我正在尝试创建一个包含坐标信息的表格，但是输出很混乱，因为有些坐标有分数和负号。有没有办法格式化它以获得更好的外观？
\documentclass{article}
\begin{document}
\centering
\begin{tabular}{cccc}
\hline
Test Sample & Output 1 & Output 2 & Classification \\ \hline
$(-\frac{14}{3},-\frac{8}{3})$ & +1 & -1 & 1 \\
$(-\frac{11}{3},\frac{1}{3})$ & +1 & -1 & 1 \\
$(-\frac{7}{3},-\frac{11}{3})$ & +1 & -1 & 1 \\
$(-\frac{4}{3},\frac{8}{3})$ & +1 & +1 & 3 \\
$(-\frac{2}{3},\frac{1}{3})$ & +1 & -1 & 1 \\
$(\frac{2}{3},-2)$ & +1 & -1 & 1 \\
$(\frac{4}{3},3)$ & +1 & +1 & 3 \\
$(\frac{7}{3},0)$ & +1 & -1 & 1 \\
$(4,\frac{13}{3})$ & +1 & +1 & 3 \\
$(5,1)$ & -1 & - & 2 \\ \hline
\end{tabular}
\end{document}
答案1
编辑：
考虑的是@Mico 评论。从中建议使用appto={@{},@{}}哪个 unfirtunatela 无法按预期工作，由 `appto={,;} 模拟。Mico，非常感谢你！
使用带有库的tabularray包：booktabssiunitx
\documentclass{article}
\usepackage{tabularray}
\UseTblrLibrary{booktabs, siunitx}
\begin{document}
\begin{table}[ht]
\sisetup{retain-explicit-plus}
\begin{tblr}{colspec = { Q[r, mode=math] @{} Q[l, mode=math] %
*{2}{Q[c, si={table-format=-1.0}]}
Q[c, si={table-format= 1.0}]},
cell{2-Z}{1} = {preto={\bigl(}, appto={,\;}},
cell{2-Z}{2} = {appto={\bigr)}},
row{1} = {guard, mode=text},
rowsep = 3pt,
}
\toprule
\SetCell[c=2]{c} Test Sample
& & Output 1 & Output 2 & Classification \\
\midrule
-\frac{14}{3} & -\frac{8}{3} & +1 & -1 & 1 \\
-\frac{11}{3} & \frac{1}{3} & +1 & -1 & 1 \\
-\frac{7}{3} & -\frac{11}{3} & +1 & -1 & 1 \\
-\frac{4}{3} & \frac{8}{3} & +1 & +1 & 3 \\
-\frac{2}{3} & \frac{1}{3} & +1 & -1 & 1 \\
\frac{2}{3} & -2 & +1 & -1 & 1 \\
\frac{4}{3} & 3 & +1 & +1 & 3 \\
\frac{7}{3} & 0 & +1 & -1 & 1 \\
4 & \frac{13}{3} & +1 & +1 & 3 \\
5 & 1 & -1 & {{{--}}} & 2 \\
\bottomrule
\end{tblr}
\end{table}
\end{document}
附录：
使用nccmath包中的中等大小分数和构造\left(,right.和\left.,\right)来表示括号。由于它们，逗号周围的间距现在比以前更正确：
\documentclass{article}
\usepackage{tabularray}
\UseTblrLibrary{booktabs, siunitx}
\usepackage{nccmath}
\begin{document}
\begin{table}[ht]
\sisetup{retain-explicit-plus}
\begin{tblr}{colspec = { Q[r, mode=math] @{} Q[l, mode=math] %
*{2}{Q[c, si={table-format=-1.0}]}
Q[c, si={table-format= 1.0}]},
cell{2-Z}{1} = {preto={\left(}, appto={\,,\right.}},
cell{2-Z}{2} = {preto={\left.}, appto={\right)}},
row{1} = {guard, mode=text},
}
\toprule
\SetCell[c=2]{c} Test Sample
& & Output 1 & Output 2 & Classification \\
\midrule
-\mfrac{14}{3} & -\mfrac{8}{3} & +1 & -1 & 1 \\
-\mfrac{11}{3} & \mfrac{1}{3} & +1 & -1 & 1 \\
-\mfrac{7}{3} & -\mfrac{11}{3}& +1 & -1 & 1 \\
-\mfrac{4}{3} & \mfrac{8}{3} & +1 & +1 & 3 \\
