如何对齐表格中矩阵中的负十进制数，就像对齐siunitx表格中的数字一样（我使用的是类 elsarticle 选项：twocolumn）
\documentclass[5p,twocolumn]{elsarticle}
\usepackage{booktabs,tabularx}
\usepackage{amsmath}
\begin{document}
\begin{table}
\caption{Algorithm 3 Illustrative calculation for Numerical Example 2}
\label{tab:Example 2.2}
% \centering
\footnotesize%scriptsize
\setlength\tabcolsep{1pt}
\setlength\arraycolsep{2pt}
\begin{tabular}{ll}
\toprule
\normalfont Algorithm steps
& Numerical computation\\
\midrule
Input & Given\\
& $ \mathbf{U}|_{\theta=0.5} = \begin{bmatrix} -1 & 1 & 0.2298 & 0.5000 & 0.7702 & 0.1250 \end{bmatrix}^T$\\
& $\mathbf{U}'_{\theta}|_{\theta =0.5} = \begin{bmatrix} -2 & 2 & 0.8415 & 1 & -0.8415 & 0.7500 \end{bmatrix}^T$\\
& $ \mathbf{v} = \begin{bmatrix} -2.7063 & 1.0000 & 0.2298 & 0.5000 & 0.7702 & 0.1250 \end{bmatrix}^T$\\
& $ \mathbf{u} = \begin{bmatrix} -0.8905 & 0.3290 & 0.0756 & 0.1645 & 0.2534 & 0.0411 \end{bmatrix}^T$\\
\midrule
Step 1: & Compute $\|\mathbf{U}\|'_{\theta} = 2.4257$\\
Step 3: & Compute $\mathbf{v}'_{\theta} =$ \\
& $\begin {bmatrix}
-4.4257 & 2.0000 & 0.8415 & 1.0000 & -0.8415 & 0.7500
\end{bmatrix}^T$\\
Step 7: & Compute $\|\mathbf{v}'_{\theta}\| = 4.6451$\\
Step 8: & Compute $\mathbf{u}'_{\theta} = $ \\
& $\begin {bmatrix}
-0.0952 & 0.1552 & 0.1613 & 0.0776 & -0.6642 & 0.1839
\end{bmatrix}^T$\\
Step 9: & Compute $ \mathbf{Q}'_{\theta} = $\\
& $\begin{bmatrix}
-0.3390 & 0.3390 & 0.3017 & 0.1695 & -1.1348 & 0.3354\\
0.3390 & 0.2042 & -0.1296 & -0.1021 & 0.3585 & -0.1338\\
0.3017 & -0.1296 & -0.0488 & -0.0648 & 0.0178 & -0.0411\\
0.1695 & -0.1021 & -0.0648 & -0.0511 & 0.1792 & -0.0669\\
-1.1348 & 0.3585 & 0.0187 & 0.1792 & 0.6733 & -0.0386\\
0.3354 & -0.1338 & -0.0411 & -0.0669 & -0.0386 & -0.0303
\end{bmatrix}$\\
Step 10: & Extract $\mathbf{Q}'_{\theta} = \mathbf{Q}'_{\theta}(1,1:6)= $\\
&$\begin{bmatrix} -0.3390 & 0.3390 & 0.3017 & 0.1695 & -1.1348 & 0.3354 \end{bmatrix}^T$\\
Step 11: & $\mathbf{R}'_{\theta} = \begin{bmatrix} 2.4257 \end{bmatrix}$ \\
\midrule
Output & $\mathbf{R}'|_{\theta = 0.5} = \begin{bmatrix} 2.4257 \end{bmatrix}$\\
\midrule
Test & Accuracy of the computations:\\
& $\|(\mathbf{U}^T\mathbf{U})'_{\theta}|_{\theta=0.5} - (\mathbf{R}^T\mathbf{R})'_{\theta}|_{\theta=0.5}\|=1.7764\cdot10^{-15}$\\
\bottomrule
\end{tabular}
\end{table}
\end{document}
答案1
由于所有数字的位数完全相同，只是有无减号不同，因此最简单的方法是使用包bmatrix*中的环境右对齐矩阵的列mathtools。请参阅答案这个问题以了解更多对齐矩阵元素的方法。
\documentclass[5p,twocolumn]{elsarticle}
\usepackage{booktabs,tabularx}
\usepackage{amsmath}
\usepackage{mathtools}
\begin{document}
\footnotesize
\setlength\tabcolsep{1pt}
\setlength\arraycolsep{2pt}
\begin{tabular}{ll}
Step 9: & Compute $ \mathbf{Q}'_{\theta} = $\\
& $\begin{bmatrix*}[r]
-0.3390 & 0.3390 & 0.3017 & 0.1695 & -1.1348 & 0.3354\\
0.3390 & 0.2042 & -0.1296 & -0.1021 & 0.3585 & -0.1338\\
0.3017 & -0.1296 & -0.0488 & -0.0648 & 0.0178 & -0.0411\\
0.1695 & -0.1021 & -0.0648 & -0.0511 & 0.1792 & -0.0669\\
-1.1348 & 0.3585 & 0.0187 & 0.1792 & 0.6733 & -0.0386\\
