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在矩阵内画线

tag-icon • tag-icon tikz-pgf tikz-matrix
在矩阵内画线

我有以下代码来绘制matrix：

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{matrix}
\begin{document}
  \begin{tikzpicture}[cell/.style={rectangle,draw=black}, nodes in empty cells]
  \matrix[
  matrix of math nodes,
  row sep =-\pgflinewidth,
  column sep = -\pgflinewidth,
  nodes={anchor=center, minimum width=2cm, cell},
  column 1/.style = {nodes={minimum width=1cm, fill=mlightgray}},
  column 2/.style = {nodes={minimum width=3.2cm}},
  column 3/.style = {nodes={minimum width=3.2cm}},
  row 1/.style = {nodes={text height=1.3ex, text depth=0, fill=mlightgray}},
  row 2/.style = {text height=1.3ex, text depth=0},
  row 3/.style = {text height=1.3ex, text depth=0},
  row 4/.style = {text height=1.3ex, text depth=0},
  ] 
  {   &  \text{1} & \text{2}  \\
    \text{1} & & \\
    \text{2} & & \\
    \text{3} & & \\
  };
\end{tikzpicture}
\end{document}

现在，我想在单元格之间画线来实现以下内容（由生成pspicture，但我想将所有内容转换为 tikz）：

在此处输入图片描述

是否可以？

PS：我不需要一个通用的解决方案，只要能实现这个数字就足够了。

答案1

你可以给矩阵加上标签，然后使用(m-1-2)语法来绘制。这意味着矩阵的第一行和第二列(m)（即1-2条目的索引）。

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{matrix}
\begin{document}
  \begin{tikzpicture}[cell/.style={rectangle,draw=black}, nodes in empty cells]
  \matrix[
  matrix of math nodes,
  row sep =-\pgflinewidth,
  column sep = -\pgflinewidth,
  nodes={anchor=center, minimum width=2cm, cell},
  column 1/.style = {nodes={minimum width=1cm, fill=lightgray}},
  column 2/.style = {nodes={minimum width=3.2cm}},
  column 3/.style = {nodes={minimum width=3.2cm}},
  row 1/.style = {nodes={text height=1.3ex, text depth=0, fill=lightgray}},
  row 2/.style = {text height=1.3ex, text depth=0},
  row 3/.style = {text height=1.3ex, text depth=0},
  row 4/.style = {text height=1.3ex, text depth=0},
  ](m)
  {   &  1 & 2  \\
    1 & \bullet & \bullet \\
    2 & \bullet & \bullet \\
    3 & \bullet & \bullet \\
  };
  \draw[-latex] (m-4-3.center) -- (m-3-2.center);
  \draw[-latex] (m-4-3.center) -- (m-3-3.center);
  \draw[-latex] (m-3-2.center) -- (m-2-2.center);
  \draw[-latex] (m-3-3.center) -- (m-2-2.center);
\end{tikzpicture}
\end{document}

在此处输入图片描述

答案2

全自动 PSTricks 解决方案：

\documentclass{article}

\usepackage{multido}
\usepackage{pstricks}
\usepackage{xfp}

% settings
\psset{
  dimen = m,
  fillcolor = gray!60
}

% parameters
\def\Columns{2}
\def\ColumnWidth{3}
\def\Rows{3}
\def\RowHeight{0.7}