-\mfrac{2}{3} & \mfrac{1}{3} & +1 & -1 & 1 \\
\mfrac{2}{3} & -2\vphantom{\mfrac{1}{3}}
& +1 & -1 & 1 \\
\mfrac{4}{3} & 3\vphantom{\mfrac{1}{3}}
& +1 & +1 & 3 \\
\mfrac{7}{3} & 0\vphantom{\mfrac{1}{3}}
& +1 & -1 & 1 \\
4\vphantom{\mfrac{1}{3}}
& \mfrac{13}{3} & +1 & +1 & 3 \\
5 & 1 & -1 & {{{--}}} & 2 \\
\bottomrule
\end{tblr}
\end{table}
\end{document}
答案2
如果放大寄存器，输出就不会混乱\baselineskip。例如：
\def\crl{\crcr\noalign{\medskip\hrule\medskip}}
\vbox{\baselineskip=17pt \halign{&\ \hfil$#$\hfil\ \cr\crl
\rm Test\ Sample & \rm Output 1 & \rm Output 2 & \rm Classification \crl
(-{14\over3},-{8\over3}) & +1 & -1 & 1 \cr
(-{11\over3},{1\over3}) & +1 & -1 & 1 \cr
(-{7\over3},-{11\over3}) & +1 & -1 & 1 \cr
(-{4\over3}, {8\over3}) & +1 & +1 & 3 \cr
(-{2\over3}, {1\over3}) & +1 & -1 & 1 \cr
( {2\over3},-2) & +1 & -1 & 1 \cr
( {4\over3},3) & +1 & +1 & 3 \cr
({7\over3},0) & +1 & -1 & 1 \cr
(4, {13\over3}) & +1 & +1 & 3 \cr
(5,1) & -1 & - & 2 \crl
}}
答案3
由于表格的所有内容（标题行除外）都是数学内容，我建议您使用array环境而不是tabular环境。然后，将array环境括在\[和\]指令中，以便在（显示）数学模式下水平居中材料。
我进一步建议您 (a) 使用内联分数符号而不是\frac符号和 (b) 使用包的宏booktabs而不是\hline。
\documentclass{article}
\usepackage{array,booktabs}
\begin{document}
\[
\begin{array}{@{} r @{{},{}} l ccc @{}} % '@{{},{}}' forces ',' to have type 'mathpunct'
\toprule
\multicolumn{2}{@{}c}{\mbox{Test Sample}} & \mbox{Output 1} & \mbox{Output 2} & \mbox{Classification} \\
\midrule
(-14/3 & -8/3) & +1 & -1 & 1 \\
(-11/3 & \phantom{+}1/3) & +1 & -1 & 1 \\
(-7/3 &-11/3) & +1 & -1 & 1 \\
(-4/3 & \phantom{+}8/3) & +1 & +1 & 3 \\
(-2/3 & \phantom{+}1/3) & +1 & -1 & 1 \\
\addlinespace
(2/3 & -2) & +1 & -1 & 1 \\
(4/3 & \phantom{+}3) & +1 & +1 & 3 \\
(7/3 & \phantom{+}0) & +1 & -1 & 1 \\
(4 & 13/3) & +1 & +1 & 3 \\
(5 & \phantom{+}1) & -1 & -\phantom{1} & 2 \\
\bottomrule
\end{array}
\]
\end{document}
答案4
我会使用左对齐和内联分数。
\documentclass{article}
\usepackage{booktabs}
\usepackage{siunitx}
\begin{document}
\begin{table}[htp]
\centering
\begin{tabular}{
@{}
>{\vphantom{$\Big|$}}l
*{2}{S[table-format=-1.0,retain-explicit-plus]}
S[table-format=1.0]
@{}
}
\toprule
\multicolumn{1}{@{}l}{Test Sample}
& {Output 1}
& {Output 2}
& {Classification} \\
\midrule
$(-14/3,-8/3)$ & +1 & -1 & 1 \\
$(-11/3,1/3)$ & +1 & -1 & 1 \\
$(-7/3,-11/3)$ & +1 & -1 & 1 \\
$(-4/3,8/3)$ & +1 & +1 & 3 \\
$(-2/3,1/3)$ & +1 & -1 & 1 \\
$(2/3,-2)$ & +1 & -1 & 1 \\
$(4/3,3)$ & +1 & +1 & 3 \\
$(7/3,0)$ & +1 & -1 & 1 \\
$(4,13/3)$ & +1 & +1 & 3 \\
$(5,1)$ & -1 & -1 & 2 \\
\bottomrule
\end{tabular}
\caption{Some descriptive caption}
\end{table}
\end{document}
