2

First: This is similar to "Count the number of elements of each size in a list" and "Position of largest element in a list" but it's not the same.

Code + question:

\documentclass{beamer}

\usepackage{lmodern}

\def\elevA{6}
\def\elevB{0}
\def\elevC{0}
\def\elevD{3}
\def\elevE{0}
\def\elevF{2}
\def\elevG{1}
\def\elevH{1}
\def\elevI{1}
\def\elevJ{5}
\def\elevK{0}
\def\elevL{3}
\def\elevM{7}
\def\elevN{3}
\def\elevO{1}
\def\elevP{1}
\def\elevQ{0}
\def\elevR{0}
\def\elevS{0}
\def\elevT{2}
\def\elevU{1}
\def\elevV{2}
\def\elevW{0}
\def\elevX{2}
\def\elevY{1}
\def\elevZ{1}
\def\elevAa{4}
\def\elevAb{6}
\def\elevAc{1}
\def\elevAd{1}
\def\elevAe{6}
\def\elevAf{2}
\def\elevAg{0}
\def\elevAh{3}
\def\elevAi{2}
\def\elevAj{1}
\def\elevAk{0}
\def\elevAl{0}
\def\elevAm{0}
\def\elevAn{0}
\def\elevAo{0}
\def\elevAp{4}

\begin{document}

\begin{frame}
What number among the ones defined in the input occurs most often?\\[\baselineskip]
(I know that the answer is `$0$' in this case but how can I get \LaTeX{} to calculate it?)
\end{frame}

\end{document}

output

Note: All the values are defined beforehand via the \def method.

Community
  • 1
Svend Tveskæg
  • 31,033

3 Answers3

2

Assuming the possible values are from 0 to 9. Can be easily extended for cases the possible values not known in advance. But there is no point in aiming for the most general.

The context from the earlier questions linked to in the post contains all necessary structure, but we must do some stuff from scratch here.

\documentclass{beamer}

\usepackage{lmodern}

\def\elevA{6}
\def\elevB{0}
\def\elevC{0}
\def\elevD{3}
\def\elevE{0}
\def\elevF{2}
\def\elevG{1}
\def\elevH{1}
\def\elevI{1}
\def\elevJ{5}
\def\elevK{0}
\def\elevL{3}
\def\elevM{7}
\def\elevN{3}
\def\elevO{1}
\def\elevP{1}
\def\elevQ{0}
\def\elevR{0}
\def\elevS{0}
\def\elevT{2}
\def\elevU{1}
\def\elevV{2}
\def\elevW{0}
\def\elevX{2}
\def\elevY{1}
\def\elevZ{1}
\def\elevAa{4}
\def\elevAb{6}
\def\elevAc{1}
\def\elevAd{1}
\def\elevAe{6}
\def\elevAf{2}
\def\elevAg{0}
\def\elevAh{3}
\def\elevAi{2}
\def\elevAj{1}
\def\elevAk{0}
\def\elevAl{0}
\def\elevAm{0}
\def\elevAn{0}
\def\elevAo{0}
\def\elevAp{4}

\makeatletter
\@tfor\svend@x:=0123456789\do{\@namedef{svend@\svend@x}{0}}

\@tfor\svend@x:=\elevA
\elevB
\elevC
\elevD
\elevE
\elevF
\elevG
\elevH
\elevI
\elevJ
\elevK
\elevL
\elevM
\elevN
\elevO
\elevP
\elevQ
\elevR
\elevS
\elevT
\elevU
\elevV
\elevW
\elevX
\elevY
\elevZ
\elevAa
\elevAb
\elevAc
\elevAd
\elevAe
\elevAf
\elevAg
\elevAh
\elevAi
\elevAj
\elevAk
\elevAl
\elevAm
\elevAn
\elevAo
\elevAp\do{\expandafter\edef\csname svend@\svend@x\endcsname
            {\the\numexpr\csname svend@\svend@x\endcsname+\@ne}}