\begin{document}

\begin{pspicture}(\fpeval{\Columns*\ColumnWidth+1},\fpeval{(\Rows+1)*\RowHeight})
  \multido{\r = 0+\RowHeight}{\fpeval{\Rows+1}}{%
    \psframe[fillstyle = solid](0,\r)(1,\fpeval{\r+\RowHeight})%
  }
  \multido{
           \rE = 0+\RowHeight,
           \rF = \fpeval{0.5*\RowHeight}+\RowHeight,
           \iB = \Rows+-1
          }{\Rows}{%
    \multido{
             \rA = 1+\ColumnWidth,
             \rB = \fpeval{0.5*\ColumnWidth+1}+\ColumnWidth,
             \rC = \fpeval{0.5*\RowHeight}+\RowHeight,
             \rD = \fpeval{1.5*\RowHeight}+-\RowHeight,
             \iA = 1+1
            }{\Columns}{%
      \psframe[fillstyle = solid]%
        (\rA,\fpeval{\Rows*\RowHeight})%
        (\fpeval{\rA+\ColumnWidth},\fpeval{(\Rows+1)*\RowHeight})
      \rput(\rB,\fpeval{(\Rows+0.5)*\RowHeight}){$\iA$}
      \psframe(\rA,\rE)(\fpeval{\rA+\ColumnWidth},\fpeval{\rE+\RowHeight})
      \rput(\rB,\rF){\psdot}}
    \rput(0.5,\rF){$\iB$}}
  \psline{->}(\fpeval{1+(\Columns-0.5)*\ColumnWidth},\fpeval{0.5*\RowHeight})%
             (\fpeval{1+0.5*\ColumnWidth},\fpeval{(\Rows-1.5)*\RowHeight})
  \psline{->}(\fpeval{1+0.5*\ColumnWidth},\fpeval{(\Rows-1.5)*\RowHeight})%
             (\fpeval{1+0.5*\ColumnWidth},\fpeval{(\Rows-0.5)*\RowHeight})
  \psline{->}(\fpeval{1+(\Columns-0.5)*\ColumnWidth},\fpeval{0.5*\RowHeight})%
             (\fpeval{1+(\Columns-0.5)*\ColumnWidth},\fpeval{1.5*\RowHeight})
  \psline{->}(\fpeval{1+(\Columns-0.5)*\ColumnWidth},\fpeval{1.5*\RowHeight})%
             (\fpeval{1+0.5*\ColumnWidth},\fpeval{(\Rows-0.5)*\RowHeight})
\end{pspicture}

\end{document}

输出

您所要做的就是改变参数的值，绘图就会相应地调整。

更新

如果您需要宏，可以按如下方式操作：

\documentclass{article}

\usepackage{multido}
\usepackage{pstricks}
\usepackage{xfp}

% settings
\psset{
  dimen = m,
  fillcolor = gray!60
}

% \Table[<cell width>,<cell height>]{<number of columns}{<number of rows>}
\def\Table[#1,#2]#3#4{%
\begin{pspicture}(\fpeval{#3*#1+1},\fpeval{(#4+1)*#2})
  \multido{\r = 0+#2}{\fpeval{#4+1}}{%
    \psframe[fillstyle = solid](0,\r)(1,\fpeval{\r+#2})%
  }
  \multido{
           \rE = 0+#2,
           \rF = \fpeval{0.5*#2}+#2,
           \iB = #4+-1
          }{#4}{%
    \multido{
             \rA = 1+#1,
             \rB = \fpeval{0.5*#1+1}+#1,
             \rC = \fpeval{0.5*#2}+#2,
             \rD = \fpeval{1.5*#2}+-#2,
             \iA = 1+1
            }{#3}{%
      \psframe[fillstyle = solid]%
        (\rA,\fpeval{#4*#2})%
        (\fpeval{\rA+#1},\fpeval{(#4+1)*#2})
      \rput(\rB,\fpeval{(#4+0.5)*#2}){$\iA$}
      \psframe(\rA,\rE)(\fpeval{\rA+#1},\fpeval{\rE+#2})
      \rput(\rB,\rF){\psdot}}
    \rput(0.5,\rF){$\iB$}}
  \psline{->}(\fpeval{1+(#3-0.5)*#1},\fpeval{0.5*#2})%
             (\fpeval{1+0.5*#1},\fpeval{(#4-1.5)*#2})
  \psline{->}(\fpeval{1+0.5*#1},\fpeval{(#4-1.5)*#2})%
             (\fpeval{1+0.5*#1},\fpeval{(#4-0.5)*#2})
  \psline{->}(\fpeval{1+(#3-0.5)*#1},\fpeval{0.5*#2})%
             (\fpeval{1+(#3-0.5)*#1},\fpeval{1.5*#2})
  \psline{->}(\fpeval{1+(#3-0.5)*#1},\fpeval{1.5*#2})%
             (\fpeval{1+0.5*#1},\fpeval{(#4-0.5)*#2})
\end{pspicture}}

\begin{document}

\Table[3,0.7]{2}{3}

\end{document}

\end{document}

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