34

maybe I'm stupid, I looked at the many examples in which somehow shows the work with the coordinates, but I don't understand it. Simple example - I have: 

\begin{tikzpicture}
   \coordinate [label=left:$D$] (D) at (0.3,0.5);
\end{tikzpicture}

Now I would like to use x-part of my coordinate in the next \draw command, but I don't know how to simply extract it.

\draw[name path=my_line, gray] (**x-part of D coordinate**,0) node[below, red]{$c$} -- (0.75,1.1);

I hope that's my question is obvious. Thanks!!

JardaFait
  • 3,922
  • look up the let keyword – daleif Aug 06 '12 at 22:21
  • See http://tex.stackexchange.com/questions/18389/tikz-node-at-same-x-coordinate-as-another-node-but-specified-y-coordinate or http://tex.stackexchange.com/questions/33703/extract-x-y-coordinate-of-an-arbitrary-point-in-tikz – Torbjørn T. Aug 06 '12 at 22:23

3 Answers3

33

You can use the let syntax (See Section 14.15 The Let Operation of the manual); another option (suggested by percusse in a comment) is to use the |- syntax (page 131 of the manual). An example with both possibilities:

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{calc,intersections}

\begin{document}


\begin{tikzpicture}
  \coordinate [label=left:$D$] (D) at (0.3,0.5);
  \draw[name path=my_line, gray] let \p1=(D) in (\x1,0) node[below, red] {$c$} -- (0.75,1.1);
\end{tikzpicture}


\begin{tikzpicture}
  \coordinate [label=left:$D$] (D) at (0.3,0.5);
  \draw (D |- {{(0,0)}}) node[below, red] {$c$} -- (0.75,1.1);
\end{tikzpicture}


\end{document}

Quoting from the manual:

the meaning of (p-|q) is ''the intersection of a vertical line through p and a horizontal line through q''.

enter image description here

PatrickT
  • 2,923
Gonzalo Medina
  • 505,128
23

You can also adapt the solution from Extract x, y coordinate of an arbitrary point in TikZ and invoke \ExtractCoordinate{D} before you need the coordinates of this point and then use \XCoord or \YCoord where you need the value.

The code below produces output identical to Gonzalo Medina's answer.

Code:

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{calc}

%% https://tex.stackexchange.com/questions/86897/recover-scaling-factor-in-tikz
\newcommand\getscale[1]{%
  \begingroup
    \pgfgettransformentries{\scaleA}{\scaleB}{\scaleC}{\scaleD}{\whatevs}{\whatevs}%
    \pgfmathsetmacro{#1}{sqrt(abs(\scaleA\scaleD-\scaleB*\scaleC))}%
    \expandafter
  \endgroup
  \expandafter\def\expandafter#1\expandafter{#1}%
}


\makeatletter
% https://tex.stackexchange.com/questions/33703/extract-x-y-coordinate-of-an-arbitrary-point-in-tikz
\newdimen@XCoord
\newdimen@YCoord
\newdimen\XCoord
\newdimen\YCoord
\newcommand*{\ExtractCoordinate}[1]{%
    \getscale{@scalefactor}%
    \path [transform canvas] (#1); \pgfgetlastxy{@XCoord}{@YCoord};%
    \pgfmathsetlength{\XCoord}{@XCoord/@scalefactor}%
    \pgfmathsetlength{\YCoord}{@YCoord/@scalefactor}%
}
\makeatother


\begin{document}
\begin{tikzpicture}[scale=2]
   \coordinate (D) at (0.3,0.5);


\ExtractCoordinate{D}
  \fill [red] (D) circle (1pt);
  \draw[gray] (\XCoord,0) node[below, red] {$c$} -- (0.75,1.1);
\end{tikzpicture}
\end{document}
Peter Grill
  • 223,288
  • Unless I am doing something wrong, this method doesn't seem to work with scaling. I.e. if we pass a parameter 'scale=0.5' to \begin{tikzpicture}, the resulting coordinates \XCoord, \YCoord seem off. I am trying to take the coordinates of a regular polygon defined as \node[draw=none, minimum size=4cm, regular polygon, regular polygon sides = 8] (a) {}; And then taking \ExtractCoordinate{a.corner 3}, for example, produces a wrong coordinate with scaling (but a correct one without scaling). – mathreader Apr 27 '22 at 12:26
  • 1
    @mathreader: Solution updated to handle scale= issue. Thanks for pointing that out. – Peter Grill Apr 27 '22 at 21:23
19

For the sake of completeness and future reference, if you have a point (a) defined, you can define a new point (b) which depends on the coordinates of (a) using the let as follows:

\path let \p1 = (a) in coordinate (b) at (2*\x1,\y1/2);

Since \coordinate is an alias for \path coordinate you have to use the above code if you want to use let for the definition of a coordinate.

Dror
  • 22,613