0.3354 & -0.1338 & -0.0411 & -0.0669 & -0.0386 & -0.0303
\end{bmatrix*}$
\end{tabular}
\end{document}
答案2
对于负数的对齐，解决方案与bmatrix*from相同。此外，我建议定义一个新的浮点数 ( )，它有自己的计数器；此外，我还改进了多个下标的对齐，并定义了一个命令，带有星号版本，可自动根据其内容调整其大小。mathtoolsalgonorm
\documentclass[5p,twocolumn]{elsarticle}
\usepackage{booktabs,tabularx, float, caption}
\usepackage{mathtools}
\DeclarePairedDelimiter{\abs}{\lvert}{\rvert}
\DeclarePairedDelimiter{\norm}{\lVert}{\rVert}
\floatstyle{plaintop}
\newfloat{algo}{tbp}{loa}{}
\floatname{algo}{Algorithm}
\usepackage{lipsum}
\begin{document}
\lipsum[1]
\setcounter{algo}{2}
\begin{algo}
\captionsetup{font=footnotesize, labelfont=bf, singlelinecheck=off}
\caption{Illustrative calculation for Numerical Example 2}
\label{alg:Example 2.2}
% \centering
\footnotesize%scriptsize
\setlength\tabcolsep{1pt}
\setlength\arraycolsep{2pt}
\delimitershortfall=-1sp
%\setlength{\extrarowheight}{3pt}
\begin{tabular}{p{15mm}l@{}}
\toprule
\multicolumn{2}{l}{\normalfont Algorithm steps.\quad Numerical computation}\\
\midrule
Input & Given\\
& $ \mathbf{U}|_{\theta=0.5} = \begin{bmatrix} -1 & 1 & 0.2298 & 0.5000 & 0.7702 & 0.1250 \end{bmatrix}^T$\\
& $\mathbf{U}'_{\theta\smash{\bigm|}\theta=0.5} = \begin{bmatrix} -2 & 2 & 0.8415 & 1 & -0.8415 & 0.7500 \end{bmatrix}^T$\\
& $ \mathbf{v} = \begin{bmatrix} -2.7063 & 1.0000 & 0.2298 & 0.5000 & 0.7702 & 0.1250 \end{bmatrix}^T$\\
& $ \mathbf{u} = \begin{bmatrix} -0.8905 & 0.3290 & 0.0756 & 0.1645 & 0.2534 & 0.0411 \end{bmatrix}^T$\\
\midrule
Step 1: & Compute $\norm*{\mathbf{U}}'_{\theta} = 2.4257$\\
\addlinespace
Step 3: & Compute $\mathbf{v}'_{\theta} =$ \\
& $\begin {bmatrix}
-4.4257 & 2.0000 & 0.8415 & 1.0000 & -0.8415 & 0.7500
\end{bmatrix}^T$\\
\addlinespace
Step 7: & Compute $\norm{\mathbf{v}'_{\theta}}= 4.6451$\\
\addlinespace
Step 8: & Compute $\mathbf{u}'_{\theta} = $ \\
& $\begin {bmatrix}
-0.0952 & 0.1552 & 0.1613 & 0.0776 & -0.6642 & 0.1839
\end{bmatrix}^T$\\
\addlinespace
Step 9: & Compute $ \mathbf{Q}'_{\theta} = $\\
\multicolumn{2}{r}{$\begin{bmatrix*}[r]
-0.3390 & 0.3390 & 0.3017 & 0.1695 & -1.1348 & 0.3354\\
0.3390 & 0.2042 & -0.1296 & -0.1021 & 0.3585 & -0.1338\\
0.3017 & -0.1296 & -0.0488 & -0.0648 & 0.0178 & -0.0411\\
0.1695 & -0.1021 & -0.0648 & -0.0511 & 0.1792 & -0.0669\\
-1.1348 & 0.3585 & 0.0187 & 0.1792 & 0.6733 & -0.0386\\
0.3354 & -0.1338 & -0.0411 & -0.0669 & -0.0386 & -0.0303
\end{bmatrix*}$}\\
\addlinespace
Step 10: & Extract $\mathbf{Q}'_{\theta} = \mathbf{Q}'_{\theta}(1,1:6)= $\\
&$\begin{bmatrix} -0.3390 & 0.3390 & 0.3017 & 0.1695 & -1.1348 & 0.3354 \end{bmatrix}^T$\\
\addlinespace
Step 11: & $\mathbf{R}'_{\theta} = \begin{bmatrix} 2.4257 \end{bmatrix}$ \\
\midrule
\addlinespace
Output & $\mathbf{R}'|_{\theta = 0.5} = \begin{bmatrix} 2.4257 \end{bmatrix}$\\
\midrule
\addlinespace
Test & Accuracy of the computations:\\
& $\norm*{(\mathbf{U}^T\mathbf{U})'_{\theta\smash{\bigm|}\theta=0.5} - (\mathbf{R}^T\mathbf{R})'_{\theta\smash{\bigm|}\theta=0.5}}=1.7764\cdot10^{-15}$\\
\bottomrule
\end{tabular}
\end{algo}
\end{document}
