LaTeX 테이블 라인이 연결되지 않았습니다

Question

내 테이블의 선이 연결되어 있지 않습니다.
모든 선을 연결해야 하는 테이블이 필요합니다.

테이블의 LaTeX 코드는 다음과 같습니다.

\begin{table}[]
\centering
\resizebox{\textwidth}{!}{%
\begin{tabular}{|c|ccccccccccccc|}
\hline
\multirow{2}{*}{Constant Value of the real part} & \multicolumn{13}{c|}{Binary representation of real part} \\ \cline{2-14} 
 & \multicolumn{1}{c|}{20} & \multicolumn{2}{c|}{2-1} & \multicolumn{1}{c|}{2-2} & \multicolumn{1}{c|}{2-3} & \multicolumn{1}{c|}{2-4} & \multicolumn{1}{c|}{2-5} & \multicolumn{1}{c|}{2-6} & \multicolumn{1}{c|}{2-7} & \multicolumn{1}{c|}{2-8} & \multicolumn{1}{c|}{2-9} & \multicolumn{1}{c|}{2-10} & 2-11 \\ \hline
0.9988 & \multicolumn{2}{c|}{1} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{-1} & \multicolumn{1}{c|}{0} & 1 \\ \hline
0.9952 & \multicolumn{2}{c|}{1} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{-1} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{-1} & 0 \\ \hline
0.9892 & \multicolumn{2}{c|}{1} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{-1} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{1} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{0} & 1 \\ \hline
0.9808 & \multicolumn{2}{c|}{1} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{-1} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{-1} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{0} & 0 \\ \hline
0.9697 & \multicolumn{2}{c|}{1} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{-1} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{1} & 0 \\ \hline
0.9569 & \multicolumn{2}{c|}{1} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{-1} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{1} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{1} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{0} & -1 \\ \hline
0.9415 & \multicolumn{2}{c|}{1} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{-1} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{1} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{0} & 0 \\ \hline
0.9239 & \multicolumn{2}{c|}{1} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{-1} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{-1} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{1} & \multicolumn{1}{c|}{0} & 0 \\ \hline
0.9040 & \multicolumn{2}{c|}{1} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{-1} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{1} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{-1} & \multicolumn{1}{c|}{0} & -1 \\ \hline
0.8819 & \multicolumn{2}{c|}{1} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{-1} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{1} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{-1} & 0 \\ \hline
0.8577 & \multicolumn{2}{c|}{1} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{-1} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{-1} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{-1} & \multicolumn{1}{c|}{0} & 0 \\ \hline
0.8315 & \multicolumn{2}{c|}{1} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{-1} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{1} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{1} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{-1} & 0 \\ \hline
0.8032 & \multicolumn{2}{c|}{1} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{-1} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{-1} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{-1} & \multicolumn{1}{c|}{0} & 0 \\ \hline
0.7730 & \multicolumn{2}{c|}{1} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{-1} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{1} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{-1} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{0} & 1 \\ \hline
0.7410 & \multicolumn{2}{c|}{1} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{-1} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{-1} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{-1} & \multicolumn{1}{c|}{0} & 1 \\ \hline
0.7071 & \multicolumn{2}{c|}{1} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{-1} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{1} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{1} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{0} & 0 \\ \hline
\end{tabular}%
}
\end{table}

Answer 1

문제는 코드에서 두 번째 열에 사용 \multicolumn{2}{}{}하지만 두 번째 행에서는 세 번째 열에 사용한다는 것입니다. 물론 이것은 정렬되지 않습니다.

나는 당신이 제시하는 코드가 읽기 어렵고 유지 관리하기 어려운 코드를 생성하지 않는 일부 테이블 생성기에 의해 생성되었다고 가정합니다. 따라서 코드를 정리하고 불필요한 \multicolumn명령을 모두 제거하는 것이 좋습니다. 그러면 상황이 훨씬 쉬울 수 있으며 \resizebox글꼴 크기가 왜곡되므로 나쁜 습관을 사용하여 표 형식의 크기를 조정할 필요가 없습니다 .

