☰

如何使表格数据顶部对齐

tag-icon • tag-icon tables tabularx tabulary
如何使表格数据顶部对齐

我有这个表格数据，我想让数据与顶部对齐，而不是像这里显示的那样居中。

%#%
\begin{frame}{frame 1}
    \begin{tabular}[t]{l | l | l}
    \parbox{0.3\textwidth}{
    \footnotesize
    
    \tcboxmath[colframe=nus-blue,colback=white,boxrule=0.5pt,arc=4pt,left=6pt,right=6pt,top=6pt,bottom=6pt,boxsep=0pt,width=\@tempdima]{C_{ij}^{*}}
        \begin{align*}
            \textcolor{nus-orange}{C_{33}^0} + \textcolor{nus-orange}{C_{33}^1} &= C_{33}^* = \rho V_p^2,\\[1ex]
            \textcolor{nus-orange}{C_{44}^0} + \textcolor{nus-orange}{C_{44}^1} &= C_{44}^* = \rho V_s^2.
        \end{align*}
    }
    &
    \parbox{0.3\textwidth}{
    \footnotesize
    
    \tcboxmath[colframe=nus-blue,colback=white,boxrule=0.5pt,arc=4pt,left=6pt,right=6pt,top=6pt,bottom=6pt,boxsep=0pt,width=\@tempdima]{C_{ij}^{0}}
        \begin{align*}
            \textcolor{nus-orange}{C_{11}^0} = \textcolor{nus-orange}{C_{33}^0} &= \lambda + 2 \mu,\\[1ex]
            \textcolor{nus-orange}{C_{66}^0} = \textcolor{nus-orange}{C_{44}^0} &= \mu.
        \end{align*}
        \footnotesize
        \boxed{\lambda = K^{HS+} - \frac{2}{3} \mu^{HS+},}
        \boxed{\mu = \mu^{HS+}.}
    }
    &
    \parbox{0.3\textwidth}{
    \footnotesize
    
    \tcboxmath[colframe=nus-blue,colback=white,boxrule=0.5pt,arc=4pt,left=6pt,right=6pt,top=6pt,bottom=6pt,boxsep=0pt,width=\@tempdima]{C_{ij}^{1}}
        \begin{align*}
            \textcolor{nus-orange}{C_{33}^{1}} &= -\frac{(\lambda + 2 \mu)^2}{\mu} \textcolor{nus-blue}{d_c^{33}} U_3, \\[1ex]
            \textcolor{nus-orange}{C_{44}^{1}} &= -\mu \textcolor{nus-blue}{d_c^{44}} U_1, \\[1ex]
            U_1 &= \frac{16(\lambda + 2 \mu)}{3(3 \lambda + 4 \mu)},\\[1ex]
            U_3 &= \frac{4(\lambda + 2 \mu)}{3(\lambda + \mu)(1 + K)},\\[1ex]
            K   &= \frac{K_f(\lambda + 2 \mu)}{\pi \textcolor{nus-blue}{\alpha} \mu(\lambda + \mu)}.
        \end{align*}
    }
    \end{tabular}

\end{frame}

答案1

像这样？

由于您没有提供 MWE，我发明了颜色定义。

\documentclass{beamer}
\usepackage{tabularx}
\usepackage[most]{tcolorbox}
\colorlet{nus-orange}{yellow!50!red}
\colorlet{nus-blue}{white!50!blue}
\tcbset{colframe=nus-blue,colback=white,boxrule=0.5pt,arc=4pt,left=6pt,right=6pt,top=6pt,bottom=6pt,boxsep=0pt,width=\@tempdima}

\begin{document}
\begin{frame}{frame 1}\scriptsize
    \begin{tabularx}{\linewidth}{@{}X|X|X@{}}
        \tcboxmath{C_{ij}^{*}}\medskip
        
        $\begin{aligned}
            \textcolor{nus-orange}{C_{33}^0} + \textcolor{nus-orange}{C_{33}^1} &= C_{33}^* = \rho V_p^2,\\[1ex]
            \textcolor{nus-orange}{C_{44}^0} + \textcolor{nus-orange}{C_{44}^1} &= C_{44}^* = \rho V_s^2.
        \end{aligned}$
    &
    \tcboxmath{C_{ij}^{0}}\medskip
    
        $\begin{aligned}
            \textcolor{nus-orange}{C_{11}^0} = \textcolor{nus-orange}{C_{33}^0} &= \lambda + 2 \mu,\\[1ex]
            \textcolor{nus-orange}{C_{66}^0} = \textcolor{nus-orange}{C_{44}^0} &= \mu.
        \end{aligned}$
        \boxed{\lambda = K^{HS+} - \frac{2}{3} \mu^{HS+},}
        \boxed{\mu = \mu^{HS+}.}%
    &
    \tcboxmath{C_{ij}^{1}}\medskip
    
