[Tex/LaTex] Arranging rows and columns [Tabular]

tables

Hello there, here is my table but I can't figure out how to arrange (x,y,z) coordinates in a good looking(i mean they are not in a straight line 🙂 ) and also every box , they start appear center-north, for good looking, I want to center-center 🙂 sorry for bad language. thanks advance.

Here is my code, sorry for not arranging it, I haven't get used to TeX.SE yet 🙂

\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|}
  \hline
  Step(k) & I.strategy & II.strategy & $\,$ Return $\,$ & Total Return &$\quad$ $g_{i}^r$ $\quad$ &$\quad$ $g_{k}^c$ $\quad$ & $\,$ $U_k$ $\,$ & $\,$ $V_k$  $\,$ & $\displaystyle \frac{minU_k}{k}$ & $ \displaystyle \frac{maxV_k}{k}$ \\[2ex] \hline
   1 & $I_1$ & $II_1$ & 9 & 9 & (9,2,-1) & (9,-9,6) & (9,2,{\color{blue}-1})     & ({\color{red}9},-9,6)     & -1              & 9               \\[1.5ex] \hline
   2 & $I_1$ & $II_3$ &-1 & 8 & (9,2,-1) & (-1,0,2) & (18,4,{\color{blue}-2})    & ({\color{red}8},-9,8)     & -1              & 4               \\[1.5ex] \hline 
   3 & $I_1$ & $II_3$ &-1 & 7 & (9,2,-1) & (-1,0,2) & (27,6,{\color{blue}-3})    & (7,-9,{\color{red}10})    & -1              & $\frac{10}{3}$     \\[1.5ex] \hline
   4 & $I_3$ & $II_3$ & 2 & 9 & (6,-6,2) & (-1,0,2) & (33,0,{\color{blue}-1})    & (6,-9,{\color{red}12})    & $-\frac{1}{4}$  & $3$              \\[1.5ex] \hline
   5 & $I_3$ & $II_3$ & 2 &11 & (6,-6,2) & (-1,0,2) & (39,{\color{blue}-6},1)    & (5,-9,{\color{red}14})    & $-\frac{6}{5}$  & $\frac{14}{5}$     \\[1.5ex] \hline 
   6 & $I_3$ & $II_2$ &-6 & 5 & (6,-6,2) & (2,9,-6) & (45,{\color{blue}-12},3)   & (7,0,{\color{red}8})      & $-2$            & $\frac{4}{3}$     \\[1.5ex] \hline
   7 & $I_3$ & $II_2$ &-6 &-1 & (6,-6,2) & (2,9,-6) & (51,{\color{blue}-18},5)   & ({\color{red}9},9,2)      & $-\frac{18}{7}$ & $\frac{9}{7}$     \\[1.5ex] \hline
   8 & $I_1$ & $II_2$ & 2 & 1 & (9,2,-1) & (2,9,-6) & (60,{\color{blue}-16},4)   & (11,{\color{red}18},-4)   & $-2$            & $\frac{9}{4}$     \\[1.5ex] \hline 
   9 & $I_2$ & $II_2$ & 9 &10 & (-9,9,0) & (2,9,-6) & (51,{\color{blue}-7},4)    & (13,{\color{red}17},-10)  & $-\frac{7}{9}$  & $\frac{17}{9}$     \\[1.5ex] \hline
  10 & $I_2$ & $II_2$ & 9 &19 & (-9,9,0) & (2,9,-6) & (42,{\color{blue}2},4)     & (15,{\color{red}26},-16)  & $\frac{1}{5}$   & $\frac{13}{5}$     \\[1.5ex] \hline 
  11 & $I_2$ & $II_2$ & 9 &28 & (-9,9,0) & (2,9,-6) & (33,11,{\color{blue}4})    & (17,{\color{red}35},-22)  & $\frac{4}{11}$  & $\frac{35}{11}$     \\[1.5ex] \hline
  12 & $I_2$ & $II_3$ & 0 &28 & (-9,9,0) & (-1,0,2) & (24,20,{\color{blue}4})    & (16,{\color{red}35},-20)  & $\frac{1}{3}$   & $\frac{35}{12}$     \\[1.5ex] \hline 
  13 & $I_2$ & $II_3$ & 0 &28 & (-9,9,0) & (-1,0,2) & (15,29,{\color{blue}4})    & (15,{\color{red}35},-18)  & $\frac{4}{13}$  & $\frac{35}{13}$     \\[1.5ex] \hline
  14 & $I_2$ & $II_3$ & 0 &28 & (-9,9,0) & (-1,0,2) & (6,38,{\color{blue}4})     & (14,{\color{red}35},-16)  & $\frac{4}{14}$  & $\frac{35}{14}$     \\[1.5ex] \hline
  15 & $I_2$ & $II_3$ & 0 &28 & (-9,9,0) & (-1,0,2) & ({\color{blue}-3},47,4)    & (13,{\color{red}35},-14)  & $-\frac{3}{5}$  & $\frac{7}{3}$     \\[1.5ex] \hline
\end{tabular}

