3

I am drawing a figure which has many nodes and many lines connecting the nodes. I want every line connecting two nodes to start from "north" and to end to "south". So, essentially, my code is something like

\begin{tikzpicture}
 \node (a) at (0,0) {};
 \node (b) at (0,1) {};
 \draw (a.north) -- (b.south);
\end{tikzpicture}

with many nodes definitions and many draws. I want to define once and for all that all line should start from "north" and end to "south", so I tried something like

 every path/.style={parent anchor=north, child anchor=south}

but it doesn't work. The question is: is there a way to specify in the template that any line should start from "north" and end to "south" so avoiding the need to write this in every draw?

| improve this question | |
  • Why not using \draw (a) -- (b);? – percusse Nov 3 '17 at 13:11
  • You could use a loop to draw the lines, \foreach \i/\j in {a/b, c/d, e/f} {\draw (\i.north) -- (\j.south);}, is that an option? – Torbjørn T. Nov 3 '17 at 13:20
  • @percusse If the nodes aren't at the same x-coordinate, that won't work. (I assumed that's the case here, even if the example doesn't show it.) – Torbjørn T. Nov 3 '17 at 13:22
  • @percusse: I chose a bad example, my mistake: I have many nodes and they are not one on top of the other. Please read: \node (a) at (0,0) and \node (b) at (1,1) – brad Nov 3 '17 at 13:26
  • @Torbiorn: that is a nice suggestion, thank you. The only problem is that I want to add colors to some of the nodes. – brad Nov 3 '17 at 13:29
4

You can use a custom to path 

\documentclass[tikz]{standalone}
\begin{document}
\begin{tikzpicture}[ns/.style={
    to path={(\tikztostart.north) -- (\tikztotarget.south)\tikztonodes (\tikztotarget)}
}]
 \node[draw] (a) at (0,0) {};
 \node[draw] (b) at (1,1) {};
 \node[draw] (c) at (2,-1){};
 \draw (a) to [ns] (b) to[ns] (c) to[ns] (a);
\end{tikzpicture}
\end{document}

enter image description here

| improve this answer | |
  • This answer is closer to my original question. All in all, in my original code was enough to substitute \draw (a.north) -- (b.south); with \draw (a.north) to (b.south); to have the style every path/.style={parent anchor=north, child anchor=south} performing what I meant. Yet, I do not understand the difference. – brad Nov 3 '17 at 14:10
  • @brad For that you should better use every to – percusse Nov 3 '17 at 14:11
4

You could use macros. I should add that using one letter macro names is a bad idea. Also, I'm surprised I didn't have to use \pgfextra.

\documentclass{standalone}
\usepackage{tikz}

\newcommand{\n}[1]{(#1.north)}
\newcommand{\s}[1]{(#1.south)}

\begin{document}

\begin{tikzpicture}
 \node (a) at (0,0) {};
 \node (b) at (0,1) {};
 \draw \n{a} -- \s{b};
\end{tikzpicture}

\end{document}
| improve this answer | |
  • When the parser is not looking for something particular, say ( or [, it will expand things as it walks on the code. Otherwise you have to pause the parser with \pgfextra – percusse Nov 3 '17 at 14:09
3

You could use a loop to draw the lines. To specify different colours for the lines, you can have a third loop variable, and leave that empty where you want the default colour (black, unless you say otherwise).

output of code

\documentclass[border=5mm]{standalone}
\usepackage{tikz}
\begin{document}
\begin{tikzpicture}
 \node (a) at (0,0) {};
 \node (b) at (0,1) {};
 \node (c) at (0.2,0) {};
 \node (d) at (0.2,1) {};
 \node (e) at (0.4,0) {};
 \node (f) at (0.4,1) {};

\foreach \i/\j/\clr in {
   a/b/,
   c/d/red,
   e/f/}
  \draw [\clr] (\i.north) -- (\j.south);
\end{tikzpicture}
\end{document}
| improve this answer | |
  • I upvoted your answer, even if eventually I accepted the other one, because anyway this one is a nice workaround which can be useful in other occasions. – brad Nov 3 '17 at 14:13

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