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I have some geometry that I need to clip. I would like to use the built in node shapes like diamonds, stars, etc...

http://www.texample.net/tikz/examples/node-shapes/

Basically I use standard tikz draw commands to generate the picture I want but then I want to clip them according to some built in shape(or even custom shape).

\documentclass{article}

\usepackage{tikz}
\begin{document}
\scrollmode


\begin{tikzpicture}

  \begin{scope}
    \clip (0,0) diamond (1cm);
    \fill[black] (0cm,1cm) rectangle (-1cm, -1cm);
  \end{scope}

\end{tikzpicture}
\end{document}

I need someway to specify the size of the clip shape also.

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1 Answer

up vote 13 down vote accepted

Normally \node does not accept the clip option for its path use but the lower level PGF version does. It's a little bit more laborous but essentially the same idea.

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{shapes.geometric}
\begin{document}
\begin{tikzpicture}

  \begin{scope}
    %\pgfset{shape aspect=0.5} Uncomment this and remove minimum size for this option
    \pgfset{minimum size=3cm,inner sep=2mm}
    \pgfnode{diamond}{center}{}{nodename}{\pgfusepath{stroke,clip}}
    \fill[black] (-1cm,-1cm) rectangle (1cm, 1cm);
  \end{scope}

\end{tikzpicture}
\end{document}

enter image description here

There is a problem here about the bounding box. The actual bounding box is computed by the size of the rectangle if too large even if it is clipped. So one should do another trick after everything is drawn outside the scope, say with an additional node via [use as bounding box].

update (altermundus)

\fbox{\begin{tikzpicture}
    \begin{scope}
        [local bounding box=bb]  \node  [draw,shape=diamond,minimum size=3cm,inner sep=2mm]{};  
    \end{scope}
      \pgfset{minimum size=3cm,inner sep=2mm}
      \pgfnode{diamond}{center}{}{nodename}{\pgfusepath{stroke,clip}}
      \fill[black] (-3cm,-1cm) rectangle (3cm, 1cm);
      \pgfresetboundingbox
       \useasboundingbox (bb.south west) rectangle  (bb.north east);
  \end{tikzpicture}} 

enter image description here


EDIT: Regarding the position of the clipping node...

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{shapes.callouts,shapes.geometric}
\begin{document}
\begin{tikzpicture}
\draw[top color= black!50] (-2,-2) rectangle (5,4);
\begin{scope}   
    \pgfset{minimum width=5cm,minimum height=3cm}
    \pgfsetlinewidth{1mm}
    \pgftransformshift{\pgfpoint{1.5cm}{1.5cm}}
    \pgfnode{cloud callout}{center}{}{nodename}{\pgfusepath{stroke,clip}}
    %Cleaning up the mess we caused
    \pgftransformreset
    \pgfsetlinewidth{0.4pt} 
    \pgfset{minimum width=1pt,minimum height=1pt}
    % Back to drawing
    \fill[white] (-2cm,-2cm) rectangle (5cm,4cm);
    \fill[red] (0cm,0cm) rectangle (1.5cm, 1.5cm);
    \fill[yellow] (2cm,2cm) circle (1cm);
    \node[fill=blue,rotate=90,isosceles triangle,draw,minimum height=1.5cm] at (2.5cm,1cm) {};
\end{scope}
\node[align=center,draw,anchor=north west] (a) at (nodename.pointer) {Geometric\\Thinker};
\end{tikzpicture}
\end{document}

enter image description here

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@Altermundus Thanks for the update! – percusse Apr 25 '12 at 9:33
How do you change the position? I've tried using \pgfxy and \pgfpoint for center but I get errors. – Uiy Apr 26 '12 at 7:29
I think I didn't understand. You only need \pgf... commands to build the node but the rest can keep TikZ syntax. Is it the rectangle's position relative to the diamond that you want to change or you want to shift the scope ? – percusse Apr 26 '12 at 8:00
When I try it in my own code a diamond is formed at the origin but the part I want to clip is no where near the origin. It looks more like it just puts a diamond at the origin rather than clip anything but I basically cut and pasted your code. All I did was wrap my plotting code(which just plots some circle) with the scope env. and put the pgf set amd node of your code right below the begin. – Uiy Apr 26 '12 at 8:17
For example, change your code's rectangle position and you'll end up with an empty region if it is outside the clip region. (e.g., change the (-1cm, -1cm) to (2cm,2cm) and you'll not end up with anything visible) – Uiy Apr 26 '12 at 8:19
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