I suggest a more open look to the table and also some tricks for simplifying the input.
However, I believe that the slashed form 10/3
would render better than \frac{10}{3}
.
\documentclass{article}
\usepackage[landscape,a4paper]{geometry}
\usepackage{array,booktabs}
\usepackage{amsmath}
\usepackage{xcolor}
\newcommand{\bn}[1]{% blue number
\textcolor{blue}{#1}%
}
\newcommand{\rn}[1]{% red number
\textcolor{red}{#1}%
}
\begin{document}
\begin{table}
\centering
\renewcommand{\arraystretch}{1.2}
\addtolength{\tabcolsep}{-.2pt}
\begin{tabular}{
*{3}{c}
*{8}{>{$}c<{$}}
}
\toprule
Step($k$) &
Strategy I &
Strategy II &
\multicolumn{1}{c}{Return} &
\multicolumn{1}{c}{Total Return} &
g_{i}^r &
g_{k}^c &
U_k &
V_k &
\dfrac{\min U_k}{k} &
\dfrac{\max V_k}{k} \\
\midrule
1 & I$_1$ & II$_1$ & 9 & 9 & (9,2,-1) & (9,-9,6) & (9,2,\bn{-1}) & (\rn{9},-9,6) & -1 & 9 \\
2 & I$_1$ & II$_3$ &-1 & 8 & (9,2,-1) & (-1,0,2) & (18,4,\bn{-2}) & (\rn{8},-9,8) & -1 & 4 \\
3 & I$_1$ & II$_3$ &-1 & 7 & (9,2,-1) & (-1,0,2) & (27,6,\bn{-3}) & (7,-9,\rn{10}) & -1 & \frac{10}{3} \\
4 & I$_3$ & II$_3$ & 2 & 9 & (6,-6,2) & (-1,0,2) & (33,0,\bn{-1}) & (6,-9,\rn{12}) & -\frac{1}{4} & 3 \\
5 & I$_3$ & II$_3$ & 2 &11 & (6,-6,2) & (-1,0,2) & (39,\bn{-6},1) & (5,-9,\rn{14}) & -\frac{6}{5} & \frac{14}{5} \\
6 & I$_3$ & II$_2$ &-6 & 5 & (6,-6,2) & (2,9,-6) & (45,\bn{-12},3) & (7,0,\rn{8}) & -2 & \frac{4}{3} \\
7 & I$_3$ & II$_2$ &-6 &-1 & (6,-6,2) & (2,9,-6) & (51,\bn{-18},5) & (\rn{9},9,2) & -\frac{18}{7} & \frac{9}{7} \\
8 & I$_1$ & II$_2$ & 2 & 1 & (9,2,-1) & (2,9,-6) & (60,\bn{-16},4) & (11,\rn{18},-4) & -2 & \frac{9}{4} \\
9 & I$_2$ & II$_2$ & 9 &10 & (-9,9,0) & (2,9,-6) & (51,\bn{-7},4) & (13,\rn{17},-10) & -\frac{7}{9} & \frac{17}{9} \\
10 & I$_2$ & II$_2$ & 9 &19 & (-9,9,0) & (2,9,-6) & (42,\bn{2},4) & (15,\rn{26},-16) & \frac{1}{5} & \frac{13}{5} \\
11 & I$_2$ & II$_2$ & 9 &28 & (-9,9,0) & (2,9,-6) & (33,11,\bn{4}) & (17,\rn{35},-22) & \frac{4}{11} & \frac{35}{11} \\
12 & I$_2$ & II$_3$ & 0 &28 & (-9,9,0) & (-1,0,2) & (24,20,\bn{4}) & (16,\rn{35},-20) & \frac{1}{3} & \frac{35}{12} \\
13 & I$_2$ & II$_3$ & 0 &28 & (-9,9,0) & (-1,0,2) & (15,29,\bn{4}) & (15,\rn{35},-18) & \frac{4}{13} & \frac{35}{13} \\
14 & I$_2$ & II$_3$ & 0 &28 & (-9,9,0) & (-1,0,2) & (6,38,\bn{4}) & (14,\rn{35},-16) & \frac{4}{14} & \frac{35}{14} \\
15 & I$_2$ & II$_3$ & 0 &28 & (-9,9,0) & (-1,0,2) & (\bn{-3},47,4) & (13,\rn{35},-14) & -\frac{3}{5} & \frac{7}{3} \\
\bottomrule
\end{tabular}
\end{table}
\end{document}
Best Answer
colortbl fairly arbitrarily gives row color precedence over column color, if you wish to reverse that decision you just need to change the order of
\CT@column@color
and\CT@row@color
You can do this by applying a patch macro as below: