[Tex/LaTex] Booktabs: Vertically align to top for p{…}-formatted columns

booktabstables

I have a table that looks like this.

The latex code for the column definitions is the following:

\begin{tabular}{lp{4cm}cccccc}

How can I align the text in the second column with property p{4cm} vertically to the top in order to make the rows smaller?

The code below shows a MWE that produces exactly the error i mean:

\documentclass[a4paper, 8pt]{extarticle}

\usepackage{booktabs}

\usepackage[margin=0.5cm,landscape]{geometry}

\usepackage{mathtools}
\usepackage{amsmath}
\usepackage{amssymb}

\newcommand{\cW}{\mathcal{W}}
\newcommand{\N}{\mathbb{N}}

\newcommand{\Prob}[1]{P\left[#1\right]}
\newcommand{\Exp}[1]{Exp\left[#1\right]}
\newcommand{\Var}[1]{Var\left[#1\right]}
\newcommand{\riid}{\stackrel{\text{i.i.d}}{\sim}}
\newcommand{\set}[1]{\left\{ #1 \right\}}
\newcommand{\dset}[2]{\left\{ #1 \ \middle| \ #2 \right\}}

\begin{document}


\begin{tabular}{lp{4cm}cccccc}
\toprule
\textbf{Verteilung $\mu_X$}&
\textbf{Beschrieb von $X$} &
\textbf{$X=$} & 
\textbf{$\cW(X)=$} & 
\begin{tabular}{c}
\textbf{Gewichtsfunktion}\\
\textbf{$p_X(k)=\Prob{X=k}=$}
\end{tabular} &
\begin{tabular}{c}
\textbf{Verteilungsfunktion}\\
\textbf{$F_X(t)=\Prob{X\leq t}=$}
\end{tabular} &
\textbf{$\Exp{X}=$} & 
\textbf{$\Var{X}=$}
\\
\midrule
\vdots
\\
\hline
\begin{tabular}{c}
Binomial\\
$\displaystyle X\sim Bin(n,p)$
\end{tabular} 
& 
Anzahl der Erfolge $X$ bei $n$ unabhängigen 0-1-Experimenten mit
Erfolgsparameter $p$.
&
\begin{tabular}{c}
$\displaystyle \sum_{i=1}^{n} I_{A_i} =\sum_{i=1}^{n}Y_i$ \\
Sind die ZV $Y_1,\ldots,Y_n\riid Be(p)$, \\
so ist $X:=Y_1+\cdots+Y_n\sim Bin(n,p)$.
\end{tabular}
&
$\displaystyle \set{0,1,\ldots,n}$ 
&
$\displaystyle \binom{n}{k}p^k(1-p)^{n-k}$ 
&
$\displaystyle \sum_{k=0}^t \binom{n}{k}p^k(1-p)^{n-k}$ 
&
$np$ 
&
$np(1-p)$ 
\\
\hline
\begin{tabular}{c}
Geometrisch\\
$\displaystyle X\sim Geom(p)$ \\
\\
``gedächtnislos''
\end{tabular} &
Wartezeit $X$ auf den ersten Erfolg bei einer unendlichen Folge von
0-1-Experimenten mit Erfolgsparameter $p$. Spezialfall von $NB(1,p)$ 
&
\begin{tabular}{c}
$\inf\dset{i\in\N}{A_i\text{ tritt ein}}$\\
$=\inf\dset{n\in\N}{Y_i=1}$
\end{tabular} &
$\set{1,2,3,\ldots}$ &
$(1-p)^{k-1}p$ &
$1-(1-p)^t$ &
$\displaystyle \frac{1}{p}$ &
$\displaystyle \frac{1-p}{p^2}$
\\\hline
\vdots
\\\bottomrule
\end{tabular}


\end{document}

Best Answer

The problem is all the nested \begin{tabular}{c} inside your outer table, tabular are by default vertically centred which is what you show in your image, You could use \begin{tabular}[t]{c} so the inner tabular are top aligned (or better, \begin{tabular}[t]{@{}c@{}} so they don't add horizontal padding, but just removing all the inner tabular would be preferable.

I fixed several other errors in the example, undefined commands and using math italic for operators such as Exp.

