我想使用一些迭代控制流程来简化以下 LaTeX 代码。
\begin{sidewaystable}
\caption{A glance of images}
\centering
\begin{tabular}{| c ||c| c| c |c| c|| c |c| c|c|c| }
\hline
\backslashbox{Theme}{Class} &\multicolumn{5}{|c|}{Class 0} & \multicolumn{5}{|c|}{Class 1} \\
\hline
\hline
1 &
\includegraphics[scale=2]{../../results/1/0_1.eps}
&\includegraphics[scale=2]{../../results/1/0_2.eps}
&\includegraphics[scale=2]{../../results/1/0_3.eps}
&\includegraphics[scale=2]{../../results/1/0_4.eps}
&\includegraphics[scale=2]{../../results/1/0_5.eps}
&\includegraphics[scale=2]{../../results/1/1_1.eps}
&\includegraphics[scale=2]{../../results/1/1_2.eps}
&\includegraphics[scale=2]{../../results/1/1_3.eps}
&\includegraphics[scale=2]{../../results/1/1_4.eps}
&\includegraphics[scale=2]{../../results/1/1_5.eps} \\
\hline
... % similarly for 2, 3, ..., 22
\hline
23 &
\includegraphics[scale=2]{../../results/23/0_1.eps}
&\includegraphics[scale=2]{../../results/23/0_2.eps}
&\includegraphics[scale=2]{../../results/23/0_3.eps}
&\includegraphics[scale=2]{../../results/23/0_4.eps}
&\includegraphics[scale=2]{../../results/23/0_5.eps}
&\includegraphics[scale=2]{../../results/23/1_1.eps}
&\includegraphics[scale=2]{../../results/23/1_2.eps}
&\includegraphics[scale=2]{../../results/23/1_3.eps}
&\includegraphics[scale=2]{../../results/23/1_4.eps}
&\includegraphics[scale=2]{../../results/23/1_5.eps} \\
\hline
\end{tabular}
\end{sidewaystable}
我了解到forloop package提供for
循环。但我不知道如何将其应用到我的案例中?或者其他不通过forloop的方法?
如果我还想简单地另一个类似的情况,唯一的区别是目录不是从 1, 2, 到 23 运行,而是以某种任意顺序运行,例如 3, 2, 6, 9,... ,甚至是诸如 dira、dirc、dird、dirb 之类的字符串列表,...那么如何将 LaTeX 代码放入循环中呢?
最佳答案
您可以使用pgf提供的工具pgffor
包。基本语法是:
\foreach \n in {0,...,22}{do something}
值得注意的是,这个 for 循环不限于整数,例如:
\foreach \n in {apples,burgers,cake}{Let's eat \n.\par}
关于latex - LaTeX 中的迭代,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/2561791/