How to find the widest of several nodes with Tikz

Question

I have several nodes and I need to find which one is the widest.

This example draws a vertical line in through the center of the widest node, but I know which node to choose.

\documentclass[12pt,a4paper]{article}
\usepackage{tikz}
\begin{document}
\begin{tikzpicture}
    \foreach \name[count = \n] in {peter, wellingon, john, william}{
        \node[draw, anchor = west] at (0, -\n) (name-\n) {\name};
    }
    \draw let \p1 = (name-1.center) in (\x1, 0) -- (\x1, -5);
\end{tikzpicture}
\end{document}

Can I make this choice automatically? I could't find a way to get a coordinate and store it in a global macro for later use. The let operator gives easy access to coordinates, but only inside a path, as far as I know.

There's How can I get the height of a TikZ node and store it in a length? - TeX - LaTeX Stack Exchange which should be sufficient... — user202729, Dec 22 '21 at 20:16
Or https://tex.stackexchange.com/questions/521098/how-to-calculate-the-size-of-a-tikz-node#comment1317732_521098 ? But using fit or something may be simpler as you don't need to compute the maximum manually — user202729, Dec 22 '21 at 20:17

antshar · Answer 1 · 2021-12-23T14:33:59.990

Edit:

You can box all the content by placing everything in scope. Adding local bounding box option allows you to name it and use as a regular node.

\documentclass[12pt,a4paper]{article}
\usepackage{tikz}
\begin{document}
\begin{tikzpicture}
\begin{scope}[local bounding box=foo]
    \foreach \name[count = \n] in {peter, wellingon, john, william}{
        \node[draw, anchor = west] at (0, -\n) (name-\n) {\name};
    }
\end{scope}
\draw ([yshift=20pt]foo.north) -- ([yshift=-20pt]foo.south);
\end{tikzpicture}
\end{document}

Thanks to @sgmoye for this remark.

Original post:

One can be solved using fit library so that you will able to create a phantom box (node) with the side length of the biggest node.

\documentclass[12pt,a4paper]{article}
\usepackage{tikz}
\usetikzlibrary{fit}
\begin{document}
\begin{tikzpicture}
\begin{scope}[local bounding box=foo]
    \foreach \name[count = \n] in {peter, wellingon, john, william}{
        \node[draw, anchor = west] at (0, -\n) (name-\n) {\name};
    }
\end{scope}
\nodefit=(foo),inner sep=0pt{};
\draw[shorten >=-20pt,shorten <=-20pt] (bar.south) -- (bar.north);
\end{tikzpicture}
\end{document}

Actually, you can replace all of the drawing in your proposal with \draw ([yshift=20pt]foo.north) -- ([yshift=-20pt]foo.south); I had not known about local bounding box=... possibility -- thank you for that. And no need for the \usetikzlibrary{fit}. — sgmoye, Dec 23 '21 at 10:57
@sgmoye your're right. Great remark, I edited the answer. Thanks! — antshar, Dec 23 '21 at 14:34

score 1 · Answer 2 · answered Dec 23 '21 at 15:55

The answer by @antshar is far and away the best approach. However, you did ask how to find the longest node, and this does answer that question. This is a far less elegant solution and is not really recommended. I give it only FYI. The origins for this code can be found at Color two columns separately in a matrix though variations on it can be found in a number of places. Output is the same.

\documentclass[12pt,a4paper]{article}
\usepackage{tikz}
\newlength{\tempx}
\newlength{\longest}
\begin{document}
\begin{tikzpicture}
    \foreach \name[count = \n] in {peter, wellingon, john, william}{%
        \node[draw, anchor = west] at (0, -\n) (name-\n) {\name};
        \pgfextractx{\tempx}
            {\pgfpointdiff{\pgfpointanchor{name-\n}{west}}{\pgfpointanchor{name-\n}{east}}}
            \ifdim\tempx>\longest \global\longest=\tempx\fi%% foreach is local
    }%
    \draw ([xshift=\longest/2,yshift=-20pt]current bounding box.south west) --
        ([xshift=\longest/2,yshift=20pt]current bounding box.north west);
\end{tikzpicture}
\end{document}

How to find the widest of several nodes with Tikz

2 Answers2

Edit:

Original post: