Aligning table of data using dcolumn

Question

I have a table with one index column and three columns of data. I would like the table have the data aligned using the decimal point. Following the answer to this question I try the following code:

\documentclass{article}
\usepackage{booktabs,subcaption,amsfonts,dcolumn} 
\newcolumntype{d}[1]{D..{#1}}
\newcommand\mc[1]{\multicolumn{1}{c}{#1}} % handy shortcut macro
\begin{document}

\begin{table}
\centering\small
\begin{tabular}[t]{@{} l c *{3}{d{1.11}} @{}}
\toprule
 \mc{$n$} & \mc{$\nu=1/3$} & \mc{$\nu=1/5$} & \mc{$\nu=1/7$} \\
\midrule
   1 &         -2.6448(2) &         -3.5613(2) &         -3.7317(2) \\
   2 &          1.0022(1) &          2.8438(1) &          3.4644(1) \\
   3 &          0.0613(4) &         -1.8256(4) &         -3.0903(4) \\
   4 &         -0.4103(4) &          0.7144(4) &          2.4534(4) \\
   5 &          0.3928(6) &          0.1675(6) &         -1.5920(6) \\
   6 &         -0.2573(5) &         -0.6854(5) &          0.6627(5) \\
   7 &          0.1225(5) &          0.8790(5) &          0.1630(5) \\
\bottomrule
\end{tabular}
\caption{Expansion coefficients for various filling factors.}
\label{tab:LaughlinScaledPairCorrCoeffs}
\end{table}

However there are a couple of problems: the header row is not aligned with the data rows, the numbers in the first data column are not aligned following the decimal point, and the third data column has more space than the others.

Answer 1

Some suggestions

• Use d{2.7} instead of d{1.11} for the numerical columns. Don't forget to set aside some space for the - (minus) symbols -- hence the number 2 ahead of the decimal marker. The widths of ( and ) are about half that of the numerals; hence the suggestion to use 7 for the number of digits after the decimal marker.

• Don't encase $n$ in the very first header cell in \mc, as its column type is l.

• Get rid of the unused c column

\documentclass{article}
\usepackage{booktabs,dcolumn} 
\newcolumntype{d}[1]{D..{#1}}
\newcommand\mc[1]{\multicolumn{1}{c}{#1}} % handy shortcut macro
\begin{document}

\begin{table}
\centering
%\small
\begin{tabular}{@{} l *{3}{d{2.7}} @{}}
\toprule
 $n$ & \mc{$\nu=1/3$} & \mc{$\nu=1/5$} & \mc{$\nu=1/7$} \\
\midrule
   1 &         -2.6448(2) &         -3.5613(2) &         -3.7317(2) \\
   2 &          1.0022(1) &          2.8438(1) &          3.4644(1) \\
   3 &          0.0613(4) &         -1.8256(4) &         -3.0903(4) \\
   4 &         -0.4103(4) &          0.7144(4) &          2.4534(4) \\
   5 &          0.3928(6) &          0.1675(6) &         -1.5920(6) \\
   6 &         -0.2573(5) &         -0.6854(5) &          0.6627(5) \\
   7 &          0.1225(5) &          0.8790(5) &          0.1630(5) \\
\bottomrule
\end{tabular}
\caption{Expansion coefficients for various filling factors.}
\label{tab:LaughlinScaledPairCorrCoeffs}
\end{table}

\end{document}

Answer 2

Same code with minor modification

\documentclass{article}
\usepackage{booktabs,subcaption,amsfonts,dcolumn} 
\newcolumntype{d}[1]{D..{#1}}
\newcommand\mc[1]{\multicolumn{1}{r}{#1}} % handy shortcut macro

\begin{document}
\begin{table}
\centering\small
\begin{tabular}[t]{r l r*{3}{d{1.11}}}
\toprule
\mc{$n$} & \mc{$\nu=1/3$} & \mc{$\nu=1/5$} & \mc{$\nu=1/7$} \\
\midrule
 1 &    \mc{-2.6448(2) } &   \mc{ -3.5613(2)} &    \mc{ -3.7317(2)} \\
 2 &     \mc{1.0022(1) } &   \mc{  2.8438(1) }&     \mc{ 3.4644(1) }\\
 3 &    \mc{ 0.0613(4)} &    \mc{ -1.8256(4) }&    \mc{  -3.0903(4) }\\
 4 &    \mc{ -0.4103(4) } &    \mc{ 0.7144(4)} &    \mc{    2.4534(4) }\\
 5 &     \mc{ 0.3928(6) } &    \mc{ 0.1675(6) }&     \mc{  -1.5920(6)} \\
 6 &     \mc{0.2573(5) } &  \mc{  -0.6854(5) }&    \mc{  0.6627(5)} \\
 7 &     \mc{ 0.1225(5)}  &     \mc{ 0.8790(5)} &    \mc{ 0.1630(5) }\\
 \bottomrule
 \end{tabular}
\caption{Expansion coefficients for various filling factors.}
\label{tab:LaughlinScaledPairCorrCoeffs}
\end{table}
\end{document}