\def\svend@Max {-1}
\def\svend@Min {"7FFFFFF}

\@tfor\svend@x:=0123456789\do{%
  \ifnum\csname svend@\svend@x\endcsname>\z@
      \ifnum\csname svend@\svend@x\endcsname>\svend@Max\relax
        \edef\MostFrequent {\svend@x}%
        \edef\svend@Max {\csname svend@\svend@x\endcsname }%
      \fi
      \ifnum\csname svend@\svend@x\endcsname<\svend@Min\relax
        \edef\LessFrequent {\svend@x}%
        \edef\svend@Min {\csname svend@\svend@x\endcsname }%
      \fi
  \fi
}

\makeatother



\begin{document}

\begin{frame}
What number among the ones defined in the input occurs most often?\\
(I got \LaTeX{} to compute that the smallest among them is `$\MostFrequent$')\\[\baselineskip]

And what number among the ones defined occurs less often ?\\
(I got \LaTeX{} to compute that the smallest among them is `$\LessFrequent$')

\end{frame}

\end{document}

enter image description here

  • @SvendTveskæg let's hope it fits well with all the other macros you are gathering step by step ;-). Side remark: in \def\svend@Min {"7FFFFFF} there should be an additional F. But this doesn't matter really as "7FFFFFF is 134217727 and I doubt your input will have more than 134217727 occurrences for a given value ! –  Mar 22 '15 at 17:29
2

A small addition to the answer you sadly unaccepted last time:-)

\documentclass{beamer}

\usepackage[danish]{babel}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage{lmodern}
\usepackage{booktabs}
\usepackage{siunitx}
\usepackage{array}

\def\most{0}
\def\mostnum{0}

\def\zz\ignorespaces#1{%
\expandafter\xdef\csname zz#1\endcsname{%
\expandafter\ifx\csname zz#1\endcsname\relax
1%
\else
\the\numexpr\csname zz#1\endcsname+1\relax
\fi}%
\ifnum\mostnum<\csname zz#1\endcsname
  \gdef\most{#1}%
   \global\expandafter\let\expandafter\mostnum\csname zz#1\endcsname
\fi
#1}

\begin{document}

\def\elevA{6}
\def\elevB{0}
\def\elevC{0}
\def\elevD{3}
\def\elevE{0}
\def\elevF{2}
\def\elevG{1}
\def\elevH{1}
\def\elevI{1}
\def\elevJ{5}
\def\elevK{0}
\def\elevL{3}
\def\elevM{7}
\def\elevN{3}
\def\elevO{1}
\def\elevP{1}
\def\elevQ{0}
\def\elevR{0}
\def\elevS{0}
\def\elevT{2}
\def\elevU{1}
\def\elevV{2}
\def\elevW{0}
\def\elevX{2}
\def\elevY{1}
\def\elevZ{1}
\def\elevAa{4}
\def\elevAb{6}
\def\elevAc{1}
\def\elevAd{1}
\def\elevAe{6}
\def\elevAf{2}
\def\elevAg{0}
\def\elevAh{3}
\def\elevAi{2}
\def\elevAj{1}
\def\elevAk{0}
\def\elevAl{0}
\def\elevAm{0}
\def\elevAn{0}
\def\elevAo{0}
\def\elevAp{4}

\begin{frame}

\begin{table}
\centering
\small
  \begin{tabular}{
    *{14}{>\zz c}
  }
   \toprule
    \elevA  & \elevB  & \elevC  & \elevD  & \elevE  & \elevF  & \elevG  &
    \elevH  & \elevI  & \elevJ  & \elevK  & \elevL  & \elevM  & \elevN    \\[0.5ex]
    \elevO  & \elevP  & \elevQ  & \elevR  & \elevS  & \elevT  & \elevU  &
    \elevV  & \elevW  & \elevX  & \elevY  & \elevZ  & \elevAa & \elevAb   \\[0.5ex]
    \elevAc & \elevAd & \elevAe & \elevAf & \elevAg & \elevAh & \elevAi &
    \elevAj & \elevAk & \elevAl & \elevAm & \elevAn & \elevAo & \elevAp   \\
   \bottomrule
  \end{tabular}
\end{table}

{\count0=0
\loop
\the\count0:\csname zz\the\count0\endcsname\endgraf
\ifnum\count0<8
\advance\count0 1
\repeat
}


\most\ occurred the most times (\mostnum)
\end{frame}

\end{document}
David Carlisle
  • 757,742
  • Great. Actually, I'll accept your answer to the other question once again. With this additional code, your answers give the most homogeneous code all in all. :) – Svend Tveskæg Mar 22 '15 at 17:05
  • 1
    Reusing code from (even one's own) earlier answers is cheating ;-) ! –  Mar 22 '15 at 17:31
2

Here's the extended version for also counting the number that appeared the most after the last \countappearances declaration.

