[Tex/LaTex] Multi-column problem

horizontal alignmentmulticolumntables

Consider the following table. First I have troubles to get the multicolumn-elements in the first row of the table limited to a maximum width (I tried it here with the command C{0.75} but obviously it does not work). The second issue: I would like to have the column-pair "orig." and "norm." centered to each of its multicolumn-elements.

\documentclass{article}
\usepackage{multirow}
\usepackage[ngerman]{babel}
\usepackage{booktabs}
\usepackage{colortbl}

\definecolor{LightGray}{gray}{.8}


\begin{document}
\newcolumntype{C}[1]{>{\centering\arraybackslash}m{#1}}


\begin{table}[htbp]
   \centering
   \begin{tabular}{C{1.7cm}C{0.75cm}C{0.75cm}C{0.75cm}C{0.75cm}C{0.75cm}C{0.75cm}C{0.75cm}}

      \toprule
    &
    \multicolumn{2}{c}{This is a long text for two rows} &
    \multicolumn{2}{c}{This is a long text for three rows} &
    \multicolumn{2}{c}{This is a long Text} \\

    \multirow{-2}{*}{Row 1} &
    org. & norm. &
    org. & norm. &
    org. & norm. &
    \multirow{-2}{*}{Result}\\

     \midrule

    1   &   0   &   0   &   1   &   0.14    &   7360    &   1   &   0.58    \\
    2   &   0.42    &   0.97    &   7   &   1   &   3420    &   0.46    &   0.85    \\
    3   &   0.42    &   0.97    &   7   &   1   &   3420    &   0.46    &   0.85    \\
    4   &   0.42    &   0.98    &   6   &   0.86    &   3610    &   0.49    &   0.80    \\
    5   &   0.43    &   0.99    &   6   &   0.86    &   4100    &   0.56    &   0.82    \\
    6   &   0.36    &   0.83    &   5   &   0.71    &   4280    &   0.58    &   0.71    \\
    7   &   0.36    &   0.83    &   5   &   0.71    &   4520    &   0.61    &   0.73    \\
    8   &   0.43    &   1   &   6   &   0.86    &   4000    &   0.54    &   0.82    \\
    9   &   0.27    &   0.63    &   3   &   0.43    &   3920    &   0.53    &   0.54    \\
    10  &   0.32    &   0.74    &   5   &   0.71    &   100 &   0.01    &   0.59    \\

      \bottomrule

   \end{tabular}
   \caption{caption}
   \label{tab:Example}
\end{table}

\end{document}

Best Answer

I'm not entirely sure I understood your question (in part because there are some references to rows that may actually be columns). Notice that I've chosen to dispense with the \multirow statements as they result in their arguments being place in somewhat odd locations (in my view, that is). I've also added some \cmidrule instructions in the header area of the table. Anyway, here's what I've come up with:

\documentclass{article}
\usepackage[ngerman]{babel}
\usepackage{multirow,booktabs,colortbl,tabularx}
\definecolor{LightGray}{gray}{.8}


\begin{document}
\newcolumntype{C}[1]{>{\centering\arraybackslash}p{#1}}

\begin{table}[htbp]
\centering
   \begin{tabular}{C{1.7cm} *{7}{C{0.75cm}} }
\toprule
    Column 1 &
    \multicolumn{2}{p{2cm}}{This is a long text for two columns} &
    \multicolumn{3}{p{3cm}}{This is a long text for three columns} &
    This is a long text & 
    Result\\
\cmidrule(lr){2-3} \cmidrule(lr){4-6} \cmidrule(lr){7-7}
    &
    org. & norm. &
    org. & norm. &
    org. & norm. &\\
\midrule
  1   &   0   &   0   &   1   &   0.14        & 7360  &   1   &   0.58    \\        
  2   &   0.42    &   0.97    &   7   &   1   & 3420  & 0.46  &   0.85    \\        
  3   &   0.42    &   0.97    &   7   &   1   & 3420  & 0.46  &   0.85    \\         
  4   &   0.42    &   0.98    &   6   & 0.86  & 3610  & 0.49  &   0.80    \\        
  5   &   0.43    &   0.99    &   6   & 0.86  & 4100  & 0.56  &   0.82    \\        
  6   &   0.36    &   0.83    &   5   & 0.71  & 4280  & 0.58  &   0.71    \\        
  7   &   0.36    &   0.83    &   5   & 0.71  & 4520  & 0.61  &   0.73    \\        
  8   &   0.43    &   1   &   6   &   0.86    & 4000  & 0.54  &   0.82    \\        
  9   &   0.27    &   0.63    &   3   & 0.43  & 3920  & 0.53  &   0.54    \\        
  10  &   0.32    &   0.74    &   5   & 0.71  &  100  & 0.01  &   0.59    \\    
\bottomrule
\end{tabular}
   \caption{caption}
   \label{tab:Example}
\end{table}

