Defining a command to generate a sequence

Question

I'm trying to define a command to make a list of subscripted \varphi characters. Let's call it \phiseq. I want to be able to use \phiseq{1,2,3} in the body of my document to expand to $\varphi_1,\varphi_2,\varphi_3$ . Any ideas? I've tried several experiments with \@for, but none of them have worked.

score 7 · Accepted Answer · answered Jun 18 '13 at 16:36

7

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\documentclass{article}

\makeatletter
\def\phiseq#1{$\let\comma\@empty\@for\tmp:=#1\do{\comma\varphi_{\tmp}\def\comma{,}}$}
\makeatother

\begin{document}

\phiseq{1,2,3}

\end{document}

answered Jun 18 '13 at 16:36

David Carlisle

757,742

Apparently I need to learn more about \makeatletter! Thanks! – Kellvyn Jun 18 '13 at 16:45

score 5 · Answer 2 · answered Jun 18 '13 at 16:57

You can define a fairly general command with expl3.

\documentclass{article}
\usepackage{xparse}
\ExplSyntaxOn
\cs_new_protected:Npn \kellvyn_print_seq:nnnn #1 #2 #3 #4
 {% #1 = input delimiter
  % #2 = output delimiter
  % #3 = symbol
  % #4 = list
  % store the list in a sequence
  \seq_set_split:Nnn \l_kellvyn_args_seq { #1 } { #4 }
  % clear the auxiliary sequence
  \seq_clear:N \l_kellvyn_print_seq
  % process the argument list
  \seq_map_inline:Nn \l_kellvyn_args_seq
   {
    \tl_if_eq:nnTF { ##1 } { \dots }
     {% if the item is \dots, store it unchanged
      \seq_put_right:Nn \l_kellvyn_print_seq { \dots }
     }
     {% otherwise, store "symbol_{item}"
      \seq_put_right:Nn \l_kellvyn_print_seq { #3\sb{##1} }
     }
   }
  % print the sequence
  \seq_use:Nnnn \l_kellvyn_print_seq { #2 } { #2 } { #2 }
 }
\NewDocumentCommand{\phiseq}{m}
 {
  \kellvyn_print_seq:nnnn { , } { , }{ \varphi } { #1 }
 }
\NewDocumentCommand{\printseq}{O{,}mm}
 {
  \kellvyn_print_seq:nnnn { , } { #1 } { #2 } { #3 }
 }
\ExplSyntaxOff
\begin{document}
$\phiseq{1,2,3}$

$\phiseq{1,2,\dots,n}$

$\printseq{x}{1,2,\dots,k-1,k}$

$\printseq[;]{y}{1,2,3,4}$
\end{document}

With \phiseq you get your original request, but with the bonus that a \dots input will produce dots in the output. The \printseq command is more general, the optional argument is the output delimiter. I've used also a parameter for the input delimiter for possible extensions (it's not more difficult to have it, so better be as general as possible).

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Boris · Answer 3 · 2013-06-18T16:41:12.737

2

This works for me:

\documentclass{article}
\makeatletter
\newif\if@phiseqfirst
\def\phiseq#1{%
  \@phiseqfirsttrue
  $\@for\ind:=#1\do{
    \if@phiseqfirst\else,\fi
    \varphi_{\ind}
  \@phiseqfirstfalse}$}
\makeatother

\begin{document}

\phiseq{1,2,3,4}
\end{document}

The \@phiseqfirst magic is here to get rid of the spurious comma in the end (thanks to @GonzaloMedina for noting it).

edited Jun 18 '13 at 16:41

answered Jun 18 '13 at 16:33

Boris

38,129

1

In the output, there will be a spurious comma after the last term. – Gonzalo Medina Jun 18 '13 at 16:35
Thanks. Did not see it. I've added some code to prevent it. – Boris Jun 18 '13 at 16:41
You can use the kernel provided \if@tempswa, since \@for doesn't use it. – egreg Jun 18 '13 at 20:27

score 2 · Answer 4 · answered Jun 18 '13 at 20:12

Yet another solution with pgffor.

\documentclass[varwidth]{standalone}
\usepackage{pgffor}
\def\seq#1#2{%
  \foreach \i[count = \j] in {#2}{%
    \unless\ifnum\j=1\relax, \fi$#1_{\i}$}}
\def\phiseq#1{\seq{\phi}{#1}}
\begin{document}
\seq{\psi}{1,...,5}\par
\phiseq{10,...,20}
\end{document}

enter image description here

Defining a command to generate a sequence

4 Answers4