\documentclass{article}
\usepackage[margin=2cm]{geometry}
\usepackage{multirow}

\begin{document}
    
\begin{tabular}{ | *{13}{c|} }
\hline
\multirow{2}{2.5cm}{\centering Constant Value of the real part} & 
    \multicolumn{12}{c|}{Binary representation of real part} \\ \cline{2-13} 
 & 20 & 2-1 & 2-2 & 2-3 & 2-4 & 2-5 & 2-6 & 2-7 & 2-8 & 2-9 & 2-10 & 2-11 \\ \hline
0.9988 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 & 0 & 1 \\ \hline
0.9952 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 & 0 & -1 & 0 \\ \hline
0.9892 & 1 & 0 & 0 & 0 & 0 & 0 & -1 & 0 & 1 & 0 & 0 & 1 \\ \hline
0.9808 & 1 & 0 & 0 & 0 & 0 & 0 & -1 & 0 & -1 & 0 & 0 & 0 \\ \hline
0.9697 & 1 & 0 & 0 & 0 & 0 & -1 & 0 & 0 & 0 & 0 & 1 & 0 \\ \hline
0.9569 & 1 & 0 & 0 & 0 & -1 & 0 & 1 & 0 & 1 & 0 & 0 & -1 \\ \hline
0.9415 & 1 & 0 & 0 & 0 & -1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ \hline
0.9239 & 1 & 0 & 0 & 0 & -1 & 0 & -1 & 0 & 0 & 1 & 0 & 0 \\ \hline
0.9040 & 1 & 0 & 0 & -1 & 0 & 1 & 0 & 0 & 0 & -1 & 0 & -1 \\ \hline
0.8819 & 1 & 0 & 0 & -1 & 0 & 0 & 0 & 1 & 0 & 0 & -1 & 0 \\ \hline
0.8577 & 1 & 0 & 0 & -1 & 0 & 0 & -1 & 0 & 0 & -1 & 0 & 0 \\ \hline
0.8315 & 1 & 0 & -1 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & -1 & 0 \\ \hline
0.8032 & 1 & 0 & -1 & 0 & 0 & 0 & 0 & -1 & 0 & -1 & 0 & 0 \\ \hline
0.7730 & 1 & 0 & -1 & 0 & 0 & 1 & 0 & -1 & 0 & 0 & 0 & 1 \\ \hline
0.7410 & 1 & 0 & -1 & 0 & 0 & 0 & 0 & -1 & 0 & -1 & 0 & 1 \\ \hline
0.7071 & 1 & 0 & -1 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 \\ \hline
\end{tabular}
    
\end{document}

빼기 기호를 적절하게 조판하려면 관련 셀에서 수학 모드를 사용하는 것이 좋습니다(힌트를 제공한 Mico에게 감사드립니다).

\documentclass{article}
\usepackage[margin=2cm]{geometry}
\usepackage{array, multirow}
\newcolumntype{C}{>{$}c<{$}}

\begin{document}

\begin{tabular}{ | *{13}{C|} }
\hline
\multirow{2}{2.5cm}{\centering Constant Value of the real part} & 
    \multicolumn{12}{c|}{Binary representation of real part} \\ \cline{2-13} 
 & \textrm{2-0} & \textrm{2-1} & \textrm{2-2} & \textrm{2-3} & \textrm{2-4} & \textrm{2-5} & \textrm{2-6} & \textrm{2-7} & \textrm{2-8} & \textrm{2-9} & \textrm{2-10} & \textrm{2-11} \\ \hline
0.9988 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 & 0 & 1 \\ \hline
0.9952 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 & 0 & -1 & 0 \\ \hline
0.9892 & 1 & 0 & 0 & 0 & 0 & 0 & -1 & 0 & 1 & 0 & 0 & 1 \\ \hline
0.9808 & 1 & 0 & 0 & 0 & 0 & 0 & -1 & 0 & -1 & 0 & 0 & 0 \\ \hline
0.9697 & 1 & 0 & 0 & 0 & 0 & -1 & 0 & 0 & 0 & 0 & 1 & 0 \\ \hline
0.9569 & 1 & 0 & 0 & 0 & -1 & 0 & 1 & 0 & 1 & 0 & 0 & -1 \\ \hline
0.9415 & 1 & 0 & 0 & 0 & -1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ \hline
0.9239 & 1 & 0 & 0 & 0 & -1 & 0 & -1 & 0 & 0 & 1 & 0 & 0 \\ \hline
0.9040 & 1 & 0 & 0 & -1 & 0 & 1 & 0 & 0 & 0 & -1 & 0 & -1 \\ \hline
0.8819 & 1 & 0 & 0 & -1 & 0 & 0 & 0 & 1 & 0 & 0 & -1 & 0 \\ \hline
0.8577 & 1 & 0 & 0 & -1 & 0 & 0 & -1 & 0 & 0 & -1 & 0 & 0 \\ \hline
0.8315 & 1 & 0 & -1 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & -1 & 0 \\ \hline
0.8032 & 1 & 0 & -1 & 0 & 0 & 0 & 0 & -1 & 0 & -1 & 0 & 0 \\ \hline
0.7730 & 1 & 0 & -1 & 0 & 0 & 1 & 0 & -1 & 0 & 0 & 0 & 1 \\ \hline
0.7410 & 1 & 0 & -1 & 0 & 0 & 0 & 0 & -1 & 0 & -1 & 0 & 1 \\ \hline
0.7071 & 1 & 0 & -1 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 \\ \hline
\end{tabular}

