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表格内矩阵的垂直居中

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表格内矩阵的垂直居中

我试图将表格内容垂直和水平居中。表格包含Bmatrix和bmatrix元素。只有第一列完全居中。第二列仅水平居中，尽管我使用设置了两列m{...}：

\begin{table}[h]
\centering
\begin{tabular}{|>{\centering\arraybackslash}m{5cm}|>{\centering\arraybackslash}m{5cm}|}
\hline
\textbf{Set Representation} & \textbf{Binary Representation} \\ \hline
$U =  \begin{Bmatrix} 1 & 2 & 3 & 4 \end{Bmatrix}$          & $U = \begin{bmatrix} 1 & 1 & 1 & 1 \end{bmatrix} = 15$                \\[.5cm] \hline
$S_1 = \begin{Bmatrix} 1 & 2 \end{Bmatrix}$                & $S_1 = \begin{bmatrix} 0 & 0 & 1 & 1 \end{bmatrix} = 3$              \\[.5cm] \hline
$S_2 = \begin{Bmatrix} 1 \end{Bmatrix}$                  &  $S_2 = \begin{bmatrix} 0 & 0 & 0 & 1 \end{bmatrix} = 1$              \\[.5cm] \hline
$S_3 = \begin{Bmatrix} 1 & 3 \end{Bmatrix}$             & $S_3 = \begin{bmatrix} 0 & 1 & 0 & 1 \end{bmatrix} = 5$              \\[.5cm] \hline
$S_4 = \begin{Bmatrix} 4 \end{Bmatrix}$                  & $S_4 = \begin{bmatrix} 1 & 0 & 0 & 0 \end{bmatrix} = 8$              \\[.5cm] \hline
\end{tabular}%\quad}
\caption{\textbf{Conversion from set to binary representation}}\label{table:setToBin}
\end{table}

结果如下： 右列未垂直居中。

答案1

借助以下makecell包：

在此处输入图片描述

\documentclass{article}
\usepackage{amsmath}
\usepackage{makecell}
\usepackage[skip=1ex, font=bf]{caption}

\begin{document}
    \begin{table}[htb]
\[\setcellgapes{5pt}\makegapedcells % <--- for vertical space around cells contents
\begin{array}{|c|c|}
\hline
\textbf{Set Representation} & \textbf{Binary Representation}      \\
    \hline
U =  \begin{Bmatrix} 1 & 2 & 3 & 4 \end{Bmatrix}
    &   U = \begin{bmatrix} 1 & 1 & 1 & 1 \end{bmatrix} = 15    \\
    \hline
S_1 = \begin{Bmatrix} 1 & 2 \end{Bmatrix}
    &   S_1 = \begin{bmatrix} 0 & 0 & 1 & 1 \end{bmatrix} = 3   \\
    \hline
S_2 = \begin{Bmatrix} 1 \end{Bmatrix}
    &   S_2 = \begin{bmatrix} 0 & 0 & 0 & 1 \end{bmatrix} = 1    \\
    \hline
S_3 = \begin{Bmatrix} 1 & 3 \end{Bmatrix}
    &   S_3 = \begin{bmatrix} 0 & 1 & 0 & 1 \end{bmatrix} = 5   \\
    \hline
S_4 = \begin{Bmatrix} 4 \end{Bmatrix}
    &   S_4 = \begin{bmatrix} 1 & 0 & 0 & 0 \end{bmatrix} = 8   \\
    \hline
\end{array}%
\]
    \caption{Conversion from set to binary representation}
\label{table:setToBin}
    \end{table}
\end{document}

答案2

根据答案：这里您可以使用@StevenB.Segletes包\belowbaseline的命令stackengine：

\documentclass[12pt,a4paper]{article}       

\usepackage{amsmath}
\usepackage{array}
\usepackage{stackengine}
\begin{document}