\documentclass{beamer}

\usepackage[danish]{babel}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage{lmodern}
\usepackage{booktabs}
\usepackage{siunitx}
\usepackage{xparse}

\ExplSyntaxOn
\prop_new:N \g_svend_elev_prop
\prop_gput:Nnn \g_svend_elev_prop {A} {6}
\prop_gput:Nnn \g_svend_elev_prop {B} {0}
\prop_gput:Nnn \g_svend_elev_prop {C} {0}
\prop_gput:Nnn \g_svend_elev_prop {D} {3}
\prop_gput:Nnn \g_svend_elev_prop {E} {0}
\prop_gput:Nnn \g_svend_elev_prop {F} {2}
\prop_gput:Nnn \g_svend_elev_prop {G} {1}
\prop_gput:Nnn \g_svend_elev_prop {H} {1}
\prop_gput:Nnn \g_svend_elev_prop {I} {1}
\prop_gput:Nnn \g_svend_elev_prop {J} {5}
\prop_gput:Nnn \g_svend_elev_prop {K} {0}
\prop_gput:Nnn \g_svend_elev_prop {L} {3}
\prop_gput:Nnn \g_svend_elev_prop {M} {7}
\prop_gput:Nnn \g_svend_elev_prop {N} {3}
\prop_gput:Nnn \g_svend_elev_prop {O} {1}
\prop_gput:Nnn \g_svend_elev_prop {P} {1}
\prop_gput:Nnn \g_svend_elev_prop {Q} {0}
\prop_gput:Nnn \g_svend_elev_prop {R} {0}
\prop_gput:Nnn \g_svend_elev_prop {S} {0}
\prop_gput:Nnn \g_svend_elev_prop {T} {2}
\prop_gput:Nnn \g_svend_elev_prop {U} {1}
\prop_gput:Nnn \g_svend_elev_prop {V} {2}
\prop_gput:Nnn \g_svend_elev_prop {W} {0}
\prop_gput:Nnn \g_svend_elev_prop {X} {2}
\prop_gput:Nnn \g_svend_elev_prop {Y} {1}
\prop_gput:Nnn \g_svend_elev_prop {Z} {1}
\prop_gput:Nnn \g_svend_elev_prop {Aa} {4}
\prop_gput:Nnn \g_svend_elev_prop {Ab} {6}
\prop_gput:Nnn \g_svend_elev_prop {Ac} {1}
\prop_gput:Nnn \g_svend_elev_prop {Ad} {1}
\prop_gput:Nnn \g_svend_elev_prop {Ae} {6}
\prop_gput:Nnn \g_svend_elev_prop {Af} {2}
\prop_gput:Nnn \g_svend_elev_prop {Ag} {0}
\prop_gput:Nnn \g_svend_elev_prop {Ah} {3}
\prop_gput:Nnn \g_svend_elev_prop {Ai} {2}
\prop_gput:Nnn \g_svend_elev_prop {Aj} {1}
\prop_gput:Nnn \g_svend_elev_prop {Ak} {0}
\prop_gput:Nnn \g_svend_elev_prop {Al} {0}
\prop_gput:Nnn \g_svend_elev_prop {Am} {0}
\prop_gput:Nnn \g_svend_elev_prop {An} {0}
\prop_gput:Nnn \g_svend_elev_prop {Ao} {0}
\prop_gput:Nnn \g_svend_elev_prop {Ap} {4}