\end{document}

---

또 다른 접근 방식은 패키지를 사용 booktabs하고 수직 규칙을 완전히 제거하는 것입니다.

\documentclass{article}
\usepackage[margin=2cm]{geometry}
\usepackage{array, multirow, booktabs}
\newcolumntype{C}{>{$}c<{$}}

\begin{document}

\begin{tabular}{ *{13}{C} }
\toprule
 & \multicolumn{12}{c}{Binary representation of real part} \\ \cmidrule{2-13} 
\multirow{-2}{2.5cm}{\centering Constant Value of the real part} 
 & \textrm{2-0} & \textrm{2-1} & \textrm{2-2} & \textrm{2-3} & \textrm{2-4} & \textrm{2-5} & \textrm{2-6} & \textrm{2-7} & \textrm{2-8} & \textrm{2-9} & \textrm{2-10} & \textrm{2-11} \\ 
\midrule
0.9988 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 & 0 & 1 \\ 
0.9952 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 & 0 & -1 & 0 \\ 
0.9892 & 1 & 0 & 0 & 0 & 0 & 0 & -1 & 0 & 1 & 0 & 0 & 1 \\ 
0.9808 & 1 & 0 & 0 & 0 & 0 & 0 & -1 & 0 & -1 & 0 & 0 & 0 \\ 
0.9697 & 1 & 0 & 0 & 0 & 0 & -1 & 0 & 0 & 0 & 0 & 1 & 0 \\ 
0.9569 & 1 & 0 & 0 & 0 & -1 & 0 & 1 & 0 & 1 & 0 & 0 & -1 \\ 
0.9415 & 1 & 0 & 0 & 0 & -1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 
0.9239 & 1 & 0 & 0 & 0 & -1 & 0 & -1 & 0 & 0 & 1 & 0 & 0 \\ 
0.9040 & 1 & 0 & 0 & -1 & 0 & 1 & 0 & 0 & 0 & -1 & 0 & -1 \\ 
0.8819 & 1 & 0 & 0 & -1 & 0 & 0 & 0 & 1 & 0 & 0 & -1 & 0 \\ 
0.8577 & 1 & 0 & 0 & -1 & 0 & 0 & -1 & 0 & 0 & -1 & 0 & 0 \\ 
0.8315 & 1 & 0 & -1 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & -1 & 0 \\ 
0.8032 & 1 & 0 & -1 & 0 & 0 & 0 & 0 & -1 & 0 & -1 & 0 & 0 \\ 
0.7730 & 1 & 0 & -1 & 0 & 0 & 1 & 0 & -1 & 0 & 0 & 0 & 1 \\ 
0.7410 & 1 & 0 & -1 & 0 & 0 & 0 & 0 & -1 & 0 & -1 & 0 & 1 \\ 
0.7071 & 1 & 0 & -1 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 \\ 
\bottomrule
\end{tabular}

\end{document}

Answer 2

tabular*테이블이 텍스트 블록 내부에 맞는지 또는 동등하게 테이블의 너비 가 \textwidth. 또한 12개 데이터 열의 너비가 동일한지 확인하고 암시된 소수점 표시에 숫자를 정렬했습니다. 아, 그리고 모든 수직 규칙을 없애고 훨씬 적은 수의 수평 규칙을 사용하겠습니다.

