18

What is the best way to draw braces between tick marks? It would be convenient to assign nodes at tick marks, is this doable?

The best method for placement that I have right now is a guess-and-check.

MWE:

\documentclass{article}

\usepackage{tikz}
\usetikzlibrary{patterns,decorations.pathreplacing}
\usepackage{pgfplots}
\pgfplotsset{compat=1.10}

\begin{document}

\begin{tikzpicture}
\begin{axis}[
    axis x line = middle,
    axis y line = middle,
    xticklabels={}
    ]
\addplot [domain=-1:5] { -x * (x-4) };
\end{axis}
\draw [thick,decoration={brace,mirror},decorate] (1.2,3) -- (5.7,3) node[midway,below,yshift=-.1cm] {4};
\end{tikzpicture}

\end{document}

MWEpicture

2
  • Update: In my actual implementation (not the MWE), my brace fell out of the graphing area and was clipped off. However, I did not want to use clip=false, since my graph would have extended out as well. To fix all this, I used after end axis/.code={\draw [decoration={brace},decorate] (A) -- (B) node {yadda}; in the argument of \begin{axis}, and \coordinate (A) at (axis cs:pi,0);, \coordinate (B) at (axis cs:5*pi/3,0); just before \end{axis}. I hope this helps anyone else who might run into the same situation!
    – nivk
    Commented May 14, 2014 at 6:51
  • (Special thanks to Jake's answer in Annotate Plots in TikZ/PGFplots.)
    – nivk
    Commented May 14, 2014 at 6:59

1 Answer 1

23

Below I present three levels of improvement:

  1. You can move the \draw inside the axis environment and use the axis cs coordinate system which gives you access to the coordinates of the plot; also, you can use the raise option to shift the brace:

    \documentclass{article}
    \usepackage{pgfplots}
    \usetikzlibrary{patterns,decorations.pathreplacing}
    \pgfplotsset{compat=1.10}
    
    \begin{document}
    
    \begin{tikzpicture}
    \begin{axis}[
        axis x line = middle,
        axis y line = middle,
        xticklabels={}
        ]
    \addplot [domain=-1:5] { -x * (x-4) };
    \draw [thick,decoration={brace,mirror,raise=5pt},decorate] 
      (axis cs:0,0) --
        node[below=7pt] {4} 
      (axis cs:4,0);
    \end{axis}
    \end{tikzpicture}
    
    \end{document}
    

    output of solution 1

  2. You don't have to know beforehand the coordinates for the intersection points; you can let the intersections library do the calculations for you:

    \documentclass{article}
    \usepackage{pgfplots}
    \usetikzlibrary{patterns,decorations.pathreplacing,intersections}
    \pgfplotsset{compat=1.10}
    
    \begin{document}
    
    \begin{tikzpicture}
    \begin{axis}[
        axis x line = middle,
        axis y line = middle,
        xticklabels={},
        samples=100
        ]
    \addplot[name path=curve,domain=-1:5] { -x * (x-4) };
    \addplot[name path=line,domain=-1:5,forget plot] {0};
    \draw[
      name intersections={of=curve and line, by={a,b}},
      thick,
      decoration={brace,mirror,raise=5pt},decorate
    ] 
      (a) --
        node[below=7pt] {4} 
      (b);
    \end{axis}
    \end{tikzpicture}
    
    \end{document}
    
  3. Finally you can also use the calc library and the lengthconvert package, for example, so the length (4) is also automatically calculated:

    \documentclass{article}
    \usepackage{lengthconvert}
    \usepackage{pgfplots}
    \usetikzlibrary{patterns,decorations.pathreplacing,intersections,calc}
    \pgfplotsset{compat=1.10}
    
    \begin{document}
    
    \begin{tikzpicture}
    \begin{axis}[x=1cm,
        axis x line = middle,
        axis y line = middle,
        xticklabels={},
        samples=100
        ]
    \addplot[name path=curve,domain=-1:5] { -x * (x-4) };
    \addplot[name path=line,domain=-1:5,forget plot] {0};
    \draw[
      name intersections={of=curve and line, by={a,b}},
      decoration={brace,mirror,raise=5pt},
      decorate
    ]
      let \p1 = ($ (a) - (b) $),
          \n2 = {veclen(\x1,\y1)}
      in
      (a) --
        node[below=7pt] {\Convert[precision=1,number-only=true]{\n2}} 
      (b);
    \end{axis}
    \end{tikzpicture}
    
    \end{document}
    
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .