tikz 图片周围的括号

Question

为什么下面代码中的括号比tikzpicture更大？

\documentclass{article}

\usepackage[utf8]{inputenc}
\usepackage{tikz}

\begin{document}

\newcommand{\nada}[2] {\fill[lightgray] (#1,#2) circle (0.5);}
\newcommand{\tr}[2]   {\fill[blue] (#1,#2) circle (0.5);}
\newcommand{\ts}[2]   {\fill[red] (#1,#2) circle (0.5);}

\centering

\begin{tikzpicture}[x=4mm,y=4mm]
\foreach \x in {1,...,7} \tr{\x}{0};
\foreach \x in {7,...,9} \ts{\x}{0};
\fill[green] (10,0) circle (0.5);
\end{tikzpicture}

Traditional validation

\vspace{5mm}

\[E\left[
\begin{tikzpicture}[x=4mm,y=4mm]
\foreach \x in {1,...,4}  \tr{\x}{0};
\foreach \x in {5,...,6}  \ts{\x}{0};
\foreach \x in {7,...,10} \nada{\x}{0};
\foreach \x in {1,...,2}  \nada{\x}{1};
\foreach \x in {3,...,6}  \tr{\x}{1};
\foreach \x in {7,...,8}  \ts{\x}{1};
\foreach \x in {9,...,10} \nada{\x}{1};
\foreach \x in {1,...,4}  \nada{\x}{2};
\foreach \x in {5,...,8}  \tr{\x}{2};
\foreach \x in {9,...,10} \ts{\x}{2};
\end{tikzpicture}
\right]\]

Sliding window

\vspace{5mm}

\[E\left[
\begin{tikzpicture}[x=4mm,y=4mm]
\foreach \x in {1,...,7}  \tr{\x}{0};
\foreach \x in {8,...,10} \ts{\x}{0};
\foreach \x in {1,...,4}  \tr{\x}{1};
\foreach \x in {5,...,7}  \ts{\x}{1};
\foreach \x in {8,...,10} \tr{\x}{1};
\foreach \x in {1,...,4}  \ts{\x}{2};
\foreach \x in {5,...,10} \tr{\x}{2};
\end{tikzpicture}
\right]\]

$k$-fold ($k=3$)

\end{document}

Answer 1

您还可以简单地调整图片相对于基线的位置，而无需改变周围的数学环境。这yshift=.5ex是针对圆圈略低于基线这一事实的调整：

\documentclass{article}
\usepackage[utf8]{inputenc}
\usepackage{tikz}

\begin{document}

\newcommand{\nada}[2] {\fill[lightgray] (#1,#2) circle (0.5);}
\newcommand{\tr}[2]   {\fill[blue] (#1,#2) circle (0.5);}
\newcommand{\ts}[2]   {\fill[red] (#1,#2) circle (0.5);}

\centering

\begin{tikzpicture}[x=4mm,y=4mm]
\foreach \x in {1,...,7} \tr{\x}{0};
\foreach \x in {7,...,9} \ts{\x}{0};
\fill[green] (10,0) circle (0.5);
\end{tikzpicture}

Traditional validation

\vspace{5mm}

\[E\left[
\begin{tikzpicture}[x=4mm,y=4mm,baseline={([yshift=-.5ex]current bounding box.center)}]
\foreach \x in {1,...,4}  \tr{\x}{0};
\foreach \x in {5,...,6}  \ts{\x}{0};
\foreach \x in {7,...,10} \nada{\x}{0};
\foreach \x in {1,...,2}  \nada{\x}{1};
\foreach \x in {3,...,6}  \tr{\x}{1};
\foreach \x in {7,...,8}  \ts{\x}{1};
\foreach \x in {9,...,10} \nada{\x}{1};
\foreach \x in {1,...,4}  \nada{\x}{2};
\foreach \x in {5,...,8}  \tr{\x}{2};
\foreach \x in {9,...,10} \ts{\x}{2};
\end{tikzpicture}
\right]\]

Sliding window

\vspace{5mm}

\[E\left[
\begin{tikzpicture}[x=4mm,y=4mm,baseline={([yshift=-.5ex]current bounding box.center)}]
\foreach \x in {1,...,7}  \tr{\x}{0};
\foreach \x in {8,...,10} \ts{\x}{0};
\foreach \x in {1,...,4}  \tr{\x}{1};
\foreach \x in {5,...,7}  \ts{\x}{1};
\foreach \x in {8,...,10} \tr{\x}{1};
\foreach \x in {1,...,4}  \ts{\x}{2};
\foreach \x in {5,...,10} \tr{\x}{2};
\end{tikzpicture}
\right]\]

$k$-fold ($k=3$)

\end{document}

Answer 2

您可以先将图片的基线设置为位于中间；然后使用gathered环境（的更高级别版本\vcenter）来修复居中。

\documentclass{article}

\usepackage[utf8]{inputenc}
\usepackage{amsmath}
\usepackage{tikz}

\begin{document}

\newcommand{\nada}[2] {\fill[lightgray] (#1,#2) circle (0.5);}
\newcommand{\tr}[2]   {\fill[blue] (#1,#2) circle (0.5);}
\newcommand{\ts}[2]   {\fill[red] (#1,#2) circle (0.5);}

\centering

\begin{tikzpicture}[x=4mm,y=4mm]
\foreach \x in {1,...,7} \tr{\x}{0};
\foreach \x in {7,...,9} \ts{\x}{0};
\fill[green] (10,0) circle (0.5);
\end{tikzpicture}

