请考虑下表：
\documentclass[10pt,letterpaper,oneside]{article}
\usepackage[dvips,top=1.2in,bottom=0.65in,left=1.15in,right=1.15in,includefoot]{geometry}
\usepackage{booktabs}
\begin{document}
\begin{center}
\begin{tabular}{p{4.2cm}p{1.5cm}p{5.5cm}p{3.25cm}}
\toprule
\emph{\centering observed frequencies of each detection history} & \emph{\centering detections per site} & \emph{equivalent detection histories} ($h_i$) & \emph{probability} \\
\midrule
\multicolumn{1}{c} {10}& \multicolumn{1}{c}4 &\texttt{1111}& $\psi p^4$\\
\multicolumn{1}{c}{20}& \multicolumn{1}{c}3 &\texttt{1110, 1101, 1011, 0111}& $\psi p^3(1-p)$\\
\multicolumn{1}{c}{30}&\multicolumn{1}{c}2 &\texttt{1100, 1001, 1010, 0011, 0101}& $\psi p^2(1-p)^2$\\
\multicolumn{1}{c}{20}& \multicolumn{1}{c}1 &\texttt{1000, 0100, 0010, 0001}& $\psi p(1-p)^3$\\
\multicolumn{1}{c}{20}& \multicolumn{1}{c}0 &\texttt{0000} & $\psi (1-p)^4 + (1-\psi)$\\
\bottomrule
\end{tabular}
\end{center}
\end{document}
我正在尝试弄清楚如何：
- 将列标题居中（而不是默认的左对齐），以及
- 如何强制单行标题（例如示例中的第 3 列和第 4 列）对齐在中间或标题“单元格”的底部。
我能弄清楚如何生成多行列标题的唯一方法是使用“段落”，但如果我这样做，我无论如何也想不出如何使列标题文本居中，或者将其对齐到顶部以外的任何位置。
答案1
如果所有行都不超过一行，那么使用多行列并没有什么实际好处。在这些情况下，l和c是最简单的。
对于其他表格，您可以使用array代替m{}进行p{}垂直居中和>{\centering\arraybackslash}进行水平居中。此外，请考虑使用tabularx以避免需要手动计算正确的宽度。对于您的示例中的表格，表格对于页面来说太宽了。
\documentclass{article}
\usepackage{array,tabularx}
\usepackage[top=1.2in,bottom=0.65in,left=1.15in,right=1.15in,includefoot]{geometry}
\usepackage{booktabs}
\newcolumntype{C}[1]{>{\centering\arraybackslash}m{#1}}
\newcolumntype{Y}{>{\centering\arraybackslash}X}
\renewcommand{\tabularxcolumn}[1]{m{#1}}
\begin{document}
\begin{center}
\begin{tabularx}{\linewidth}{YC{1.5cm}ll}
\toprule
\emph{observed frequencies of each detection history} & \emph{detections per site} & \multicolumn{1}{c}{\emph{equivalent detection histories} ($h_i$)} & \multicolumn{1}{c}{\emph{probability}} \\
\midrule
10& 4 &\texttt{1111}& $\psi p^4$\\
20& 3 &\texttt{1110, 1101, 1011, 0111}& $\psi p^3(1-p)$\\
30& 2 &\texttt{1100, 1001, 1010, 0011, 0101}& $\psi p^2(1-p)^2$\\
20& 1 &\texttt{1000, 0100, 0010, 0001}& $\psi p(1-p)^3$\\
20& 0 &\texttt{0000} & $\psi (1-p)^4 + (1-\psi)$\\
\bottomrule
\end{tabularx}
\end{center}
\end{document}
答案2
\documentclass[10pt,letterpaper,oneside]{article}
\usepackage[dvips,
top=1.2in,bottom=0.65in,left=1.15in,right=1.15in,
includefoot]{geometry}
\usepackage{array,
booktabs}
\newcommand\mch[2]{\multicolumn{1}{>{\centering\arraybackslash\itshape}b{#1}}{#2}}
\begin{document}
\begin{center}
\begin{tabular}{c c
>{\ttfamily}l
>{$}l<{$}
}
\toprule
\mch{3.5cm}{observed frequencies of each detection history}
& \mch{1.5cm}{detections per site}
& \mch{5.5cm}{equivalent detection histories ($h_i$)}
& \mch{3cm}{probability} \\
\midrule
10 & 4 & 1111 & \psi p^4 \\
20 & 3 & 1110, 1101, 1011, 0111 & \psi p^3(1-p) \\
3 & 2 & 1100, 1001, 1010, 0011, 0101& \psi p^2(1-p)^2 \\
20 & 1 & 1000, 0100, 0010, 0001 & \psi p(1-p)^3 \\
20 & 0 & 0000 & \psi (1-p)^4 + (1-\psi) \\
\bottomrule
\end{tabular}
\end{center}
\end{document}
列宽由新命令（您可以选择其他名称）确定，该命令\mch有两个参数：第一个是列宽，第二个是单元格的内容。内容居中并以斜体字体显示。
答案3
一个简单的解决方案是使用makecell，它允许换行和使用和命令对l, r, c列中的单元格进行通用格式设置。默认对齐方式是垂直和水平居中。\thead\makecell
\documentclass{article}
\usepackage[top=1.2in,bottom=0.65in,left=1.15in,right=1.15in,includefoot]{geometry}
\usepackage{booktabs}
\usepackage{makecell}
\renewcommand{\theadfont}{\normalsize\itshape}
\begin{document}
\begin{center}
\begin{tabular}{ccll}
\toprule
\thead{observed frequencies of\\each detection history} & \thead{detections\\ per site} & \thead[l]{equivalent detection histories ($h_i$)} & \thead[l]{probability} \\
\midrule
10& 4 &\texttt{1111}& $\psi p^4$\\
20& 3 &\texttt{1110, 1101, 1011, 0111}& $\psi p^3(1-p)$\\
30& 2 &\texttt{1100, 1001, 1010, 0011, 0101}& $\psi p^2(1-p)^2$\\
20& 1 &\texttt{1000, 0100, 0010, 0001}& $\psi p(1-p)^3$\\
20& 0 &\texttt{0000} & $\psi (1-p)^4 + (1-\psi)$\\
\bottomrule
\end{tabular}
\end{center}
\vspace{1cm}
\begin{center}
\renewcommand{\theadalign}{b}
\begin{tabular}{ccll}
\toprule
\thead{observed frequencies of\\each detection history} & \thead{detections\\ per site} & \thead[l]{equivalent detection histories ($h_i$)} & \thead[l]{probability} \\
\midrule
10& 4 &\texttt{1111}& $\psi p^4$\\
20& 3 &\texttt{1110, 1101, 1011, 0111}& $\psi p^3(1-p)$\\
30& 2 &\texttt{1100, 1001, 1010, 0011, 0101}& $\psi p^2(1-p)^2$\\
20& 1 &\texttt{1000, 0100, 0010, 0001}& $\psi p(1-p)^3$\\
20& 0 &\texttt{0000} & $\psi (1-p)^4 + (1-\psi)$\\
\bottomrule
\end{tabular}
\end{center}
\end{document}