\begin{table}[h]
\centering
\begin{tabular}{|>{\centering\arraybackslash}m{5cm}|>{\centering\arraybackslash}m{5cm}|}
\hline
\textbf{Set Representation} & \textbf{Binary Representation}\\ \hline
$U =  \begin{Bmatrix} 1 & 2 & 3 & 4 \end{Bmatrix}$        & \belowbaseline[-3pt]{$U = \begin{bmatrix} 1 & 1 & 1 & 1 \end{bmatrix} = 15$ }              \\[.5cm] \hline
$S_1 = \begin{Bmatrix} 1 & 2 \end{Bmatrix}$           & \belowbaseline[-3pt]{$S_1 = \begin{bmatrix} 0 & 0 & 1 & 1 \end{bmatrix} = 3$ }             \\[.5cm] \hline
$S_2 = \begin{Bmatrix} 1 \end{Bmatrix}$                  &  \belowbaseline[-3pt]{$S_2 = \begin{bmatrix} 0 & 0 & 0 & 1 \end{bmatrix} = 1$   }           \\[.5cm] \hline
$S_3 = \begin{Bmatrix} 1 & 3 \end{Bmatrix}$             &\belowbaseline[-3pt]{ $S_3 = \begin{bmatrix} 0 & 1 & 0 & 1 \end{bmatrix} = 5$  }           \\[.5cm] \hline
$S_4 = \begin{Bmatrix} 4 \end{Bmatrix}$                  & \belowbaseline[-3pt]{$S_4 = \begin{bmatrix} 1 & 0 & 0 & 0 \end{bmatrix} = 8$    }          \\[.5cm] \hline
\end{tabular}%\quad}
\caption{\textbf{Conversion from set to binary representation}}\label{table:setToBin}
\end{table}

\end{document}

在此处输入图片描述

答案3

我只能使用超大分隔符来重现矩阵

\linespread{1.667}

和

\usepackage[doublespacing]{setspace}

分隔符具有正确的大小。

您可以在第一列插入一个支柱；此外，w{c}{5cm}如果您想要固定大小，我建议使用，而不是说明m符，但我不认为这合理，因此我建议使用不同的实现booktabs。

\documentclass{article}
\usepackage{amsmath,array}
\usepackage{booktabs} % for the second table
\usepackage{caption}
\usepackage[doublespacing]{setspace}

\usepackage{lipsum} % for mock text

\captionsetup{font=bf}

\newcommand{\bs}{\vphantom{$\Big|$}}

\begin{document}

\lipsum[3]

\begin{table}[!htp]
\centering
\begin{tabular}{| >{\bs}w{c}{5cm} | w{c}{5cm} |}
\hline
\textbf{Set Representation} & \textbf{Binary Representation} \\
\hline
$U   = \begin{Bmatrix} 1 & 2 & 3 & 4 \end{Bmatrix}$ &
  $U   = \begin{bmatrix} 1 & 1 & 1 & 1 \end{bmatrix} = 15$ \\
\hline
$S_1 = \begin{Bmatrix} 1 & 2 \end{Bmatrix}$ &
  $S_1 = \begin{bmatrix} 0 & 0 & 1 & 1 \end{bmatrix} = 3$ \\
\hline
$S_2 = \begin{Bmatrix} 1 \end{Bmatrix}$ &
  $S_2 = \begin{bmatrix} 0 & 0 & 0 & 1 \end{bmatrix} = 1$ \\
\hline
$S_3 = \begin{Bmatrix} 1 & 3 \end{Bmatrix}$ &
  $S_3 = \begin{bmatrix} 0 & 1 & 0 & 1 \end{bmatrix} = 5$ \\
\hline
$S_4 = \begin{Bmatrix} 4 \end{Bmatrix}$ &
  $S_4 = \begin{bmatrix} 1 & 0 & 0 & 0 \end{bmatrix} = 8$ \\
\hline
\end{tabular}

\caption{Conversion from set to binary representation}\label{table:setToBin}

\end{table}

\begin{table}[!htp]
\centering
\begin{tabular}{ cc }
\toprule
\textbf{Set Representation} & \textbf{Binary Representation} \\
\midrule
$U   = \begin{Bmatrix} 1 & 2 & 3 & 4 \end{Bmatrix}$ &
  $U   = \begin{bmatrix} 1 & 1 & 1 & 1 \end{bmatrix} = 15$ \\
\addlinespace
$S_1 = \begin{Bmatrix} 1 & 2 \end{Bmatrix}$ &
  $S_1 = \begin{bmatrix} 0 & 0 & 1 & 1 \end{bmatrix} = 3$ \\
\addlinespace
$S_2 = \begin{Bmatrix} 1 \end{Bmatrix}$ &
  $S_2 = \begin{bmatrix} 0 & 0 & 0 & 1 \end{bmatrix} = 1$ \\
\addlinespace
$S_3 = \begin{Bmatrix} 1 & 3 \end{Bmatrix}$ &
  $S_3 = \begin{bmatrix} 0 & 1 & 0 & 1 \end{bmatrix} = 5$ \\
\addlinespace
$S_4 = \begin{Bmatrix} 4 \end{Bmatrix}$ &
  $S_4 = \begin{bmatrix} 1 & 0 & 0 & 0 \end{bmatrix} = 8$ \\
\bottomrule
\end{tabular}

\caption{Conversion from set to binary representation}\label{table:setToBin2}

\end{table}

\end{document}

\textbf最后说明：的主体中不应有\caption；这种格式决定最好在caption包的全局基础上进行。

在此处输入图片描述

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