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如何处理这个长表

tag-icon • tag-icon tables longtable
如何处理这个长表

我有一张长表，右侧有一列特别宽。我想要这种类型的输出，例子

我得到的代码运行得并不完美。

\begin{longtable}{rl}
\midrule
\midrule
\endhead
\multicolumn{2}{r}{to be continued \dots} \\
\endfoot
\bottomrule
\endlastfoot
$\tau_2$                      &=$\quad \inf\{t-\tau_1 \mid t>\tau_1, X_t \geq 0, X_{\tau_1}<0 \}$ \ \text{--} \ \text{time elasped after $\tau$ (or $\tau_1$) when $X_t$ goes back above zero} \\
$g_t^X$                 \quad &= $\quad \sup\{ s \leq t \mid \text{sign}(X_s) \neq \text{sign}(X_t)  \}$ \ \text{--} \ \text{last crossing time of 0 before time $t$} \\
$d_t^X$                 \quad &= $\quad \inf\{ s \geq t \mid \text{sign}(X_s) \neq \text{sign}(X_t)  \}$ \ \text{--} \ \text{first crossing time of 0 before time $t$} \\
\end{longtable}

我怎样才能得到如图所示的输出？如能得到任何帮助我将不胜感激。

答案1

像这样？

\documentclass{article}

\usepackage{booktabs}
\usepackage{longtable}
\usepackage{amsmath}

\begin{document}
\begin{longtable}{rp{3in}}
\midrule
\midrule
\endhead
\multicolumn{2}{r}{to be continued \dots} \\
\endfoot
\bottomrule
\endlastfoot
$\tau_2$   \quad                   =& $\inf\{t-\tau_1 \mid t>\tau_1, X_t \geq 0, X_{\tau_1}<0 \}$ -- time elasped after $\tau$ (or $\tau_1$) when $X_t$ goes back above zero \\
$g_t^X$                 \quad =& $\sup\{ s \leq t \mid \text{sign}(X_s) \neq \text{sign}(X_t)  \}$ -- last crossing time of 0 before time $t$ \\
$d_t^X$                 \quad =&$\inf\{ s \geq t \mid \text{sign}(X_s) \neq \text{sign}(X_t)  \}$ -- first crossing time of 0 before time $t$ \\
\end{longtable}
\end{document}

答案2

我不确定你的问题是什么...但也许我的回答会对你有帮助。

以下是代码：

\documentclass[a4paper,10pt]{article}
\usepackage[utf8]{inputenc}
 \usepackage{longtable}
 \usepackage{array}
 \usepackage{booktabs}
 \usepackage{amsmath}
%opening
\title{}
\author{}


\begin{document}

\maketitle

\newcolumntype{L}{>{\small\raggedright\arraybackslash}p{0.7\textwidth}}

\begin{longtable}{rcL}
\midrule
\midrule
\endhead
\multicolumn{3}{r}{to be continued \dots} \\
\endfoot
\bottomrule
\endlastfoot
$\tau_2$     &                 &=$\quad \inf\{t-\tau_1 \mid t>\tau_1, X_t \geq 0, X_{\tau_1}<0 \}$ \ \text{--} \ \text{time elasped after $\tau$ (or $\tau_1$) when $X_t$ goes back above zero} \\
\\
$g_t^X$      &            &= $\quad \sup\{ s \leq t \mid \text{sign}(X_s) \neq \text{sign}(X_t)  \}$ \ \text{--} \ \text{last crossing time of 0 before time $t$} \\
\\
$d_t^X$      &            &= $\quad \inf\{ s \geq t \mid \text{sign}(X_s) \neq \text{sign}(X_t)  \}$ \ \text{--} \ \text{first crossing time of 0 before time $t$} \\
\end{longtable}

