How can I shift a line the endpoints of which are defined by anchors like this?
\draw (i0) -- (i1);
The following approach does not work:
\draw[xshift=2pt] (i0) -- (i1);
[Tex/LaTex] Shift line connecting anchors
tikz-pgf
Related Solutions
You can add the shift by one of the following ways. Only the last one needs the calc library. In both cases you need to surround the coordinate with { } to either mask the inner [ ] or ,.
Hello\tikz[baseline={([yshift=-5pt]current bounding box.north)}]\draw(0,0) circle (10pt);\LaTeX.
Hello\tikz[baseline={($ (current bounding box.north) - (0,5pt) $)}]\draw(0,0) circle (10pt);\LaTeX.
Note that the ([<options>]coordinate) style works also in general (so does the other one). If you want to shift from the center of a node you should explicitly reference the center anchor to avoid some possible issues. Without an anchor the special border anchor is used which seems to cause issues with shifts. With baseline there might be no problem, but for drawn lines there is.
This kind of thing is done quite a fair bit in the various shapes libraries. It isn't necessarily straightforward, and someone may have found an easier way of doing it, but the following illustrates the general technique used.
\documentclass[tikz,border=5]{standalone}
\pgfkeys{/pgf/.cd,
box pins/.initial=1,
}
\makeatletter
% A very simple shape
\pgfdeclareshape{box}{
\nodeparts{}
\saveddimen\halfwidth{%
\pgfmathsetlength\pgf@x{\pgfkeysvalueof{/pgf/minimum width}/2}%
}
\saveddimen\halfheight{%
\pgfmathsetlength\pgf@x{\pgfkeysvalueof{/pgf/minimum height}/2}%
}
\savedmacro\boxpins{%
\pgfmathparse{int(\pgfkeysvalueof{/pgf/box pins})}%
\let\boxpins=\pgfmathresult
}
\savedanchor\centerpoint{\pgfpointorigin}
\anchor{center}{\centerpoint}
\backgroundpath{%
\pgfpathrectanglecorners{\pgfqpoint{-\halfwidth}{-\halfheight}}%
{\pgfqpoint{\halfwidth}{\halfheight}}
}
%
% Now add some code which will be executed
% every time a box node is created.
\pgfutil@g@addto@macro\pgf@sh@s@box{%
\pgfmathloop%
% The macro \boxpins will have been defined when code is executed
\ifnum\pgfmathcounter>\boxpins\relax%
\else%
% Only add if anchor not defined yet defined.
\pgfutil@ifundefined{pgf@anchor@box@pin \pgfmathcounter}{%
% Create anchor pin X for the shape box
\expandafter\xdef\csname pgf@anchor@box@pin \pgfmathcounter\endcsname{%
\noexpand\pgf@sh@lib@box@pinanchor{\pgfmathcounter}%
}%
}{}%
\repeatpgfmathloop%
}
}
\def\pgf@sh@lib@box@pinanchor#1{%
% All the saved dimens and macros will be available
% when this macro is called.
\ifnum#1>\boxpins\relax%
% Should issue an error.
\pgfpointorigin%
\else
\pgfpoint{\halfwidth}{\halfheight*2/(\boxpins+1)*#1-\halfheight}%
\fi
}
\begin{document}
\begin{tikzpicture}[minimum size=1cm]
\foreach \b in {1,...,5}{
\node [box, box pins=\b, draw, label=center:\b] at (0,2*\b) (box-\b) {};
\foreach \p in {1,...,\b}
\draw (2,2*\b) -- (box-\b.pin \p);
}
% box-1 does not have a pin 5 anchor
\draw [red] (2,3) -- (box-1.pin 5);
\end{tikzpicture}
\end{document}
Best Answer
Approach 1
You can work with the library
calc:Approach 2
Another approach based on the
letoperation. An example is given in the question TikZ: Node at same x-coordinate as another node, but specified y-coordinate?Approach 3
Jake mentioned another approach. You set the shift to the coordinate directly. (this method is documented in the manual section 13 "Specifying Coordinates")
Note: By using this method will work fine if you define
i0andi1with\coordinate. If you definei0andi1with\nodeyou must give an anchorThis limitation isn't relevant for the other approaches.
Complete example with result