recursive list type like definition in LaTeX

Question

I'm trying to typeset an OCaml like list and would like to define a command mylist for it. The expected result is something like below:

\mylist{}          -->     []
\mylist{aa}        -->     [aa]
\mylist{aa,bb}     -->     [aa;bb]
\mylist{aa,bb,cc}  -->     [aa;bb;cc]

I expect that , is always the typed separator (in LaTeX) but is displayed as another sign (here it is ;) in the pdf document (it's acceptable to hardcode the displayed one in mylist definition).

After searching for a while I found this answer and changed it a bit.

\newcounter{listcount}\newcounter{totalcount}%
\newcommand{\mylist}[1]{%
  \setcounter{totalcount}{0}% Reset total count
  \renewcommand*{\do}[1]{\stepcounter{totalcount}}% Reconfigure count
  \docsvlist{#1}% Count number of items
  \setcounter{listcount}{0}% Reset current item count
  \renewcommand*{\do}[1]{% Reconfigure item \do
    \stepcounter{listcount}% Next item
    ##1;\ifnum\value{listcount}<\value{totalcount}\,\fi% Print item
  }
  \left[\docsvlist{#1}\right]% Process list
}

The problem is that when the list contains >1 elements, there is a trailing ;.

\mylist{aa,bb} --> [aa;bb;]

Intuitively there is a need to test whether the element is the last one. However I seem lost in the macro definition w.r.t. listcount and the number of commas.

I also found a use of forcsvlist but the problem is similar. There seems a solution by using other packages, but a bit overkill so I tend not to use that.

Answer 1

The character ; must be shifted to be placed inside of the \ifnum\value....\fi test.

I also added a much, much, much shorter expl3 version right before \begin{document}, although the expl3 syntax might be very strange to L3 newbies...

I used the \docsvlist and \forcsvlist macros up to two years (2015) ago, but shifted then to he much more powerful expl3 syntax (although not easy to learn), especially for the multitude of lists there are in expl3, such as seq or clist or prop lists, which provide more functionality, in my point of view.

But first to the \ifnum... test:

 \ifnum\value{listcount} < \value{totalcount}
 ;\,
 \fi

just tests whether the current value of the listcount counter is smaller than the value of totalcount counter and prints a ;\, then, otherwise it doesn't.

The expl3 version is at the end and much simpler in its usage, no tests are needed (in fact, they are hidden behind \seq_use).

First enter the expl3 - syntax style with \ExplSyntaxOn and store the comma separated list values of the argument to a temporary sequence variable \l_tmpa_seq. Later display this list with \seq_use:Nn \l_tmpa_seq {;\,} -- the command takes care about the end of the list.

In order to prevent space gobbling, the wrapper command \displaylist is defined outside, see also Spaces in l3prop for the topic of spaces with expl3.

---

\documentclass{article}

\usepackage{etoolbox}

\newcounter{listcount}
\newcounter{totalcount}%
\newcommand{\mylist}[1]{%
  \setcounter{totalcount}{0}% Reset total count
  \renewcommand*{\do}[1]{\stepcounter{totalcount}}% Reconfigure count
  \docsvlist{#1}% Count number of items
  \setcounter{listcount}{0}% Reset current item count
  \renewcommand*{\do}[1]{% Reconfigure item \do
    \stepcounter{listcount}% Next item
    ##1\ifnum\value{listcount}<\value{totalcount};\,\fi% Print item
  }
  \left[\docsvlist{#1}\right]% Process list
}


\begin{document}
$\mylist{aa,bb,cc}$

$\mylist{aa}$


\end{document}

Of course $\mylist{}$ will yield [].

\documentclass{article}
\usepackage{expl3}

\ExplSyntaxOn
\newcommand{\myotherlist}[1]{%
  \seq_set_from_clist:Nn \l_tmpa_seq  {#1}
  \displaylist{\seq_use:Nn \l_tmpa_seq {;\,}}
}
\ExplSyntaxOff
\newcommand{\displaylist}[1]{%
  \left[#1\right]%
}

\begin{document}

$\myotherlist{}$

$\myotherlist{aa,bb,cc}$

$\myotherlist{aa}$

\end{document}

Answer 2

Here is an alternative solution which does not need to iterate the list twice. After the first element, the \do is replaced by \mydo:

