我想用一些迭代控制流來簡化下面的LaTeX代碼。乳膠中的迭代
\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}
我得知forloop package提供for
循環。但我不確定如何將其應用於我的案例?或者其他方法不通過forloop?
如果我也想簡單另一個類似的情況下,其中唯一的區別是該目錄不爲1,2運行,至23,但在一些任意的順序如3,2,6,9 ,...,甚至是一系列字符串,例如dira,dirc,dird,dirb,....我如何將LaTeX代碼轉換爲循環呢?
請檢查是否接受的答案仍然是最好的選擇。最高票數的答案几乎是票數的三倍!謝謝。 – Sebastian 2013-11-08 15:48:36