预定义元素集的中位数

Question

我想使用 LaTeX 3 自动计算中位数一组预定义的数据。

请考虑以下示例：

\documentclass{article}

\def\waterAa{82}
\def\waterAb{51}
\def\waterAc{144}
\def\waterAd{84}
\def\waterAe{120}
\def\waterAf{148}
\def\waterAg{148}
\def\waterAh{108}
\def\waterAi{160}
\def\waterAj{86}

\begin{document}

\noindent Consider the data set
\begin{equation}
  \{\waterAa, \waterAb, \waterAc, \waterAd, \waterAe, \waterAf, \waterAg, \waterAh, \waterAi, \waterAj\}
\end{equation}
To find the median of a data set, we first have to arrange the elements relative to their values:
\begin{equation}
  \{\waterAb, \waterAa, \waterAd, \waterAj, \waterAh, \waterAe, \waterAc, \waterAf, \waterAg, \waterAi\}
\end{equation}
Since the number of elements is $10$, the median~$M$ is the average of the fifth and sixth element:
\begin{equation}
  M = (\waterAh+\waterAe)/2 = 114.
\end{equation}

\section*{Question}
How to I \emph{automatically} calculate the median of a data set (preferably using \LaTeX 3)?

\end{document}

Answer 1

您可以使用浮点数；诀窍是在吸收它们时扩展它们的值。

\documentclass{article}

\usepackage{xparse}
\usepackage{expl3}

\ExplSyntaxOn

\NewDocumentCommand{\median}{m}
 {
  \svend_median:x { #1 }
 }

\clist_new:N \l__svend_median_clist
\int_new:N \l__svend_median_int

\cs_new_protected:Nn \svend_median:n
 {
  % set a comma separated list
  \clist_set:Nn \l__svend_median_clist { #1 }
  % sort it numerically
  \clist_sort:Nn \l__svend_median_clist
   {
    \fp_compare:nTF { ##1 > ##2 }
     { \sort_return_swapped: }
     { \sort_return_same: }
   }
  % compute the number of items
  \int_set:Nn \l__svend_median_int { \clist_count:N \l__svend_median_clist }
  \int_if_odd:nTF {\l__svend_median_int }
   {% if the number is odd, return the middle item
    \clist_item:Nn \l__svend_median_clist { (\l__svend_median_int + 1)/2 }
   }
   {% otherwise the average of the middle two elements
    \fp_eval:n
     {
      ( 
       \clist_item:Nn \l__svend_median_clist { \l__svend_median_int/2 }
       +
       \clist_item:Nn \l__svend_median_clist { \l__svend_median_int/2 + 1 }
      )/2
     }
   }
 }
\cs_generate_variant:Nn \svend_median:n { x }
\ExplSyntaxOff

\begin{document}

\def\waterAa{82}
\def\waterAb{51}
\def\waterAc{144}
\def\waterAd{84}
\def\waterAe{120}
\def\waterAf{148}
\def\waterAg{148}
\def\waterAh{108}
\def\waterAi{160}
\def\waterAj{86}


The median is \median{82,51,144,84,120,148,148,108,160,86}

The median is \median{
  \waterAa, \waterAb, \waterAc, \waterAd, \waterAe, 
  \waterAf, \waterAg, \waterAh, \waterAi, \waterAj
}

The median is \median{
  \waterAa, \waterAb, \waterAc, \waterAd, \waterAe, 
  \waterAf, \waterAg, \waterAh, \waterAi, %\waterAj
}

The median is \median{1,2,3,4}

The median is \median{1,2,3}

The median is \median{3.3,4.4,5.503,6.01}

The median is \median{3.3,4.4,5.503}

\end{document}

Answer 2

更新：l3sort在以前的版本中使用过，它的宏现在包含在内expl3。

幸运的是，对于数字序列的排序（我这里只假设整数值），有支持expl3。

其余部分只是确定索引中间位置的代码。但就目前情况而言，该代码不可扩展，因为\seq_sort:Nn不可扩展，并且\int_set:Nn也会阻止扩展。

\documentclass{article}

\usepackage{xparse}


\ExplSyntaxOn

\cs_generate_variant:Nn \seq_item:Nn {NV}

\NewDocumentCommand{\median}{m}{%
  \seq_set_from_clist:Nn \l_tmpa_seq {#1}
  \seq_sort:Nn \l_tmpa_seq {%
    \int_compare:nNnTF { ##1 } > { ##2 }
    { \sort_reversed: }
    { \sort_ordered: }
  }
  \int_set:Nn \l_tmpa_int {\seq_count:N \l_tmpa_seq }
  \int_if_odd:nTF { \l_tmpa_int } {%
    \int_set:Nn \l_tmpb_int {\l_tmpa_int / 2}
    \int_set:Nn \l_tmpa_int {\l_tmpb_int}
  }{%
    \int_set:Nn \l_tmpb_int {\l_tmpa_int / 2 }
    \int_set:Nn \l_tmpa_int {\l_tmpb_int + 1}
  }
  \fp_eval:n {(\seq_item:NV \l_tmpa_seq \l_tmpb_int + \seq_item:NV \l_tmpa_seq \l_tmpa_int)/2}
}

\ExplSyntaxOff

\begin{document}

\def\waterAa{82}
\def\waterAb{51}
\def\waterAc{144}
\def\waterAd{84}
\def\waterAe{120}
\def\waterAf{148}
\def\waterAg{148}
\def\waterAh{108}
\def\waterAi{160}
\def\waterAj{86}


The median is \median{82,51,144,84,120,148,148,108,160,86}

The median is \median{\waterAa, \waterAb, \waterAc, \waterAd, \waterAe, \waterAf, \waterAg, \waterAh, \waterAi, \waterAj}

\end{document}