\documentclass{article}
\usepackage{amsmath,booktabs,siunitx}
\newcommand\mytab[1]{\smash[b]{%
   \begin{tabular}[t]{@{}l@{}} #1 \end{tabular}}}
\newcommand\mc[1]{\multicolumn{1}{c}{#1}} % handy shortcut macro

\begin{document}

\begin{table}
\setlength\tabcolsep{0pt} % make LaTeX figure out the amount of intercolumn whitespace

\begin{tabular*}{\textwidth}{@{\extracolsep{\fill}} 
    c 
    *{12}{S[table-format=-1.0]} @{}}
\toprule
\mytab{Constant value\\of real part} & 
\multicolumn{12}{c}{Binary representation of real part} \\ 
\cmidrule{2-13} 
  & \mc{2-0} & \mc{2-1} & \mc{2-2} & \mc{2-3} 
  & \mc{2-4} & \mc{2-5} & \mc{2-6} & \mc{2-7} 
  & \mc{2-8} & \mc{2-9} & \mc{\llap{2}-10}& \mc{\llap{2}-11} \\ 
\midrule
0.9988 & 1 & 0 &  0 &  0 &  0 &  0 &  0 &  0 &  0 & -1 &  0 &  1 \\
0.9952 & 1 & 0 &  0 &  0 &  0 &  0 &  0 &  0 & -1 &  0 & -1 &  0 \\
0.9892 & 1 & 0 &  0 &  0 &  0 &  0 & -1 &  0 &  1 &  0 &  0 &  1 \\
0.9808 & 1 & 0 &  0 &  0 &  0 &  0 & -1 &  0 & -1 &  0 &  0 &  0 \\
\addlinespace
0.9697 & 1 & 0 &  0 &  0 &  0 & -1 &  0 &  0 &  0 &  0 &  1 &  0 \\
0.9569 & 1 & 0 &  0 &  0 & -1 &  0 &  1 &  0 &  1 &  0 &  0 & -1 \\
0.9415 & 1 & 0 &  0 &  0 & -1 &  0 &  0 &  0 &  1 &  0 &  0 &  0 \\
0.9239 & 1 & 0 &  0 &  0 & -1 &  0 & -1 &  0 &  0 &  1 &  0 &  0 \\
\addlinespace
0.9040 & 1 & 0 &  0 & -1 &  0 &  1 &  0 &  0 &  0 & -1 &  0 & -1 \\
0.8819 & 1 & 0 &  0 & -1 &  0 &  0 &  0 &  1 &  0 &  0 & -1 &  0 \\
0.8577 & 1 & 0 &  0 & -1 &  0 &  0 & -1 &  0 &  0 & -1 &  0 &  0 \\
0.8315 & 1 & 0 & -1 &  0 &  0 &  0 &  1 &  0 &  1 &  0 & -1 &  0 \\
\addlinespace
0.8032 & 1 & 0 & -1 &  0 &  0 &  0 &  0 & -1 &  0 & -1 &  0 &  0 \\
0.7730 & 1 & 0 & -1 &  0 &  0 &  1 &  0 & -1 &  0 &  0 &  0 &  1 \\
0.7410 & 1 & 0 & -1 &  0 &  0 &  0 &  0 & -1 &  0 & -1 &  0 &  1 \\
0.7071 & 1 & 0 & -1 &  0 &  0 &  0 &  1 &  0 &  1 &  0 &  0 &  0 \\ 
\bottomrule
\end{tabular*}

\end{table}

\end{document} 

Answer 3

나는 당신이 테이블에 대한 좋은 코드를 생성한다고 주장하는 일부 소프트웨어의 피해자인 것 같습니다. 분명히 완전히 잘못된 것입니다.\multicolumn 다음에 대한 선언을모든열에 항목을 입력합니다.

게다가 소프트웨어가 열 수를 잘못 입력했습니다.

booktabs다음은 열 사이 공간의 자동 크기 조정을 사용하여 다른 방식으로 구현한 것입니다 . 크기 조정을 피하십시오.