\documentclass[a4paper, 8pt]{extarticle}

\usepackage{booktabs}

\usepackage[margin=0.5cm,landscape]{geometry}

\usepackage{mathtools}
\usepackage{amsmath,amsfonts}

\newcommand{\cW}{\mathcal{W}}

\newcommand{\Prob}[1]{P\left[#1\right]}
\newcommand{\Exp}[1]{\operatorname{Exp}\left[#1\right]}
\newcommand{\Var}[1]{\operatorname{Var}\left[#1\right]}
\newcommand{\riid}{\stackrel{\text{i.i.d}}{\sim}}
\newcommand{\set}[1]{\left\{ #1 \right\}}
\newcommand{\dset}[2]{\left\{ #1 \ \middle| \ #2 \right\}}
\DeclareMathOperator\Bin{\mathrm{Bin}}
\newcommand\N{\mathbb{N}}
\begin{document}


\begin{tabular}{lp{4cm}cccccc}
\toprule
\textbf{Verteilung $\mu_X$}&
\textbf{Beschrieb von $X$} &
\textbf{$X=$} & 
\textbf{$\cW(X)=$} & 
\begin{tabular}[t]{@{}c@{}}
\textbf{Gewichtsfunktion}\\
\textbf{$p_X(k)=\Prob{X=k}=$}
\end{tabular} &
\begin{tabular}[t]{@{}c@{}}
\textbf{Verteilungsfunktion}\\
\textbf{$F_X(t)=\Prob{X\leq t}=$}
\end{tabular} &
\textbf{$\Exp{X}=$} & 
\textbf{$\Var{X}=$}
\\
\midrule
\vdots
\\
\hline
\begin{tabular}[t]{c}
Binomial\\
$\displaystyle X\sim \Bin(n,p)$
\end{tabular} 
& 
Anzahl der Erfolge $X$ bei $n$ unabhängigen 0-1-Experimenten mit
Erfolgsparameter $p$.
&
\begin{tabular}[t]{@{}c@{}}
$\displaystyle \sum_{i=1}^{n} I_{A_i} =\sum_{i=1}^{n}Y_i$ \\
Sind die ZV $Y_1,\ldots,Y_n\riid Be(p)$, \\
so ist $X:=Y_1+\cdots+Y_n\sim Bin(n,p)$.
\end{tabular}
&
$\displaystyle \set{0,1,\ldots,n}$ 
&
$\displaystyle \binom{n}{k}p^k(1-p)^{n-k}$ 
&
$\displaystyle \sum_{k=0}^t \binom{n}{k}p^k(1-p)^{n-k}$ 
&
$\displaystyle np$ 
&
$np(1-p)$ 
\\
\hline
\begin{tabular}[t]{c}
Geometrisch\\
$\displaystyle X\sim Geom(p)$ \\
\\
``gedächtnislos''
\end{tabular} &
Wartezeit $X$ auf den ersten Erfolg bei einer unendlichen Folge von
0-1-Experimenten mit Erfolgsparameter $p$. Spezialfall von $NB(1,p)$ &
\begin{tabular}[t]{@{}c@{}}
$\displaystyle\inf\dset{i\in\N}{A_i\text{ tritt ein}}$\\
$\displaystyle{}=\inf\dset{n\in\N}{Y_i=1}$
\end{tabular} &
$\displaystyle\set{1,2,3,\ldots}$ &
$\displaystyle(1-p)^{k-1}p$ &
$\displaystyle1-(1-p)^t$ &
$\displaystyle \frac{1}{p}$ &
%???$\displaystyle \frac{1-p}{p^2}$ &
\\\hline
\vdots
\\\bottomrule
\end{tabular}


\end{document}

Or perhaps, despite the question title, you suggest in comments you really wanted centre alignment:

\documentclass[a4paper, 8pt]{extarticle}

\usepackage{booktabs,array}

\usepackage[margin=0.5cm,landscape]{geometry}

\usepackage{mathtools}
\usepackage{amsmath,amsfonts}

\newcommand{\cW}{\mathcal{W}}

\newcommand{\Prob}[1]{P\left[#1\right]}
\newcommand{\Exp}[1]{\operatorname{Exp}\left[#1\right]}
\newcommand{\Var}[1]{\operatorname{Var}\left[#1\right]}
\newcommand{\riid}{\stackrel{\text{i.i.d}}{\sim}}
\newcommand{\set}[1]{\left\{ #1 \right\}}
\newcommand{\dset}[2]{\left\{ #1 \ \middle| \ #2 \right\}}
\DeclareMathOperator\Bin{\mathrm{Bin}}
\newcommand\N{\mathbb{N}}
\begin{document}