\documentclass{article}

\usepackage{etoolbox}

\newcommand*{\mydo}[1]{;\,#1}

\newcommand{\mylist}[1]{%
  \renewcommand*{\do}[1]{##1\let\do\mydo}%
  \left[\docsvlist{#1}\right]% Process list
}

\begin{document}

$\mylist{aa,bb,cc}$

$\mylist{aa}$

\end{document}

Answer 3

A fully expandable version:

\documentclass{article}

\makeatletter
\newcommand{\ocamllist}[1]{\left[\ocaml@list#1,,\@nil}
\def\ocaml@list#1,#2,#3\@nil{%
  \if\relax\detokenize{#2}\relax
    \expandafter\@firstoftwo % no comma
  \else
    \expandafter\@secondoftwo
  \fi
  {\if\relax\detokenize{#1}\relax\,\else#1\fi\right]}
  {#1;\ocaml@list#2,#3,\@nil}
}
\makeatother

\begin{document}

$\ocamllist{}$

$\ocamllist{aa}$

$\ocamllist{aa,bb}$

$\ocamllist{aa,bb,cc}$

\end{document}

Not expandable, but more compact:

\documentclass{article}
\usepackage{xparse}

\ExplSyntaxOn
\NewDocumentCommand{\ocamllist}{m}
 {
  \left[
  \clist_set:Nn \l_hongxu_ocamllist_clist {#1}
  \clist_if_empty:NTF \l_hongxu_ocamllist_clist
   { \, }
   { \clist_use:Nn \l_hongxu_ocamllist_clist {;} }
  \right]
 }
\clist_new:N \l_hongxu_ocamllist_clist
\ExplSyntaxOff

\begin{document}

$\ocamllist{}$

$\ocamllist{aa}$

$\ocamllist{aa,bb}$

$\ocamllist{aa,bb,cc}$

\end{document}

Answer 4

May be I have not understood the problem, but instead of getting dizzy counting the items of the list and so... why not just take the argument as one string where you must replace the commas by semicolons?

\documentclass{article}

\usepackage{xstring}
\def\mylist#1{\ensuremath{\mathbin{[\StrSubstitute{#1}{,}{;}]}}}
\begin{document}

\mylist{}           

\mylist{aa}         

\mylist{aa,bb}      

\mylist{aa,bb,cc}   

\end{document}

Answer 5

Here is an expanable solution using only TeX primitives:

\def\ocamlist#1{\ocamlistA #1,,}
\def\ocamlistA#1,{[#1\ocamlistB}
\def\ocamlistB#1,{\ifx,#1,#1]\else;#1\expandafter\ocamlistB\fi}

$\ocamlist{aa,bb}$

\end

Answer 6

Here is a listofitems approach.

\documentclass{article}
\usepackage{listofitems}
\newcommand\mylist[1]{%
  [\ifx\relax#1\relax\else
  \readlist*\MyList{#1}%
  \foreachitem\i\in\MyList{\ifnum\icnt>1;\fi\i}\fi]%
}
\begin{document}
\mylist{}

\mylist{aa}

\mylist{aa,bb}

\mylist{aa,bb,cc}
\end{document}

If the space following the punctuation is important,

\documentclass{article}
\usepackage{listofitems}
\newcommand\mylist[1]{%
  [\ifx\relax#1\relax\else
  \readlist*\MyList{#1}%
  \foreachitem\i\in\MyList{\ifnum\icnt>1; \fi\i}\fi]%
}
\begin{document}
\mylist{}

\mylist{aa}

\mylist{aa,bb}

\mylist{aa,bb,cc}
\end{document}

Answer 7

This falls under Cunning (La)TeX tricks:

[aa;bb;cc]
[aa]
[]
[aa:bb:cc:dd]

\documentclass{article}

\usepackage{etoolbox}

\newcommand{\mylist}[2][;]{%
  \def\itemdelim{\def\itemdelim{#1}}% https://tex.stackexchange.com/a/89187/5764
  \renewcommand*{\do}[1]{\itemdelim##1}%
  [\docsvlist{#2}]% Process list
}

\begin{document}

\mylist{aa,bb,cc}

\mylist{aa}

\mylist{}

\end{document}

The principle is similar to Thomas' answer and delays the setting of \itemdelim by a single iteration, \defining it as a (re)definition. This way, at it's first use, it redefines \itemdelim without setting anything. Subsequent uses (if they exist) will actually set \itemdelim.