如何在同一行中对齐不同高度的子图

Question

以下代码：

\documentclass[journal,twocolumn,10pt]{IEEEtran}
\usepackage[compress]{cite}
\usepackage{times} % Font Times New Roman throughout the document
\usepackage{subfigure}
\usepackage{epsfig}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{array}
\usepackage{algorithm}
\usepackage{algorithmic}
\usepackage{tikz,pgfplots}
\usepackage{adjustbox}
\pgfplotsset{compat=1.14}
\usepgfplotslibrary{patchplots}
\usepgfplotslibrary{fillbetween}
\usepackage{caption}
\usetikzlibrary{arrows.meta}
\usepackage{graphicx}
\usepackage{float}
\usepackage{mathtools}


\begin{document}

\begin{figure}[t]
\centering
\captionsetup{justification=centering}
\resizebox {0.8 \columnwidth}{!}{
\subfigure[Line]{\centering
\begin{tikzpicture}
\draw (0,0) node {$S= \begin{bmatrix}
     0  &   0  &   0  &   0\\
     1  &   0  &   0  &   0\\
     0  &   1  &   0  &   0\\
     0  &   0  &   1  &   0\\
     1  &   0  &   1  &   0\\
     0  &   0  &   0  &   1\\
     1  &   0  &   0  &   1\\
     0  &   1  &   0  &   1\\
    \end{bmatrix}$};
\end{tikzpicture} }
\subfigure[Square]{\centering
\begin{tikzpicture}
\draw (0,0) node {$S= \begin{bmatrix}
     0  &   0  &   0  &   0\\
     1  &   0  &   0  &   0\\
     0  &   1  &   0  &   0\\
     0  &   0  &   1  &   0\\
     1  &   0  &   1  &   0\\
     0  &   0  &   0  &   1\\
     0  &   1  &   0  &   1\\
    \end{bmatrix}$};
\end{tikzpicture}}
}\\
\resizebox {0.8 \columnwidth}{!}{
\subfigure[Full]{\centering
\begin{tikzpicture}
\draw (0,0) node {$S= \begin{bmatrix}
     0  &   0  &   0  &   0\\
     1  &   0  &   0  &   0\\
     0  &   1  &   0  &   0\\
     0  &   0  &   1  &   0\\
     0  &   0  &   0  &   1\\
\end{bmatrix}$};
\end{tikzpicture}}
\subfigure[Star]{\centering
\begin{tikzpicture}
\draw (0,0) node {  $S= \begin{bmatrix}
     0  &   0  &   0  &   0\\
     1  &   0  &   0  &   0\\
     0  &   1  &   0  &   0\\
     0  &   0  &   1  &   0\\
     0  &   1  &   1  &   0\\
     0  &   0  &   0  &   1\\
     0  &   1  &   0  &   1\\
     0  &   0  &   1  &   1\\
     0  &   1  &   1  &   1\\
    \end{bmatrix}$};
\end{tikzpicture}}}
\caption{Action rule matrices for network graphs of figure \ref{NETEXAMPLE}}
\label{ACTIVATIONFIG}
\end{figure}

\end{document}

产生以下图片：

水平对齐矩阵，将同一行中的标题设置为对齐。我想对齐同一行的矩阵，使其具有S=相同的高度（两行中对齐的 S）。

问候

Answer 1

• 问题有了新内容，所以答案也变了……
• 像这样（我仍然不知道，你喜欢获得什么，现在我猜你喜欢S在方程式中对齐子标题）：

    \documentclass[journal,twocolumn,10pt]{IEEEtran}
    \usepackage{caption, subcaption}
    \usepackage{amsmath}
    \usepackage{array}