\end{document}

结果如下：

在此处输入图片描述

编辑：

如果您愿意，可以将其更改为：

\begin{longtable}{rccL}
\midrule
\midrule
\endhead
\multicolumn{4}{r}{to be continued \dots} \\
\endfoot
\bottomrule
\endlastfoot
$\tau_2$     &                 &= &$\inf\{t-\tau_1 \mid t>\tau_1, X_t \geq 0, X_{\tau_1}<0 \}$  \text{--}  time elasped after $\tau$ (or $\tau_1$) when $X_t$ goes back above zero \\
\\
$g_t^X$      &            &=  &$\sup\{ s \leq t \mid \text{sign}(X_s) \neq \text{sign}(X_t)  \}$  \text{--}  last crossing time of 0 before time $t$ \\
\\
$d_t^X$      &            &= &$\inf\{ s \geq t \mid \text{sign}(X_s) \neq \text{sign}(X_t)  \}$  \text{--}  first crossing time of 0 before time $t$ \\
\end{longtable}

对于下一个结果：

在此处输入图片描述

答案3

我认为您不希望等号周围有这么宽的空格。无论如何，它们可以轻松自定义，稍后再见。

\documentclass{article}
\usepackage{amsmath}
\usepackage{longtable}
\usepackage{array,booktabs}

\DeclareMathOperator{\sign}{sign}

\begin{document}

\begin{longtable}{@{}>{$}r<{{}$}@{}p{.8\textwidth}@{}}
\toprule
\endhead
\multicolumn{2}{r}{to be continued} \\
\endfoot
\bottomrule
\endlastfoot
\tau_2 = & $\inf\{t-\tau_1 \mid t>\tau_1, X_t \geq 0, X_{\tau_1}<0 \}$
         -- time elapsed after $\tau$ (or $\tau_1$) when $X_t$ goes back above zero \\
g_t^X  = & $\sup\{ s \leq t \mid \sign(X_s) \neq \sign(X_t) \}$
         -- last crossing time of 0 before time $t$ \\
d_t^X  = & $\inf\{ s \geq t \mid \sign(X_s) \neq \sign(X_t)  \}$
         -- first crossing time of 0 before time $t$ \\
\end{longtable}

\end{document}

在此处输入图片描述

拥有更广阔的空间：

\documentclass{article}
\usepackage{amsmath}
\usepackage{longtable}
\usepackage{array,booktabs}

\DeclareMathOperator{\sign}{sign}

\begin{document}

\begin{longtable}{@{}>{$\thickmuskip=24mu\relax}r<{{}$}@{}p{.8\textwidth}@{}}
\toprule
\endhead
\multicolumn{2}{r}{to be continued} \\
\endfoot
\bottomrule
\endlastfoot
\tau_2 = & $\inf\{t-\tau_1 \mid t>\tau_1, X_t \geq 0, X_{\tau_1}<0 \}$
         -- time elapsed after $\tau$ (or $\tau_1$) when $X_t$ goes back above zero \\
g_t^X  = & $\sup\{ s \leq t \mid \sign(X_s) \neq \sign(X_t) \}$
         -- last crossing time of 0 before time $t$ \\
d_t^X  = & $\inf\{ s \geq t \mid \sign(X_s) \neq \sign(X_t)  \}$
         -- first crossing time of 0 before time $t$ \\
\end{longtable}

\end{document}

在此处输入图片描述

答案4

您还可以考虑避免使用任何类似表格的环境来进行此类输出，并清除源代码：

姆韦

\documentclass[a4paper,10pt]{article}
\usepackage{desclist}
\usepackage{amsmath}

\begin{document}
\begin{desclist}{\hfill\large}{\quad=\quad}[xxx] % xxx = longest label 

\item
[$\tau_2$]  
$\inf\{t-\tau_1 \mid t>\tau_1, X_t \geq 0, X_{\tau_1}<0 \}$  -- 
time elasped after $\tau$ (or $\tau_1$) when $X_t$ goes back above zero. 

\item
[$g_t^X$] 
$\sup\{ s \leq t \mid \text{sign}(X_s) \neq \text{sign}(X_t)  \}$ -- 
last crossing time of 0 before time~$t$.

\item
[$d_t^X$] 
$\inf\{ s \geq t \mid \text{sign}(X_s) \neq \text{sign}(X_t)\}$ -- 
first crossing time of 0 before time~$t$.

\end{desclist}
\end{document}

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