我想将公式制成表格,但我发现表格单元格的线条太近,这在审美上不令人满意。示例(公式复制自维基百科):
\begin{tabular}{|c|c|}
\hline
Cylindrical & $\displaystyle{{1 \over \rho}{\partial \over \partial\rho}\left(\rho {\partial f \over \partial \rho}\right)
+ {1 \over \rho^2}{\partial^2 f \over \partial \phi^2} + {\partial^2 f \over \partial z^2}}$\\\hline
Spherical & $\displaystyle{{1 \over r^2}{\partial \over \partial r}\!\left(r^2 {\partial f \over \partial r}\right)
\!+\!{1 \over r^2\!\sin\theta}{\partial \over \partial \theta}\!\left(\sin\theta {\partial f \over \partial \theta}\right)
\!+\!{1 \over r^2\!\sin^2\theta}{\partial^2 f \over \partial \phi^2}}$\\\hline
\end{tabular}
给出了
你会看到公式的顶部被线条切断了。有什么建议吗?
答案1
我过去曾通过修改\arraystretch
您可以通过添加到您的源来做到这一点:
{\renewcommand{\arraystretch}{1.2} %<- modify value to suit your needs
\begin{tabular}{|c|c|}
...
\end{tabular}
}
编辑:
实际上,经过更多的研究,似乎在单元格中存在方程式会出现意外行为,其中顶部的空间不成比例地增加(将值设置为 3 以开始获取底部的空间):
最近我开始使用该tabu
包替换所有表格(从tabular
到tabularx
到longtable
)。它还提供了一些其他控件。在这种情况下,\tabulinesep
对结果的影响更为恰当:
\documentclass[preview]{standalone}
\usepackage{tabu}
\begin{document}
{\tabulinesep=1.2mm
\begin{tabu} {|c|c|}
\hline
Cylindrical & $\displaystyle{{1 \over \rho}{\partial \over \partial\rho}\left(\rho {\partial f \over \partial \rho}\right)
+ {1 \over \rho^2}{\partial^2 f \over \partial \phi^2} + {\partial^2 f \over \partial z^2}}$\\\hline
Spherical & $\displaystyle{{1 \over r^2}{\partial \over \partial r}\!\left(r^2 {\partial f \over \partial r}\right)
\!+\!{1 \over r^2\!\sin\theta}{\partial \over \partial \theta}\!\left(\sin\theta {\partial f \over \partial \theta}\right)
\!+\!{1 \over r^2\!\sin^2\theta}{\partial^2 f \over \partial \phi^2}}$\\\hline
\end{tabu}}
\end{document}
其结果如下:
答案2
您可以使用不同的表格格式,省略不必要的垂直线;您还可以使用个人偏导数命令大大简化输入:
\documentclass{article}
\usepackage{booktabs,amsmath}
\newcommand{\dpder}[3][]{\dfrac{\partial^{#1}#2}{\partial #3}}
\begin{document}
\begin{tabular}{lc}
\toprule
Cartesian &
$\dpder[2]{f}{x^2}+\dpder[2]{f}{y^2}+\dpder[2]{f}{z^2}$
\\
\midrule
Cylindrical &
$\dfrac{1}{\rho} \dpder{}{\rho}{\left(\rho \dpder{f}{\rho}\right)}
+ \dfrac{1}{\rho^2} \dpder[2]{f}{\phi^2} + \dpder[2]{f}{z^2}$
\\
\midrule
Spherical &
$\dfrac{1}{r^2}\dpder{}{r}{\left(r^2 \dpder{f}{r}\right)}
+\dfrac{1}{r^2\sin\theta}\dpder{}{\theta}{\left(\sin\theta \dpder{f}{\theta}\right)}
+\dfrac{1}{r^2\sin^2\theta}\dpder[2]{f}{\phi^2}$
\\
\bottomrule
\end{tabular}
\end{document}
请注意如何使用{\left(...\right)}
(一对括号就足够了,而不需要显式备份)来避免“圆柱”和“球形”行中不必要的空格。还请注意\dfrac
在定义中的使用\dpder
,这避免了指定\displaystyle
(然而不是一个接受参数的命令,但是一个声明)。
由于booktabs
很少需要调整行距。
答案3
作为过时软件包的替代方案tabu
,新的tabularray
软件包rowsep
为表格提供了选项:
\documentclass{article}
\usepackage{tabularray}
\begin{document}
\begin{tblr}{|c|c|}
\hline
Cylindrical & $\displaystyle{{1 \over \rho}{\partial \over \partial\rho}\left(\rho {\partial f \over \partial \rho}\right)
+ {1 \over \rho^2}{\partial^2 f \over \partial \phi^2} + {\partial^2 f \over \partial z^2}}$\\\hline
Spherical & $\displaystyle{{1 \over r^2}{\partial \over \partial r}\!\left(r^2 {\partial f \over \partial r}\right)
\!+\!{1 \over r^2\!\sin\theta}{\partial \over \partial \theta}\!\left(\sin\theta {\partial f \over \partial \theta}\right)
\!+\!{1 \over r^2\!\sin^2\theta}{\partial^2 f \over \partial \phi^2}}$\\\hline
\end{tblr}
\bigskip
\SetTblrInner{rowsep=4pt}
\begin{tblr}{|c|c|}
\hline
Cylindrical & $\displaystyle{{1 \over \rho}{\partial \over \partial\rho}\left(\rho {\partial f \over \partial \rho}\right)
+ {1 \over \rho^2}{\partial^2 f \over \partial \phi^2} + {\partial^2 f \over \partial z^2}}$\\\hline
Spherical & $\displaystyle{{1 \over r^2}{\partial \over \partial r}\!\left(r^2 {\partial f \over \partial r}\right)
\!+\!{1 \over r^2\!\sin\theta}{\partial \over \partial \theta}\!\left(\sin\theta {\partial f \over \partial \theta}\right)
\!+\!{1 \over r^2\!\sin^2\theta}{\partial^2 f \over \partial \phi^2}}$\\\hline
\end{tblr}
\end{document}
答案4
有了,您可以使用和{NiceTabular}
,但您也有一个密钥。nicematrix
\arraystretch
\extrarowheight
array
cell-space-limits
\documentclass{article}
\usepackage{nicematrix}
\begin{document}
\begin{NiceTabular}{cc}[hvlines,cell-space-limits=3pt]
Cylindrical & $\displaystyle{{1 \over \rho}{\partial \over \partial\rho}\left(\rho {\partial f \over \partial \rho}\right)
+ {1 \over \rho^2}{\partial^2 f \over \partial \phi^2} + {\partial^2 f \over \partial z^2}}$\\
Spherical & $\displaystyle{{1 \over r^2}{\partial \over \partial r}\!\left(r^2 {\partial f \over \partial r}\right)
\!+\!{1 \over r^2\!\sin\theta}{\partial \over \partial \theta}\!\left(\sin\theta {\partial f \over \partial \theta}\right)
\!+\!{1 \over r^2\!\sin^2\theta}{\partial^2 f \over \partial \phi^2}}$\\
\end{NiceTabular}
\end{document}