line"> dp[j][<span class="number">1</span>]=s1[i<span class="number">-1</span>]==s2[j<span class="number">-1</span>]?dp[j<span class="number">-1</span>][<span class="number">0</span>]:<span class="built_in">min</span>(dp[j<span class="number">-1</span>][<span class="number">1</span>],dp[j][<span class="number">0</span>])+delta;</span><br><span class="line"> <span class="keyword">return</span> dp[n][<span class="number">1</span>] </span><br><span class="line">}</span><br></pre></td></tr></table></figure>
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<h1 id="论文题目"><a href="#论文题目" class="headerlink" title="论文题目"></a><center><font face="黑体" size="7">论文题目</font></center></h1><p><font face="宋体" size="4">中文摘要</font></p>
<p><font face="黑体" size="4"><strong>摘要</strong>：</font><font face="宋体" size="4"> 摘要内容，150-300字。</font></p>
<p><font face="宋体" size="4">中文关键词</font></p>
<p><font face="黑体" size="4">关键词：</font><font face="宋体" size="4">关键词1，关键词2，关键词3。</font></p>
<h2 id="一、引言"><a href="#一、引言" class="headerlink" title="一、引言"></a><font face="黑体" size="5">一、引言</font></h2><font face="宋体" size="4">
<h2 id="二、相关研究"><a href="#二、相关研究" class="headerlink" title="二、相关研究"></a><font face="黑体" size="5">二、相关研究</font></h2><font face="宋体" size="4">
<h3 id="一-研究方法"><a href="#一-研究方法" class="headerlink" title="(一) 研究方法"></a><font face="黑体" size="5">(一) 研究方法</font></h3><font face="宋体" size="4">
<h3 id="二-实验结果"><a href="#二-实验结果" class="headerlink" title="(二) 实验结果"></a><font face="黑体" size="5">(二) 实验结果</font></h3><font face="宋体" size="4">
<h2 id="三、讨论"><a href="#三、讨论" class="headerlink" title="三、讨论"></a><font face="黑体" size="5">三、讨论</font></h2><font face="宋体" size="4">
<h2 id="四、结论"><a href="#四、结论" class="headerlink" title="四、结论"></a><font face="黑体" size="5">四、结论</font></h2><font face="宋体" size="4">
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[1] 参考文献1<br>
[2] 参考文献2<br>
[3] 参考文献3<br>
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<h2 id="Quick-Start"><a href="#Quick-Start" class="headerlink" title="Quick Start"></a>Quick Start</h2><h3 id="Create-a-new-post"><a href="#Create-a-new-post" class="headerlink" title="Create a new post"></a>Create a new post</h3><figure class="highlight bash"><table><tr><td class="gutter"><pre><span class="line">1</span><br></pre></td><td class="code"><pre><span class="line">$ hexo new <span class="string">"My New Post"</span></span><br></pre></td></tr></table></figure>
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