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

\begin{document}

\begin{table}[htp]
\centering\small

\setlength{\tabcolsep}{0pt}% let TeX compute
\begin{tabular*}{\textwidth}{
  @{\extracolsep{\fill}}
  c
  *{12}{>{$}c<{$}}
}
\toprule
\smash{\begin{tabular}[t]{@{}c@{}} Real \\ part \end{tabular}}
& \multicolumn{12}{c}{Binary representation of real part} \\
\cmidrule{2-13}
& \text{2--0}
& \text{2--1} & \text{2--2} & \text{2--3} & \text{2--4} 
& \text{2--5} & \text{2--6} & \text{2--7} & \text{2--8} 
& \text{2--9} & \text{2--10} & \text{2--11} \\
\midrule
0.9988 & 1 & 0 &  0 &  0 &  0 &  0 &  0 &  0 &  0 & -1 &  0 &  1 \\
0.9952 & 1 & 0 &  0 &  0 &  0 &  0 &  0 &  0 & -1 &  0 & -1 &  0 \\
0.9892 & 1 & 0 &  0 &  0 &  0 &  0 & -1 &  0 &  1 &  0 &  0 &  1 \\
0.9808 & 1 & 0 &  0 &  0 &  0 &  0 & -1 &  0 & -1 &  0 &  0 &  0 \\
0.9697 & 1 & 0 &  0 &  0 &  0 & -1 &  0 &  0 &  0 &  0 &  1 &  0 \\
0.9569 & 1 & 0 &  0 &  0 & -1 &  0 &  1 &  0 &  1 &  0 &  0 & -1 \\
0.9415 & 1 & 0 &  0 &  0 & -1 &  0 &  0 &  0 &  1 &  0 &  0 &  0 \\
0.9239 & 1 & 0 &  0 &  0 & -1 &  0 & -1 &  0 &  0 &  1 &  0 &  0 \\
0.9040 & 1 & 0 &  0 & -1 &  0 &  1 &  0 &  0 &  0 & -1 &  0 & -1 \\
0.8819 & 1 & 0 &  0 & -1 &  0 &  0 &  0 &  1 &  0 &  0 & -1 &  0 \\
0.8577 & 1 & 0 &  0 & -1 &  0 &  0 & -1 &  0 &  0 & -1 &  0 &  0 \\
0.8315 & 1 & 0 & -1 &  0 &  0 &  0 &  1 &  0 &  1 &  0 & -1 &  0 \\
0.8032 & 1 & 0 & -1 &  0 &  0 &  0 &  0 & -1 &  0 & -1 &  0 &  0 \\
0.7730 & 1 & 0 & -1 &  0 &  0 &  1 &  0 & -1 &  0 &  0 &  0 &  1 \\
0.7410 & 1 & 0 & -1 &  0 &  0 &  0 &  0 & -1 &  0 & -1 &  0 &  1 \\
0.7071 & 1 & 0 & -1 &  0 &  0 &  0 &  1 &  0 &  1 &  0 &  0 &  0 \\
\bottomrule
\end{tabular*}

\end{table}

\end{document}

나는 단지 "실수부"보다 "실수부의 상수 값"이 더 명확하다고 생각하지 않습니다.

정말 케이지 테이블을 원하신다면

\documentclass{article}
\usepackage{amsmath}
\usepackage{array}

\begin{document}

\begin{table}[htp]
\centering\footnotesize

\addtolength{\tabcolsep}{-.066pt}

\begin{tabular}{
  |c|
  *{12}{>{$}c<{$}|}
}
\hline
\smash{\begin{tabular}[t]{@{}c@{}}Real \\ part \end{tabular}} &
\multicolumn{12}{c|}{Binary representation of real part} \\
\cline{2-13}
& \text{2--0}
& \text{2--1} & \text{2--2} & \text{2--3} & \text{2--4} 
& \text{2--5} & \text{2--6} & \text{2--7} & \text{2--8} 
& \text{2--9} & \text{2--10} & \text{2--11} \\ \hline
0.9988 & 1 & 0 &  0 &  0 &  0 &  0 &  0 &  0 &  0 & -1 &  0 &  1 \\ \hline
0.9952 & 1 & 0 &  0 &  0 &  0 &  0 &  0 &  0 & -1 &  0 & -1 &  0 \\ \hline
0.9892 & 1 & 0 &  0 &  0 &  0 &  0 & -1 &  0 &  1 &  0 &  0 &  1 \\ \hline
0.9808 & 1 & 0 &  0 &  0 &  0 &  0 & -1 &  0 & -1 &  0 &  0 &  0 \\ \hline
0.9697 & 1 & 0 &  0 &  0 &  0 & -1 &  0 &  0 &  0 &  0 &  1 &  0 \\ \hline
0.9569 & 1 & 0 &  0 &  0 & -1 &  0 &  1 &  0 &  1 &  0 &  0 & -1 \\ \hline
0.9415 & 1 & 0 &  0 &  0 & -1 &  0 &  0 &  0 &  1 &  0 &  0 &  0 \\ \hline
0.9239 & 1 & 0 &  0 &  0 & -1 &  0 & -1 &  0 &  0 &  1 &  0 &  0 \\ \hline
0.9040 & 1 & 0 &  0 & -1 &  0 &  1 &  0 &  0 &  0 & -1 &  0 & -1 \\ \hline
0.8819 & 1 & 0 &  0 & -1 &  0 &  0 &  0 &  1 &  0 &  0 & -1 &  0 \\ \hline
0.8577 & 1 & 0 &  0 & -1 &  0 &  0 & -1 &  0 &  0 & -1 &  0 &  0 \\ \hline
0.8315 & 1 & 0 & -1 &  0 &  0 &  0 &  1 &  0 &  1 &  0 & -1 &  0 \\ \hline
0.8032 & 1 & 0 & -1 &  0 &  0 &  0 &  0 & -1 &  0 & -1 &  0 &  0 \\ \hline
0.7730 & 1 & 0 & -1 &  0 &  0 &  1 &  0 & -1 &  0 &  0 &  0 &  1 \\ \hline
0.7410 & 1 & 0 & -1 &  0 &  0 &  0 &  0 & -1 &  0 & -1 &  0 &  1 \\ \hline
0.7071 & 1 & 0 & -1 &  0 &  0 &  0 &  1 &  0 &  1 &  0 &  0 &  0 \\ \hline
\end{tabular}