Traditional validation

\vspace{5mm}

\[E\left[
\!\begin{gathered}
\begin{tikzpicture}[x=4mm,y=4mm,baseline=(current bounding box.center)]
\foreach \x in {1,...,4}  \tr{\x}{0};
\foreach \x in {5,...,6}  \ts{\x}{0};
\foreach \x in {7,...,10} \nada{\x}{0};
\foreach \x in {1,...,2}  \nada{\x}{1};
\foreach \x in {3,...,6}  \tr{\x}{1};
\foreach \x in {7,...,8}  \ts{\x}{1};
\foreach \x in {9,...,10} \nada{\x}{1};
\foreach \x in {1,...,4}  \nada{\x}{2};
\foreach \x in {5,...,8}  \tr{\x}{2};
\foreach \x in {9,...,10} \ts{\x}{2};
\end{tikzpicture}
\end{gathered}
\right]\]

Sliding window

\vspace{5mm}

\[E\left[
\!\begin{gathered}
\begin{tikzpicture}[x=4mm,y=4mm,baseline=(current bounding box.center)]
\foreach \x in {1,...,7}  \tr{\x}{0};
\foreach \x in {8,...,10} \ts{\x}{0};
\foreach \x in {1,...,4}  \tr{\x}{1};
\foreach \x in {5,...,7}  \ts{\x}{1};
\foreach \x in {8,...,10} \tr{\x}{1};
\foreach \x in {1,...,4}  \ts{\x}{2};
\foreach \x in {5,...,10} \tr{\x}{2};
\end{tikzpicture}
\end{gathered}
\right]\]

$k$-fold ($k=3$)

\end{document}

之前\!是\begin{gathered}为了修复一个“功能”。

当你这样做时\left[...\right]，栅栏会相对于“公式轴”对称绘制，该轴略高于基线。由于默认情况下，TikZ 图片的基线位于 (0,0) 坐标，因此你会得到所见的效果。

因此，第一次使用盒子中心作为基线居中几乎可以完成工作，然后使用gathered就完成了。

但是，你可以gathered通过稍微移动基线来实现精确定位，利用公式轴在基线上方的高度

\fontdimen22\textfont2

相当神秘，不是吗？;-) 因此

\[E\left[
\begin{tikzpicture}[x=4mm,y=4mm,
  baseline={([yshift=-\the\fontdimen22\textfont2]current bounding box.center)}]
\foreach \x in {1,...,4}  \tr{\x}{0};
\foreach \x in {5,...,6}  \ts{\x}{0};
\foreach \x in {7,...,10} \nada{\x}{0};
\foreach \x in {1,...,2}  \nada{\x}{1};
\foreach \x in {3,...,6}  \tr{\x}{1};
\foreach \x in {7,...,8}  \ts{\x}{1};
\foreach \x in {9,...,10} \nada{\x}{1};
\foreach \x in {1,...,4}  \nada{\x}{2};
\foreach \x in {5,...,8}  \tr{\x}{2};
\foreach \x in {9,...,10} \ts{\x}{2};
\end{tikzpicture}
\right]\]

会产生

Answer 3

您可以使用\vcenter{..}在数学模式下垂直居中材料。它需要位于内部，例如\hbox。因此整个语法将是\vcenter{\hbox{⟨contents⟩}}：

\documentclass{scrartcl}

\usepackage[utf8]{inputenc}
\usepackage{tikz}

\begin{document}

\newcommand{\nada}[2] {\fill[lightgray] (#1,#2) circle (0.5);}
\newcommand{\tr}[2]   {\fill[blue] (#1,#2) circle (0.5);}
\newcommand{\ts}[2]   {\fill[red] (#1,#2) circle (0.5);}

\centering

\begin{tikzpicture}[x=4mm,y=4mm]
\foreach \x in {1,...,7} \tr{\x}{0};
\foreach \x in {7,...,9} \ts{\x}{0};
\fill[green] (10,0) circle (0.5);
\end{tikzpicture}

Traditional validation

\vspace{5mm}

\[E\left[\vcenter{\hbox{%
\begin{tikzpicture}[x=4mm,y=4mm]
\foreach \x in {1,...,4}  \tr{\x}{0};
\foreach \x in {5,...,6}  \ts{\x}{0};
\foreach \x in {7,...,10} \nada{\x}{0};
\foreach \x in {1,...,2}  \nada{\x}{1};
\foreach \x in {3,...,6}  \tr{\x}{1};
\foreach \x in {7,...,8}  \ts{\x}{1};
\foreach \x in {9,...,10} \nada{\x}{1};
\foreach \x in {1,...,4}  \nada{\x}{2};
\foreach \x in {5,...,8}  \tr{\x}{2};
\foreach \x in {9,...,10} \ts{\x}{2};
\end{tikzpicture}%
}}\right]\]

Sliding window

\vspace{5mm}

\[E\left[\vcenter{\hbox{%
\begin{tikzpicture}[x=4mm,y=4mm]
\foreach \x in {1,...,7}  \tr{\x}{0};
\foreach \x in {8,...,10} \ts{\x}{0};
\foreach \x in {1,...,4}  \tr{\x}{1};
\foreach \x in {5,...,7}  \ts{\x}{1};
\foreach \x in {8,...,10} \tr{\x}{1};
\foreach \x in {1,...,4}  \ts{\x}{2};
\foreach \x in {5,...,10} \tr{\x}{2};
\end{tikzpicture}%
}}\right]\]

$k$-fold ($k=3$)

\end{document}