        $\begin{aligned}
            \textcolor{nus-orange}{C_{33}^{1}} &= -\frac{(\lambda + 2 \mu)^2}{\mu} \textcolor{nus-blue}{d_c^{33}} U_3, \\[1ex]
            \textcolor{nus-orange}{C_{44}^{1}} &= -\mu \textcolor{nus-blue}{d_c^{44}} U_1, \\[1ex]
            U_1 &= \frac{16(\lambda + 2 \mu)}{3(3 \lambda + 4 \mu)},\\[1ex]
            U_3 &= \frac{4(\lambda + 2 \mu)}{3(\lambda + \mu)(1 + K)},\\[1ex]
            K   &= \frac{K_f(\lambda + 2 \mu)}{\pi \textcolor{nus-blue}{\alpha} \mu(\lambda + \mu)}.
        \end{aligned}$
    \end{tabularx}
\end{frame}
\end{document}

在此处输入图片描述

答案2

立即回答你的问题，

如何使表格数据顶部对齐？

将是：用 替换所有 3 个实例\parbox{0.3\textwidth}。\parbox[t]{0.3\textwidth}但是，这将是一个笨拙的解决方案。将 3 个l类型的列更改为更为优雅p{0.3\textwidth}。

以下解决方案对p-column 方法进行了轻微调整，方法是 (a) 为第二列分配略小的宽度，为第三列分配略大的宽度，以及 (b) 使用aligned环境而不是align*环境。此外，为了确保材料适合文本块的宽度，它使用tabular*环境而不是tabular环境。总体而言，结果与@CarLaTeX 的回答。

在此处输入图片描述

\documentclass{beamer}
\usepackage[most]{tcolorbox}

\begin{document}
\begin{frame}{frame 1}
\tcbset{colframe=blue,colback=white,boxrule=0.5pt,arc=4pt,
               left=6pt,right=6pt,top=6pt,bottom=6pt,boxsep=0pt,
               width=\@tempdima}
               
\footnotesize % a single '\footnotesize' directive suffices

\setlength\tabcolsep{0pt}
\begin{tabular*}{\textwidth}{@{\extracolsep{\fill}} 
    p{0.3\textwidth} p{0.28\textwidth} p{0.32\textwidth}}

    %% column 1 
    \tcboxmath{C_{ij}^{*}}
               
    \medskip
    $\begin{aligned}
     \textcolor{orange}{C_{33}^0} + \textcolor{orange}{C_{33}^1} 
         &= C_{33}^* = \rho V_p^2, \\[2ex]
     \textcolor{orange}{C_{44}^0} + \textcolor{orange}{C_{44}^1} 
         &= C_{44}^* = \rho V_s^2.
    \end{aligned}$

    &
    
    %% column 2 
    \tcboxmath{C_{ij}^{0}}

    \medskip
    $\begin{aligned}
     \textcolor{orange}{C_{11}^0} = \textcolor{orange}{C_{33}^0} 
         &= \lambda + 2\mu, \\[2ex]
     \textcolor{orange}{C_{66}^0} = \textcolor{orange}{C_{44}^0} 
         &= \mu.
    \end{aligned}$
   
    \medskip
    \boxed{\lambda = K^{HS+} - \tfrac{2}{3} \mu^{HS+},}
   
    \smallskip
    \boxed{\mu = \mu^{HS+}.}
    
    &
    
    %% column 1 
    \tcboxmath{C_{ij}^{1}}

    \medskip
    $\begin{aligned}
     \textcolor{orange}{C_{33}^{1}} 
         %% use "\smash" to perform horizontal alignment
         &= -\smash{\frac{(\lambda + 2\mu)^2}{\mu}} \textcolor{blue}{d_c^{33}} U_3, \\[2ex]
     \textcolor{orange}{C_{44}^{1}} 
         &= -\mu \textcolor{blue}{d_c^{44}} U_1, \\[1ex]
     U_1 &= \frac{16(\lambda + 2\mu)}{3(3 \lambda + 4 \mu)},\\[1ex]
     U_3 &= \frac{4(\lambda + 2\mu)}{3(\lambda + \mu)(1 + K)},\\[1ex]
     K   &= \frac{K_f(\lambda + 2\mu)}{\pi \textcolor{blue}{\alpha} \mu(\lambda + \mu)}.
    \end{aligned}$
\end{tabular*}
\end{frame}
\end{document}

相关内容