1

Consider this table I made. It's actually much longer but for the purpose of illustration I've removed most rows.

\begin{table}[h]
\begin{tabular}{ccccccccc}
\toprule
                         &                                 &                       &                        &                                    & \multicolumn{4}{l}{Distance as determined by Algorithm} \\ \cmidrule(l){6-9} 
\multirow{-2}{*}{Leaves} & \multirow{-2}{*}{Reticulations} & \multirow{-2}{*}{$N$} & \multirow{-2}{*}{$N'$} & \multirow{-2}{*}{$d_{tail}(N,N')$} & 2.1          & 2.2          & 2.3         & 2.4         \\ \midrule
5                        & 1                               & 7                     & 8                      & 3                                  & 5            & 5            & 5           & 5           \\
\rowcolor[HTML]{EFEFEF} 
5                        & 1                               & 8                     & 9                      & 2                                  & 2            & 2            & 2           & 2           \\ \bottomrule
\end{tabular}
\end{table}

enter image description here

I don't know how to evenly distribute the cells underneath "distance as determined by Algorithm" in terms of width. I've tried using tabularx, but I couldn't get it right.

1
  • Welcome to TeX.SE. – Mico Aug 3 '20 at 21:23
1

I suggest you employ a tabularx environment, set its width to \textwidth, and assign a centered version of the X column type to 7 of the 9 columns. This setup will create an appearance of roughly equally wide columns.

enter image description here

\documentclass{article}
\usepackage[table]{xcolor}
\usepackage{booktabs,ragged2e,amsmath,tabularx}
\newcolumntype{C}{>{\Centering\hspace{0pt}}X}

\begin{document}
\begin{table}
\setlength\tabcolsep{2pt} % default: 6pt
\begin{tabularx}{\textwidth}{ cCCC c CCCC }
\toprule
Leaves & Reticulations & $N$ & $N'$ & $d_{\mathrm{tail}}(N,N')$ & 
\multicolumn{4}{>{\hsize=\dimexpr4\hsize+6\tabcolsep\relax}C}{%
     Distance as determined by algorithm} \\
\cmidrule(l){6-9}
&&&&& 2.1 & 2.2 & 2.3 & 2.4 \\ 
\midrule
5 & 1 & 7 & 8 & 3 & 5 & 5 & 5 & 5 \\
\rowcolor[HTML]{EFEFEF} 
5 & 1 & 8 & 9 & 2 & 2 & 2 & 2 & 2 \\ 
\bottomrule
\end{tabularx}
\end{table}
\end{document}
1
  • 1
    Yes, perfect! Thank you! Also thanks for welcoming me. :) – Violet Aug 3 '20 at 21:49
1

Here are two solutions, one with the new w column type, and the other with tabularx, and the help of the makecell package, which allows for line breaks in standard column types:

\documentclass{article}

\usepackage[table]{xcolor}
\usepackage{booktabs}
\usepackage{tabularx, multirow}
\usepackage{makecell}

\begin{document}

\begin{table}[h]
\centering
\begin{tabular}{*{5}{c}*{4}{wc{6mm}}}
\toprule
                         & & & & & \multicolumn{4}{c}{\makecell{Distance as determined\\ by Algorithm}} \\ \cmidrule(l){6-9}
\multirow{-2}{*}{Leaves} & \multirow{-2}{*}{Reticulations} & \multirow{-2}{*}{$N$} & \multirow{-2}{*}{$N'$} & \multirow{-2}{*}{$d_{\mbox{tail}}(N,N')$} & 2.1 & 2.2 & 2.3 & 2.4 \\ \midrule
5 & 1 & 7 & 8 & 3 & 5 & 5 & 5 & 5 \\
\rowcolor[HTML]{EFEFEF}
5 & 1 & 8 & 9 & 2 & 2 & 2 & 2 & 2 \\ \bottomrule
\end{tabular}
\vskip1cm

\begin{tabularx}{\linewidth}{*{5}{c}*{4}{>{\centering\arraybackslash}X}}
\toprule
                         & & & & & \multicolumn{4}{c}{\makecell{Distance as determined\\ by Algorithm}} \\ \cmidrule(l){6-9}
\multirow{-2}{*}{Leaves} & \multirow{-2}{*}{Reticulations} & \multirow{-2}{*}{$N$} & \multirow{-2}{*}{$N'$} & \multirow{-2}{*}{$d_{\mbox{tail}}(N,N')$} & 2.1 & 2.2 & 2.3 & 2.4 \\ \midrule
5 & 1 & 7 & 8 & 3 & 5 & 5 & 5 & 5 \\
\rowcolor[HTML]{EFEFEF}
5 & 1 & 8 & 9 & 2 & 2 & 2 & 2 & 2 \\ \bottomrule
\end{tabularx}
\end{table}

\end{document} 

enter image description here

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.