\end{table}

\end{document}

\tabcolsep문서의 텍스트 너비에 따라 약간의 조정이 필요하지 않을 수도 있습니다. 여백이 덜 관대하다면 \small.

Answer 4

@Rmano의 제안에서 힌트를 얻음

MWE

\documentclass{article}
\usepackage[margin=2cm]{geometry}
\usepackage{nicematrix, xcolor}

\usepackage{booktabs}

\begin{document}
    \begin{NiceTabular}{ *{13}{c} }
        \CodeBefore
        \rowcolor{gray!50}{1}
        \rowcolors{2}{gray!25}{white}
        \Body
        \toprule
\Block[fill=white]{2-1}{Constant Value \\of the real part}                                                                                  &\Block{1-12}{Binary representation of real part} &&&&&&&&&&&\\ 
& 20 & 2-1 & 2-2 & 2-3 & 2-4 & 2-5 & 2-6 & 2-7 & 2-8 & 2-9 & 2-10 & 2-11 \\ 
        \midrule
        0.9988 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 & 0 & 1 \\ 
        0.9952 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 & 0 & -1 & 0 \\ 
        0.9892 & 1 & 0 & 0 & 0 & 0 & 0 & -1 & 0 & 1 & 0 & 0 & 1 \\ 
        0.9808 & 1 & 0 & 0 & 0 & 0 & 0 & -1 & 0 & -1 & 0 & 0 & 0 \\ 
        0.9697 & 1 & 0 & 0 & 0 & 0 & -1 & 0 & 0 & 0 & 0 & 1 & 0 \\ 
        0.9569 & 1 & 0 & 0 & 0 & -1 & 0 & 1 & 0 & 1 & 0 & 0 & -1 \\ 
        0.9415 & 1 & 0 & 0 & 0 & -1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 
        0.9239 & 1 & 0 & 0 & 0 & -1 & 0 & -1 & 0 & 0 & 1 & 0 & 0 \\ 
        0.9040 & 1 & 0 & 0 & -1 & 0 & 1 & 0 & 0 & 0 & -1 & 0 & -1 \\ 
        0.8819 & 1 & 0 & 0 & -1 & 0 & 0 & 0 & 1 & 0 & 0 & -1 & 0 \\ 
        0.8577 & 1 & 0 & 0 & -1 & 0 & 0 & -1 & 0 & 0 & -1 & 0 & 0 \\ 
        0.8315 & 1 & 0 & -1 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & -1 & 0 \\ 
        0.8032 & 1 & 0 & -1 & 0 & 0 & 0 & 0 & -1 & 0 & -1 & 0 & 0 \\ 
        0.7730 & 1 & 0 & -1 & 0 & 0 & 1 & 0 & -1 & 0 & 0 & 0 & 1 \\ 
        0.7410 & 1 & 0 & -1 & 0 & 0 & 0 & 0 & -1 & 0 & -1 & 0 & 1 \\ 
        0.7071 & 1 & 0 & -1 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 \\ 
        \bottomrule
        
    \end{NiceTabular}
    
\end{document}