    \usepackage{lipsum}

    \begin{document}
    \lipsum[11]
    \begin{figure}[htb]
    \centering
    \captionsetup{justification=centering}
    \begin{tabular}{*{2}{m{\dimexpr0.45\linewidth-2\tabcolsep}}}
    \[
    S= \begin{bmatrix}
         0  &   0  &   0  &   0\\
         1  &   0  &   0  &   0\\
         0  &   1  &   0  &   0\\
         0  &   0  &   1  &   0\\
         1  &   0  &   1  &   0\\
         0  &   0  &   0  &   1\\
         1  &   0  &   0  &   1\\
         0  &   1  &   0  &   1\\
        \end{bmatrix}
    \]
    &
    \[
    S= \begin{bmatrix}
         0  &   0  &   0  &   0\\
         1  &   0  &   0  &   0\\
         0  &   1  &   0  &   0\\
         0  &   0  &   1  &   0\\
         1  &   0  &   1  &   0\\
         0  &   0  &   0  &   1\\
         0  &   1  &   0  &   1\\
        \end{bmatrix}
    \]                      \\
    \subcaption{Line}
    &
    \subcaption{Square}     \\
    \[
    S= \begin{bmatrix}
         0  &   0  &   0  &   0\\
         1  &   0  &   0  &   0\\
         0  &   1  &   0  &   0\\
         0  &   0  &   1  &   0\\
         0  &   0  &   0  &   1\\
        \end{bmatrix}
    \]
    &
    \[
    S= \begin{bmatrix}
         0  &   0  &   0  &   0\\
         1  &   0  &   0  &   0\\
         0  &   1  &   0  &   0\\
         0  &   0  &   1  &   0\\
         0  &   1  &   1  &   0\\
         0  &   0  &   0  &   1\\
         0  &   1  &   0  &   1\\
         0  &   0  &   1  &   1\\
         0  &   1  &   1  &   1\\
        \end{bmatrix}
    \]                      \\
    \subcaption{Full}
    &
    \subcaption{Star}
    \end{tabular}
    \caption{Action rule matrices for network graphs of figure \ref{NETEXAMPLE}}
    \label{ACTIVATIONFIG}
    \end{figure}
    \end{document}

Answer 2

• 软件包subfigure已被取代subfig

• 但更好的用法是更糟糕subcaption：

  \documentclass[journal,twocolumn,10pt]{IEEEtran}
  \usepackage{caption}
  \usepackage{tikz}
  \usetikzlibrary{arrows.meta}
  \usepackage{graphicx}
  \usepackage{subcaption}
  \usepackage{amsmath}
  
  \begin{document}
  
  \begin{figure}[htb]
  \centering
  \captionsetup{justification=centering}
  \begin{subfigure}[t]{0.45\linewidth} % with option "t" subfigure is aligned to its top
  \[
  S= \begin{bmatrix}
       0  &   0  &   0  &   0\\
       1  &   0  &   0  &   0\\
       0  &   1  &   0  &   0\\
       0  &   0  &   1  &   0\\
       1  &   0  &   1  &   0\\
       0  &   0  &   0  &   1\\
       1  &   0  &   0  &   1\\
       0  &   1  &   0  &   1\\
      \end{bmatrix}
  \]
  \caption{Square}
  \end{subfigure}\hfil
  \begin{subfigure}[t]{0.45\linewidth}
  \[
  S= \begin{bmatrix}
       0  &   0  &   0  &   0\\
       1  &   0  &   0  &   0\\
       0  &   1  &   0  &   0\\
       0  &   0  &   1  &   0\\
       1  &   0  &   1  &   0\\
       0  &   0  &   0  &   1\\
       0  &   1  &   0  &   1\\
      \end{bmatrix}
  \]
  \caption{Square}
  \end{subfigure}
  
  \begin{subfigure}[t]{0.45\linewidth}
  \[
  S= \begin{bmatrix}
       0  &   0  &   0  &   0\\
       1  &   0  &   0  &   0\\
       0  &   1  &   0  &   0\\
       0  &   0  &   1  &   0\\
       0  &   0  &   0  &   1\\
      \end{bmatrix}
  \]
  \caption{Full}
  \end{subfigure}\hfil
  \begin{subfigure}[t]{0.45\linewidth}
  \[
  S= \begin{bmatrix}
       0  &   0  &   0  &   0\\
       1  &   0  &   0  &   0\\
       0  &   1  &   0  &   0\\
       0  &   0  &   1  &   0\\
       0  &   1  &   1  &   0\\
       0  &   0  &   0  &   1\\
       0  &   1  &   0  &   1\\
       0  &   0  &   1  &   1\\
       0  &   1  &   1  &   1\\
      \end{bmatrix}
  \]
  \caption{Star}
  \end{subfigure}
  \caption{Action rule matrices for network graphs of figure \ref{NETEXAMPLE}}
  \label{ACTIVATIONFIG}
  \end{figure}
  
  \end{document}
  

这使：

这就是你要找的吗？

注意：我不明白为什么将矩阵括在tikz节点中。我对它们使用简单的显示数学环境。

编辑（1）： 如果您喜欢顶部对齐前两个矩阵，然后在第二对中删除选项[t]或添加选项[b]：

\documentclass[journal,twocolumn,10pt]{IEEEtran}
\usepackage{caption}
\usepackage{tikz}
\usetikzlibrary{arrows.meta}
\usepackage{graphicx}
\usepackage{subcaption}
\usepackage{amsmath}