\begin{tabular}{lm{4cm}cccccc}
\toprule
\textbf{Verteilung $\mu_X$}&
\textbf{Beschrieb von $X$} &
\textbf{$X=$} & 
\textbf{$\cW(X)=$} & 
\begin{tabular}[c]{@{}c@{}}
\textbf{Gewichtsfunktion}\\
\textbf{$p_X(k)=\Prob{X=k}=$}
\end{tabular} &
\begin{tabular}[c]{@{}c@{}}
\textbf{Verteilungsfunktion}\\
\textbf{$F_X(t)=\Prob{X\leq t}=$}
\end{tabular} &
\textbf{$\Exp{X}=$} & 
\textbf{$\Var{X}=$}
\\
\midrule
\vdots
\\
\hline
\begin{tabular}[c]{c}
Binomial\\
$\displaystyle X\sim \Bin(n,p)$
\end{tabular} 
& 
Anzahl der Erfolge $X$ bei $n$ unabhängigen 0-1-Experimenten mit
Erfolgsparameter $p$.
&
\begin{tabular}[c]{@{}c@{}}
$\displaystyle \sum_{i=1}^{n} I_{A_i} =\sum_{i=1}^{n}Y_i$ \\
Sind die ZV $Y_1,\ldots,Y_n\riid Be(p)$, \\
so ist $X:=Y_1+\cdots+Y_n\sim Bin(n,p)$.
\end{tabular}
&
$\displaystyle \set{0,1,\ldots,n}$ 
&
$\displaystyle \binom{n}{k}p^k(1-p)^{n-k}$ 
&
$\displaystyle \sum_{k=0}^t \binom{n}{k}p^k(1-p)^{n-k}$ 
&
$\displaystyle np$ 
&
$np(1-p)$ 
\\
\hline
\begin{tabular}[c]{c}
Geometrisch\\
$\displaystyle X\sim Geom(p)$ \\
\\
``gedächtnislos''
\end{tabular} &
Wartezeit $X$ auf den ersten Erfolg bei einer unendlichen Folge von
0-1-Experimenten mit Erfolgsparameter $p$. Spezialfall von $NB(1,p)$ &
\begin{tabular}[c]{@{}c@{}}
$\displaystyle\inf\dset{i\in\N}{A_i\text{ tritt ein}}$\\
$\displaystyle{}=\inf\dset{n\in\N}{Y_i=1}$
\end{tabular} &
$\displaystyle\set{1,2,3,\ldots}$ &
$\displaystyle(1-p)^{k-1}p$ &
$\displaystyle1-(1-p)^t$ &
$\displaystyle \frac{1}{p}$ &
%???$\displaystyle \frac{1-p}{p^2}$ &
\\\hline
\vdots
\\\bottomrule
\end{tabular}


\end{document}

Related Solutions

[Tex/LaTex] Default value for \addlinespace from booktabs

use it this way:

\documentclass{article}
\usepackage{array,booktabs}

\newcommand\myVSpace[1][10pt]{\rule[\normalbaselineskip]{0pt}{#1}}

\begin{document}

\begin{tabular}{|c|p{0.66\textwidth}|}\hline
\myVSpace\textbf{Title} & \textbf{A}\\[\defaultaddspace]
Row & B\\\hline
\myVSpace[20pt]Row & B\\\hline
\end{tabular}

\end{document}

or do not use booktabs and then \\[10pt] instead. enter image description here

The image shows how the optional argument of \rule works. I used \rule[...]{10pt}{#1} to make the box visible

[Tex/LaTex] Coloring columns in a table with colortbl and booktabs

There's no way to do this given the way booktabs inserts space between above and below its rules. As Herbert notes inthe comments, it's not clear that rules and colours are really a logical combination. So probably the best you can do is to increase the row height and set the space above and below the rules to 0.

\documentclass{article}
\usepackage{xcolor}
\usepackage{booktabs,colortbl,tabularx}

\begin{document}
% Your original table
  \begin{tabularx}{4cm}{>{\columncolor{gray}}cX>{\columncolor{gray}}c}
    \toprule
    1&2&3\\
    \midrule
    one&two&three\\
    one&two&three\\
    \cmidrule{2-3}
    one&two&three\\
    \midrule
    un&deux&trois\\
    \bottomrule
  \end{tabularx}
% same table with booktab rules but no above space and making rows bigger
\setlength{\aboverulesep}{0pt}
\setlength{\belowrulesep}{0pt}
\setlength{\extrarowheight}{.75ex}
  \begin{tabularx}{4cm}{>{\columncolor{gray}}cX>{\columncolor{gray}}c}
    \toprule
    1&2&3\\
    \midrule
    one&two&three\\
    one&two&three\\
    \cmidrule{2-3}
    one&two&three\\
    \midrule
    un&deux&trois\\
    \bottomrule
  \end{tabularx}
% same table without booktab rules
\setlength{\extrarowheight}{.75ex}
  \begin{tabularx}{4cm}{>{\columncolor{gray}}cX>{\columncolor{gray}}c}
    \hline
    1&2&3\\
    \hline
    one&two&three\\
    one&two&three\\
    \cline{2-3}
    one&two&three\\
    \hline
    un&deux&trois\\
    \hline
  \end{tabularx}
\end{document}

The middle table still looks marginally better than the third one.

three tables

Best Answer

Related Solutions

[Tex/LaTex] Default value for \addlinespace from booktabs

[Tex/LaTex] Coloring columns in a table with colortbl and booktabs

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