Best Answer

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}

Related Solutions

[Tex/LaTex] Coloring rows and columns in a tabular environment

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:

enter image description here

\documentclass[12pt]{article}
\usepackage[margin=1cm,landscape,paperwidth=3in,paperheight=3in]{geometry}
\usepackage{amsmath,booktabs,colortbl}
\usepackage[table]{xcolor}

\makeatletter
\def\tmpp#1\@addtopreamble#2#3!{%
    \tmp#2!{#1}{#3}}



\def\tmp#1\CT@column@color\CT@row@color#2!#3#4{%
\def\@classz{#3\@addtopreamble{#1\CT@row@color\CT@column@color#2}#4}}

\expandafter\tmpp\@classz!
\makeatother
\pagestyle{empty}

\begin{document}
\newcolumntype{M}{>{$}c<{$}}
\newcolumntype{P}{>{\columncolor{gray!75}}M}
\rowcolors{2}{gray!25}{}
\begin{tabular}{|MPM|}
\hline
x_1 & y_1 & z_1 \\
x_2 & y_2 & z_2 \\
x_3 & y_3 & z_3 \\
x_4 & y_4 & z_4 \\
x_5 & y_5 & z_5 \\
x_6 & y_6 & z_6 \\
\hline
\end{tabular}
\end{document}

[Tex/LaTex] Table with multiple rows and columns

Assuming you want to stay with the tabularx table type that was used in my answer to the posting Table rowspan and colspan, you may achieve your objective using the methods set forth in section 4.3 of the user guide of the tabularx package. The method described there works by adjusting the relative widths of the columns of type X. (Naturally, if your tabularx table contains only a single column of type X, its width is fully determined as a residual, viz., as the difference between the overall text width and the sum of the widths of the other columns and intercolumn spaces.)

The following example shows how to this may be done to create a table in which the first and fourth columns are 50% wider than the two middle columns. Note how the four \hsize values -- 1.2, 0.8, 0.8, and 1.2 -- sum to 4, which is the number of columns of type X (or, to be even more precise, Y, where Y is a modified form of X).

enter image description here

\documentclass{article}
\usepackage[margin=1in]{geometry}
\usepackage{multirow,tabularx}
\newcolumntype{Y}{>{\centering\arraybackslash}X}
\renewcommand{\arraystretch}{2}
\begin{document}

\emph{Original form: All columns are equally wide.}

\noindent
\begin{tabularx}{\textwidth}{|*{4}{Y|}}
\hline
\multirow{2}{*}{State of Health} 
&\multicolumn{2}{c|}{Fasting Value}&After Eating\\
\cline{2-4}
&Minimum       &Maximum &2 hours after eating\\
\hline
Healthy      &70            &100     &Less than 140\\
\hline
Pre-Diabetes &101           &126     &140 to 200\\
\hline
Diabetes     &More than 126 &N/A     &More than 200\\
\hline
\end{tabularx}

\bigskip
\emph{Modified form: Columns 1 and 4 are 50\% wider than columns 2 and 3.}

\smallskip\noindent
\begin{tabularx}{\textwidth}{|
 >{\hsize=1.2\hsize}Y| 
 >{\hsize=0.8\hsize}Y|
 >{\hsize=0.8\hsize}Y|
 >{\hsize=1.2\hsize}Y|}
\hline
\multirow{2}{*}{State of Health} 
&\multicolumn{2}{c|}{Fasting Value}&After Eating\\
\cline{2-4}
&Minimum       &Maximum &2 hours after eating\\
\hline
Healthy      &70            &100     &Less than 140\\
\hline
Pre-Diabetes &101           &126     &140 to 200\\
\hline
Diabetes     &More than 126 &N/A     &More than 200\\
\hline
\end{tabularx}
\end{document}

Best Answer

Related Solutions

[Tex/LaTex] Coloring rows and columns in a tabular environment

[Tex/LaTex] Table with multiple rows and columns

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