\begin{document}

\begin{figure}[htb]
\centering
\captionsetup{justification=centering}
\begin{subfigure}[t]{0.45\linewidth}% with option "t" subfigure is aligned t its top
\[
S= \begin{bmatrix}
     0  &   0  &   0  &   0\\
     1  &   0  &   0  &   0\\
     0  &   1  &   0  &   0\\
     0  &   0  &   1  &   0\\
     1  &   0  &   1  &   0\\
     0  &   0  &   0  &   1\\
     1  &   0  &   0  &   1\\
     0  &   1  &   0  &   1\\
    \end{bmatrix}
\]
\caption{Square}
\end{subfigure}\hfil
\begin{subfigure}[t]{0.45\linewidth}
\[
S= \begin{bmatrix}
     0  &   0  &   0  &   0\\
     1  &   0  &   0  &   0\\
     0  &   1  &   0  &   0\\
     0  &   0  &   1  &   0\\
     1  &   0  &   1  &   0\\
     0  &   0  &   0  &   1\\
     0  &   1  &   0  &   1\\
    \end{bmatrix}
\]
\caption{Square}
\end{subfigure}

\begin{subfigure}{0.45\linewidth}% without option subfigure is alignet t its center
\[
S= \begin{bmatrix}
     0  &   0  &   0  &   0\\
     1  &   0  &   0  &   0\\
     0  &   1  &   0  &   0\\
     0  &   0  &   1  &   0\\
     0  &   0  &   0  &   1\\
    \end{bmatrix}
\]
\caption{Full}
\end{subfigure}\hfil
\begin{subfigure}{0.45\linewidth}
\[
S= \begin{bmatrix}
     0  &   0  &   0  &   0\\
     1  &   0  &   0  &   0\\
     0  &   1  &   0  &   0\\
     0  &   0  &   1  &   0\\
     0  &   1  &   1  &   0\\
     0  &   0  &   0  &   1\\
     0  &   1  &   0  &   1\\
     0  &   0  &   1  &   1\\
     0  &   1  &   1  &   1\\
    \end{bmatrix}
\]
\caption{Star}
\end{subfigure}

\begin{subfigure}[b]{0.45\linewidth}% with option "b" subfigure is aligned to its bottom
\[
S= \begin{bmatrix}
     0  &   0  &   0  &   0\\
     1  &   0  &   0  &   0\\
     0  &   1  &   0  &   0\\
     0  &   0  &   1  &   0\\
     0  &   0  &   0  &   1\\
    \end{bmatrix}
\]
\caption{Full}
\end{subfigure}\hfil
\begin{subfigure}[b]{0.45\linewidth}
\[
S= \begin{bmatrix}
     0  &   0  &   0  &   0\\
     1  &   0  &   0  &   0\\
     0  &   1  &   0  &   0\\
     0  &   0  &   1  &   0\\
     0  &   1  &   1  &   0\\
     0  &   0  &   0  &   1\\
     0  &   1  &   0  &   1\\
     0  &   0  &   1  &   1\\
     0  &   1  &   1  &   1\\
    \end{bmatrix}
\]
\caption{Star}
\end{subfigure}
\caption{Action rule matrices for network graphs of figure \ref{NETEXAMPLE}}
\label{ACTIVATIONFIG}
\end{figure}

\end{document}

编辑（2） 考虑土拨鼠下面的评论，前两个矩阵也可以按如下方式对齐：

\documentclass[journal,twocolumn,10pt]{IEEEtran}
\usepackage{caption}
\usepackage{tikz}
\usetikzlibrary{arrows.meta}
\usepackage{graphicx}
\usepackage{subcaption}
\usepackage{amsmath}

\begin{document}

\begin{figure}[t]
\centering
\captionsetup{justification=centering}
\begin{subfigure}[t]{0.45\linewidth}
\[
S= \begin{bmatrix}
     0  &   0  &   0  &   0\\
     1  &   0  &   0  &   0\\
     0  &   1  &   0  &   0\\
     0  &   0  &   1  &   0\\
     1  &   0  &   1  &   0\\
     0  &   0  &   0  &   1\\
     1  &   0  &   0  &   1\\
     0  &   1  &   0  &   1\\
    \end{bmatrix}
\]
\caption{Square}
\end{subfigure}\hfil
\begin{subfigure}[t]{0.45\linewidth}
\begin{gather*}  % enable stacking phantom matrices below main
S= \begin{bmatrix}
     0  &   0  &   0  &   0\\
     1  &   0  &   0  &   0\\
     0  &   1  &   0  &   0\\
     0  &   0  &   1  &   0\\
     1  &   0  &   1  &   0\\
     0  &   0  &   0  &   1\\
     0  &   1  &   0  &   1\\
    \end{bmatrix}\\       
    \vphantom{\begin{bmatrix}
     0\\
    \end{bmatrix}}
\end{gather*}
\caption{Square}
\end{subfigure}

\end{figure}
\end{document}