Given some columns of an array as comma-separated lists, how can I construct the array?

I figure if I can extract a specific row from a comma-separated list of such columns, then I'll probably be able to do the rest, but I can't even get the following code to compile:



    \foreach \x in #2 {
        \xdef\rowans{\rowans \pgfmathparse{{\x}[#1]}\pgfmathresult}


I somehow managed to figure out how to do this once before (using various things I found here on StackExchange), but now I just can't find any answers related to what I'm looking for.

Ideally I'm looking to do this with a variable number of columns, though not necessarily with a variable number of rows.

  • Welcome to TeX.SX! – user31729 Jun 5 '15 at 16:26
  • Do not put \pgfmathparse inside an \edef/\xdef. Do \pgfmathparse{{\x}[#1]}\xdef\rowans{\rowans\pgfmathresult} instead, for example. If you just do want to typeset the result, you don't need any \xdef or \rowans you can just do that in the loop: \pgfmathprint{{\x}[#1]}. – Qrrbrbirlbel Jun 5 '15 at 16:43
  • I still get the same error when I replace the relevant section with your suggestion: "Paragraph ended before \pgffor@next was complete". – Yiab Jun 6 '15 at 12:57

I'd suggest using expl3 and xparse; you define columns with a \setcolumn command and use \getrow that gathers the items in the chosen row and produces them with a separator given in the final argument.


  \clist_if_exist:cF { g_ylab_column_#1_clist }
    \clist_new:c { g_ylab_column_#1_clist }
  \clist_gset:cn { g_ylab_column_#1_clist } { #2 }

  \seq_clear:N \l_ylab_row_seq
  \clist_map_inline:nn { #2 }
    \seq_put_right:Nx \l_ylab_row_seq
      \clist_item:cn { g_ylab_column_##1_clist } { #1 }
  \seq_use:Nn \l_ylab_row_seq { #3 }




\getrow{2}{colone,coltwo,colthree}{, }

\getrow{1}{colone,coltwo,colthree}{&} \\
\getrow{2}{colone,coltwo,colthree}{&} \\


A \setcolumn command defines a clist variable if not already existent and (globally) sets it. You'll call it by name in a \getrow command that sets a seq variable with the items in the specified row. Then it produces the sequence with the chosen separator between items.

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Note that, in expl3 counting of items generally starts from 1 (which is more logical than starting from 0).

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