\end{document}

enter image description here

Related Solutions

[Tex/LaTex] ltablex – Multi-column problem

use it this way:

\multicolumn{3}{p{\dimexpr\textwidth-2\tabcolsep\relax}}{Here is the text that is too 
long for the table. If I insert ``null'' instead of all this text, the table is fine.
However, the text I would like to put in here is about 100 words long. Repeat: Here is the
text that is too long for the table. If I insert ``null'' instead of all this text, the 
table is fine. However, the text I would like to put in here is about 100 words long. Repeat:
Here is the text that is too long for the table. If I insert ``null'' instead of all this
text, the table is fine. However, the text I would like to put in here is about 100 words
long.}\\

p{\dimexpr\textwidth-2\tabcolsep\relax}} is the linewidth except the tabular border

Tables – Adjust Table Column Width When Using Multi-Column

Rather than make the first three data columns equally wide, I would like to suggest you make all six data columns equally wide.

In the following solution, the (usable) width of the data columns is calculated so as to evenly "fill" the space below the first subheader (\textbf{Partially Structured), and then all six data columns are assigned that width.

In case you're not familiar with LaTeX column width calculations, here's a step by step guide:

The total width of the first three data columns is \widthof{\textbf{Partialy Structured}} + 2\tabcolsep. By default, LaTeX inserts whitespace padding in the amount of \tabcolsep on each side of a cell, incl. a cell that spans three "basic" colums.
The total width of the first three data columns can also be expressed as 3(usable width of equal-width columns)+6\tabcolsep.
The usable width of the equal-spaced columns may therefore be calculated as
```
( \widthof{\textbf{Partialy Structured}} - 4\tabcolsep ) / 4
```

The following screenshot shows first the table with all six data columns having equal width, followed by a version that lets the first three data columns be markedly wider than the next three. I hope you will agree that the first table looks better.

\documentclass{article} % or some other suitable document class
\usepackage{calc}  % for '\widthof' macro
\usepackage{array} % for 'w' column type
\newlength\mylen
% calculate (minimal) width of the 6 data columns:
\setlength\mylen{(\widthof{\textbf{Partially Structured}}-4\tabcolsep)/3}

\begin{document}

\begin{table}[ht!]
    \centering

    %% use \mylen as width of all 6 data columns:
    \begin{tabular}{| c | *{3}{wc{\mylen}} | *{3}{wc{\mylen}} |}
        \hline
        & \multicolumn{3}{c|}{\textbf{Partially Structured}}
        & \multicolumn{3}{c|}{\textbf{Unstructured}} \\
        \hline
        & \textbf{m1} & \textbf{m2} & \textbf{m3} 
        & \textbf{m1} & \textbf{m2} & \textbf{m3} \\
        \hline
        \textbf{n1} & x & y & z & x & y & z \\
        \textbf{n2} & x & z & z & z & x & y \\
        \textbf{n3} & x & y & x & y & z & x \\
        \hline
    \end{tabular}
    
    \bigskip\bigskip
    %% same table, but with 'c' column type for final 3 columns
    \begin{tabular}{| c | *{3}{wc{\mylen}} | ccc |}
        \hline
        & \multicolumn{3}{c|}{\textbf{Partially Structured}}
        & \multicolumn{3}{c|}{\textbf{Unstructured}} \\
        \hline
        & \textbf{m1} & \textbf{m2} & \textbf{m3} 
        & \textbf{m1} & \textbf{m2} & \textbf{m3} \\
        \hline
        \textbf{n1} & x & y & z & x & y & z \\
        \textbf{n2} & x & z & z & z & x & y \\
        \textbf{n3} & x & y & x & y & z & x \\
        \hline
    \end{tabular}
    
\end{table}
\end{document}

Best Answer

Related Solutions

[Tex/LaTex] ltablex – Multi-column problem

Tables – Adjust Table Column Width When Using Multi-Column

Related Question