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重复上一帧表格的相同标签

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重复上一帧表格的相同标签

我正在制作演示文稿，想重复使用上一帧中带有相同标签的同一张表。我尝试了很多方法，但都没有成功。附件中您可以找到框架和代码。非常感谢。

框架

\section{Conceptual Framework}
\begin{frame}
\frametitle{2. Conceptual Framework}

\begin{table}
    \setlength{\extrarowheight}{2pt}
    \begin{tabular}{cc|c|c|}
      & \multicolumn{1}{c}{} & \multicolumn{2}{c}{Player 2}\\
      & \multicolumn{1}{c}{} & \multicolumn{1}{c}{C}  & \multicolumn{1}{c}{D} \\\cline{3-4}
      \multirow{2}*{Player 1}  & C & $b-c, b-c$ & $-c, b$ \\\cline{3-4}
      & D & $b, -c$ & $0, 0$ \\\cline{3-4}
    \end{tabular} 
\caption{Prisoners' Dilemma}
\label{PD}
    \end{table}

\begin{itemize}
\item{$b$ $>$ $c$ $>$ $0$}
\item{Given $c$ $>$ $0$, it is a dominant strategy to play D}
\item{Hence the NE of this game is (D,D) with payoff ($0$,$0$)}
\item{Since (C,C) gives $b$ - $c$ $>$ $0$, the equilibrium outcome (D,D) is inefficient}
\end{itemize}   
\end{frame}

\begin{frame}
\frametitle{2. Conceptual Framework}
\begin{table}
    \setlength{\extrarowheight}{2pt}
    \begin{tabular}{cc|c|c|}
      & \multicolumn{1}{c}{} & \multicolumn{2}{c}{Player 2}\\
      & \multicolumn{1}{c}{} & \multicolumn{1}{c}{C}  & \multicolumn{1}{c}{D} \\\cline{3-4}
      \multirow{2}*{Player 1}  & C & $b-c, b-c$ & $-c, b$ \\\cline{3-4}
      & D & $b, -c$ & $0, 0$ \\\cline{3-4}
    \end{tabular}
    \renewcommand\thetable{\ref{PD}} 
    \caption{Prisoners' Dilemma}    
    \addtocounter{table}{-1}    

    \end{table}
\end{frame}

答案1

在下面的 MWE 中，我已将行移到\addtocounter{table}{-1}第二帧中的标题之前。这将导致两个帧上的两个表格都使用相同的数字进行编号。

    \documentclass{beamer}
    \usepackage{array}
    \usepackage{multirow}

    \setbeamertemplate{caption}[numbered]

    \begin{document}

    \section{Conceptual Framework}
    \begin{frame}
    \frametitle{2. Conceptual Framework}

    \begin{table}
        \setlength{\extrarowheight}{2pt}
        \begin{tabular}{cc|c|c|}
          & \multicolumn{1}{c}{} & \multicolumn{2}{c}{Player 2}\\
          & \multicolumn{1}{c}{} & \multicolumn{1}{c}{C}  & \multicolumn{1}{c}{D} \\\cline{3-4}
          \multirow{2}*{Player 1}  & C & $b-c, b-c$ & $-c, b$ \\\cline{3-4}
          & D & $b, -c$ & $0, 0$ \\\cline{3-4}
        \end{tabular} 
    \caption{Prisoners' Dilemma}
    \label{PD}
        \end{table}

    \begin{itemize}
    \item $b$ $>$ $c$ $>$ $0$
    \item Given $c$ $>$ $0$, it is a dominant strategy to play D
    \item Hence the NE of this game is (D,D) with payoff ($0$,$0$)
    \item Since (C,C) gives $b$ - $c$ $>$ $0$, the equilibrium outcome (D,D) is inefficient
    \end{itemize}   
    \end{frame}

    \begin{frame}
    \frametitle{2. Conceptual Framework}
    \begin{table}

        \setlength{\extrarowheight}{2pt}
        \begin{tabular}{cc|c|c|}
          & \multicolumn{1}{c}{} & \multicolumn{2}{c}{Player 2}\\
          & \multicolumn{1}{c}{} & \multicolumn{1}{c}{C}  & \multicolumn{1}{c}{D} \\\cline{3-4}
          \multirow{2}*{Player 1}  & C & $b-c, b-c$ & $-c, b$ \\\cline{3-4}
          & D & $b, -c$ & $0, 0$ \\\cline{3-4}
        \end{tabular}
        \addtocounter{table}{-1} %<-------------------------- 
        \caption{Prisoners' Dilemma}      
        \end{table}
    \end{frame}

    \end{document}

或者，您可能对使用覆盖在第二张幻灯片上重复表格感兴趣。这样，您只需指定一次表格代码（如果您想更改表格内容，这将更容易），并且如果您从第一张幻灯片切换到第二张幻灯片，表格将不会“跳跃”。

\documentclass{beamer}
\usepackage{array}
\usepackage{multirow}

\setbeamertemplate{caption}[numbered]

\begin{document}

\section{Conceptual Framework}
\begin{frame}
\frametitle{2. Conceptual Framework}

\onslide<1-2>{
\begin{table}
    \setlength{\extrarowheight}{2pt}
    \begin{tabular}{cc|c|c|}
      & \multicolumn{1}{c}{} & \multicolumn{2}{c}{Player 2}\\
      & \multicolumn{1}{c}{} & \multicolumn{1}{c}{C}  & \multicolumn{1}{c}{D} \\\cline{3-4}
      \multirow{2}*{Player 1}  & C & $b-c, b-c$ & $-c, b$ \\\cline{3-4}
      & D & $b, -c$ & $0, 0$ \\\cline{3-4}
    \end{tabular} 
\caption{Prisoners' Dilemma}
\label{PD}
    \end{table}}

\onslide<1-1>{
\begin{itemize}
\item $b$ $>$ $c$ $>$ $0$
\item Given $c$ $>$ $0$, it is a dominant strategy to play D
\item Hence the NE of this game is (D,D) with payoff ($0$,$0$)
\item Since (C,C) gives $b$ - $c$ $>$ $0$, the equilibrium outcome (D,D) is inefficient
\end{itemize} }  
\end{frame}


\end{document}

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