\绘制矩形 TikZ

Question

我正在尝试学习 TikZ。

我写了以下小文档

\documentclass[tikz,border=12pt]{standalone}
\usepackage{ifthen}
\usepackage{pgf, tikz}
\usetikzlibrary{graphs,calc}

\makeatletter
\newcommand*\circled[1]{\smash{\tikz[baseline=(char.base)]{
            \node[anchor=text, shape=circle,fill,draw,inner sep=0pt,text=white,minimum size=1.4em] (char) {#1\strut};}}}
\makeatother

\begin{document}

    \begin{tikzpicture}
%    \draw(4,-1) rectangle(8,4);
    \foreach \y in {1, ..., 4}
    \foreach \x in {1, ..., 8} {
            \ifthenelse{\equal{\x}{5} \AND \equal{\y}{3}}{\draw(\x,\y)rectangle(0.4cm,0.4cm)}{
            \ifthenelse{\equal{\x}{4} \AND \equal{\y}{2}}{
                \draw[thick] (\x,\y) rectangle (0.4cm,0.4cm);
                \draw(\x,\y)rectangle(0.4cm,0.4cm)}{\draw(\x,\y)circle(0.4cm)}};
    }

    \end{tikzpicture}

\end{document}

它们的输出应该是一组 4x8 的圆圈，但第二行的第 4 个圆圈和第三行的第 5 个圆圈应该被矩形代替。

我不明白为什么它们会像图片中那样。

最后，如何定义一个命令，根据另一个文档传递的参数在圆圈中写入数字。我教导我可以使用新命令 \circled，但我不明白如何使用。

请帮我！

Answer 1

我建议使用pics 来实现这一点，而不是使用ifthenelse(来实现这一点)。

\documentclass[tikz,border=12pt]{standalone}
\usepackage{tikz}

\begin{document}

\begin{tikzpicture}[pics/.cd,
circ/.style={code={\draw (0,0) circle[radius=#1];}},
circ/.default=4mm,
rect/.style={code={\draw (-#1,-#1) rectangle (#1,#1);}},
rect/.default=4mm]
%    \draw(4,-1) rectangle(8,4);
\path    foreach \Y in {1,...,4}
    {foreach [evaluate=\X as \Z using {int(10*\Y+\X)}] \X in {1,...,8} {
     (\X,\Y)
     \ifnum\Z=35    
       pic{rect}
    \else
      \ifnum\Z=24
         pic{rect}
      \else
         pic{circ}
      \fi
    \fi }};
\end{tikzpicture}
\end{document}

\initbao如果您想将列表中的数字添加到方案中，可以这样做。

\documentclass[tikz,border=12pt]{standalone}

\begin{document}


\begin{tikzpicture}[circ/.style={circle,draw,minimum size=8mm},
rect/.style={rectangle,draw,minimum size=8mm}]
\def\myarray{{22,0,0,0,0,0,0,0,0, 0,2,2,6,0,0,0,0,
   0,0,0,0,6,2,2,0, 0,0,0,0,0,0,0,0,22}}
\path    foreach \Y in {1,...,4}
    {foreach [evaluate=\X as \Z using {int(10*\Y+\X)}] \X in {1,...,8} {
     (\X,\Y)
     \ifnum\Z=35    
       node[rect] {\pgfmathparse{\myarray[(\Y-1)*8+\X-1]}\pgfmathresult}
    \else
      \ifnum\Z=24
          node[rect] {\pgfmathparse{\myarray[(\Y-1)*8+\X-1]}\pgfmathresult}
      \else
          node[circ] {\pgfmathparse{\myarray[(\Y-1)*8+\X-1]}\pgfmathresult}
      \fi
    \fi 
    }};
\end{tikzpicture}
\end{document}

或者其命令版本（顺序已改变）。

\documentclass[tikz,border=12pt]{standalone}

\newcommand{\initbao}[1]{\begin{tikzpicture}[circ/.style={circle,draw,minimum size=8mm},
rect/.style={rectangle,draw,minimum size=8mm}]
\path    foreach \Y in {1,...,4}
    {foreach [evaluate=\X as \Z using {int(10*\Y+\X)}] \X in {1,...,8} {
     (\X,\Y)
     \ifnum\Z=35    
       node[rect] {\pgfmathparse{{#1}[(4-\Y)*8+\X-1]}\pgfmathresult}
    \else
      \ifnum\Z=24
          node[rect] {\pgfmathparse{{#1}[(4-\Y)*8+\X-1]}\pgfmathresult}
      \else
          node[circ] {\pgfmathparse{{#1}[(4-\Y)*8+\X-1]}\pgfmathresult}
      \fi
    \fi 
    }};
\end{tikzpicture}}
\begin{document}

\initbao{22,0,0,0,0,0,0,0,0, 0,2,2,6,0,0,0,0,
   0,0,0,0,6,2,2,0, 0,0,0,0,0,0,0,0,22}
\end{document}

Answer 2

在使用循环和条件的地方，您可以将matrix of nodes所有节点都设为圆形，并将其中一些节点的样式设为矩形。

\documentclass[tikz,border=7mm]{standalone}
\usetikzlibrary{matrix}
\begin{document}
  \begin{tikzpicture}
    \tikzstyle{r}=[rectangle] % <-- because I like tikzstyle ;)
    \matrix[matrix of nodes, nodes in empty cells,
      column sep={1cm,between origins},row sep={1cm,between origins},
      nodes={circle,anchor=center,draw,minimum size=8mm}](M){
      & & &             &       & & &   \\
      & & &             & |[r]| & & &   \\
      & & & |[r,red]| ? &       & & &   \\
      & & &             &       & & & ! \\
    };
  \end{tikzpicture}
\end{document}