Handling loops in expl3
The command \clist_put_right:Nn
will leave \l_tmpa_tl
in the input, i.e. at the time of calling \result
to check for the equality, \l_tmpa_tl
contains 5
and \l_tmpb_tl
contains e
.
However, to get the pairwise content X/Y
, the current value of \l_tmpa_tl
and \l_tmpb_tl
must be evaluated (expanded) and stored to the \clist
variable, so use the x
type:
\clist_put_right:Nx #3 {\l_tmpa_tl/\l_tmpb_tl}
Most likely a \seq
list could be faster, but this depends on the use case.
\documentclass{article}
\usepackage{xparse}
\ExplSyntaxOn
\NewDocumentCommand{\pairwise}{mmm}{
\clist_set_eq:NN \l_tmpa_clist #1
\clist_set_eq:NN \l_tmpb_clist #2
\clist_clear:N #3
\bool_until_do:nn {\clist_if_empty_p:N \l_tmpa_clist}{
\clist_pop:NN \l_tmpa_clist \l_tmpa_tl
\clist_pop:NN \l_tmpb_clist \l_tmpb_tl
\clist_put_right:Nx #3 {\l_tmpa_tl / \l_tmpb_tl}
}
}
\ExplSyntaxOff
\begin{document}
\def\this{1,2,3,4,5}
\def\that{a,b,c,d,e}
There is \{\this\} and \{\that\}.
\pairwise{\this}{\that}{\result}
\def\expected{1/a,2/b,3/c,4/d,5/e}
Now \{\result\} is equal to \{\expected\}?
\end{document}
You have to use the value of the token list variable, not the token list.
Here's a different approach using sequences and \seq_mapthread_function:NNN
, that traverses two sequences and hands the items to a two argument function.
You can give a different output separator as optional argument to `\pairwise, so for example
\pairwise[|]{a,b}{1,2}{\result}
would define \result
to a/1|b/2
.
\documentclass{article}
\usepackage{xparse}
\ExplSyntaxOn
\NewDocumentCommand{\pairwise}{O{,}mmm}
{
\seq_set_from_clist:No \l_komarov_first_seq { #2 }
\seq_set_from_clist:No \l_komarov_second_seq { #3 }
\seq_clear:N \l_komarov_output_seq
\seq_mapthread_function:NNN
\l_komarov_first_seq
\l_komarov_second_seq
\komarov_addentry:nn
\tl_set:Nx #4 { \seq_use:Nn \l_komarov_output_seq { #1 } }
}
\seq_new:N \l_komarov_first_seq
\seq_new:N \l_komarov_second_seq
\seq_new:N \l_komarov_output_seq
\cs_generate_variant:Nn \seq_set_from_clist:Nn { No }
\cs_new_protected:Nn \komarov_addentry:nn
{
\seq_put_right:Nn \l_komarov_output_seq { #1 / #2 }
}
\ExplSyntaxOff
\begin{document}
\newcommand\this{1,2,3,4,5}
\newcommand\that{a,b,c,d,e}
There is \{\this\} and \{\that\}.
\pairwise{\this}{\that}{\result}
\newcommand\expected{1/a,2/b,3/c,4/d,5/e}
Now \{\result\} equals \{\expected\}
\pairwise{1,2,3,4,5}{a,b,c,d,e}{\newresult}
Also \{\newresult\} equals \{\expected\}
\end{document}