TeX point command

Question

I try to get my own command to work to output a point in standard german notation.

My code (adapted version of wipet's code):

\documentclass[12pt]{article}
\def\punkt#1{\left(\punktA#1,,\right)}
\def\punktA#1,{\if,#1,\else #1\vert\expandafter \punktA \fi}
\begin{document}
$\punkt{\frac{1}{\pi},2,3}$
\end{document}

Two problems arose:

1. How do I get rid of that last vertical bar after the "3"?
2. How can I automatically scale the vertical bars to the biggest input?

Answer 1

It's easier to use LaTeX's built-in comma list handling.

\documentclass[12pt]{article}
\ExplSyntaxOn
\NewDocumentCommand\punkt{m}{
\left(
\clist_use:nn{#1}{\middle|}
\right)}
\ExplSyntaxOff
\begin{document}
$\punkt{\frac{1}{\pi},2,3}$
\quad
$\punkt{\frac{\frac{a}{b}}{\frac{a}{b}},2,3}$
\end{document}

As Mico notes, it would look better with some space:

\NewDocumentCommand\punkt{m}{
\left(\,
\clist_use:nn{#1}{\;\middle|\;}
\,\right)}

Answer 2

The code is wrong to begin with (the original one, I mean), because it uses \if,#1, which is not a safe test for emptiness of #1. The problem is that \if expands tokens; in your case it works, because the expansion of \frac doesn't begin with a comma. But similar code might give problems if used in different contexts.

You want to first process the first item in the comma separated list and then start the recursion, where the \middle| is placed before the other items.

\documentclass[12pt]{article}
\def\punkt#1{\left(\punktA#1,,\right)}
\def\punktA#1,{#1\punktB}
\def\punktB#1,{%
  \if\relax\detokenize{#1}\relax % safe test for emptiness
    % do nothing in the true case
  \else
    ,\middle|,#1% delimiter and item
    \expandafter\punktB % restart the recursion
  \fi
}
\begin{document}
$\punkt{}$
$\punkt{a}$
$\punkt{a,b}$
$\punkt{\frac{1}{\pi},2,3}$
\end{document}

You see that it also works with an empty argument.

Of course, the \clist_use:nn approach suggested by David is much simpler and requires almost no thinking.

With a more natural input syntax:

\documentclass[12pt]{article}
\ExplSyntaxOn
\NewDocumentCommand{\punkt}{m}
 {
  \left(
  \seq_set_split:Nnn \l_tmpa_seq { | } { #1 }
  \seq_use:Nn \l_tmpa_seq { ,\middle|, }
  \right)
 }
\ExplSyntaxOff
\begin{document}
$\punkt{}$
$\punkt{a}$
$\punkt{a|b}$
$\punkt{\frac{1}{\pi} | 2 | 3}$
\end{document}

The output is the same. Spaces around | are ignored.

Answer 3

For use in practice I suggest sticking to the answer provided by David Carlisle.

But maybe—as a moot point/as an "academical thing"—you wish to know how you could implement a loop \PunktA yourself which does not append a vertical bar behind the last item of your comma-list:

You can have \punktA process two arguments instead of just one.
The first argument is undelimited/nested between {...} and denotes the tokens to prepend to the comma-list item processed in this iteration. In the first iteration it is empty. In subsequent iterations it contains \;\middle|\;.
The second argument is comma-delimited and denotes the comma-list item processed in this iteration:

\documentclass[12pt]{article}
\def\punkt#1{\left(,\punktA{}#1,,,\right)}
\def\gobble#1{}
\def\punktA#1#2,{\if,#2,\expandafter\gobble\else #1#2\expandafter\punktA\fi{;\middle|;}}
\begin{document}
$\punkt{\frac{1}{\pi},2,3}$
\end{document}

---

Probably this does the trick, also:

\documentclass[12pt]{article}
\def\separator{;\middle|;}
\def\punkt#1{\left(,\punktA\empty#1,,,\right)}
\def\punktA#1,{\ifx#1\separator\else #1\expandafter\punktA\expandafter\separator\fi}
\begin{document}
$\punkt{\frac{1}{\pi},2,3}$
\end{document}