\prop_new:N \g_svend_count_prop
\prop_new:N \g_svend_count_zero_prop
\prop_gput:Nnn \g_svend_count_zero_prop { 0 } { 0 }
\prop_gput:Nnn \g_svend_count_zero_prop { 1 } { 0 }
\prop_gput:Nnn \g_svend_count_zero_prop { 2 } { 0 }
\prop_gput:Nnn \g_svend_count_zero_prop { 3 } { 0 }
\prop_gput:Nnn \g_svend_count_zero_prop { 4 } { 0 }
\prop_gput:Nnn \g_svend_count_zero_prop { 5 } { 0 }
\prop_gput:Nnn \g_svend_count_zero_prop { 6 } { 0 }
\prop_gput:Nnn \g_svend_count_zero_prop { 7 } { 0 }
\prop_gset_eq:NN \g_svend_count_prop \g_svend_count_zero_prop

\tl_new:N \l_svend_number_tl
\tl_new:N \l_svend_count_tl
\tl_new:N \l_svend_mostnum_tl
\tl_new:N \l_svend_mostcount_tl

\DeclareExpandableDocumentCommand{\elev}{m}
 {
  \svend_get_item:n { #1 }
 }

\NewDocumentCommand{\countappearances}{}
 {
  % now \elev will also count
  \cs_set_eq:NN \elev \svend_get_item_count:n
  % reinitialize the counter property list
  \prop_set_eq:NN \g_svend_count_prop \g_svend_count_zero_prop
 }

\NewDocumentCommand{\getmost}{m}
 {
  \tl_set:Nn \l_svend_mostnum_tl { 0 }
  \tl_set:Nx \l_svend_mostcount_tl { \prop_item:Nn \g_svend_count_prop { 0 } }
  \int_step_inline:nnnn { 1 } { 1 } { 7 }
   {
    \int_compare:xT { \prop_item:Nn \g_svend_count_prop { ##1 } > \l_svend_mostcount_tl }
     {
      \tl_set:Nn \l_svend_mostnum_tl { ##1 }
      \tl_set:Nx \l_svend_mostcount_tl { \prop_item:Nn \g_svend_count_prop { ##1 } }
     }
   }
  \tl_set_eq:NN \mostnum \l_svend_mostnum_tl
  \tl_set_eq:NN \mostcount \l_svend_mostcount_tl
  #1
 }

\cs_generate_variant:Nn \int_compare:nT { x }

\DeclareExpandableDocumentCommand{\HowMany}{m}
 {
  \prop_item:Nn \g_svend_count_prop { #1 }
 }

\cs_new:Npn \svend_get_item:n #1
 {
  \prop_item:Nn \g_svend_elev_prop { #1 }
 }

\cs_new_protected:Npn \svend_get_item_count:n #1
 {
  \tl_set:Nx \l_svend_number_tl { \svend_get_item:n { #1 } }
  % print the entry
  \tl_use:N \l_svend_number_tl
  % get the current count
  \tl_set:Nx \l_svend_count_tl
   {
    \prop_item:NV \g_svend_count_prop \l_svend_number_tl
   }
  % advance the count by 1
  \tl_set:Nx \l_svend_count_tl { \int_to_arabic:n { \l_svend_count_tl + 1 } }
  % update the property
  \prop_gput:NVV \g_svend_count_prop \l_svend_number_tl \l_svend_count_tl
 }

\cs_generate_variant:Nn \prop_item:Nn { NV }
\cs_generate_variant:Nn \prop_gput:Nnn { NVV }

\ExplSyntaxOff

\begin{document}

\begin{frame}

\begin{table}
\centering
\small
\countappearances

\begin{tabular}{*{14}{c}}
\toprule
\elev{A}  & \elev{B}  & \elev{C}  & \elev{D}  & \elev{E}  & \elev{F}  & \elev{G}  &
\elev{H}  & \elev{I}  & \elev{J}  & \elev{K}  & \elev{L}  & \elev{M}  & \elev{N}    \\[0.5ex]
\elev{O}  & \elev{P}  & \elev{Q}  & \elev{R}  & \elev{S}  & \elev{T}  & \elev{U}  &
\elev{V}  & \elev{W}  & \elev{X}  & \elev{Y}  & \elev{Z}  & \elev{Aa} & \elev{Ab}   \\[0.5ex]
\elev{Ac} & \elev{Ad} & \elev{Ae} & \elev{Af} & \elev{Ag} & \elev{Ah} & \elev{Ai} &
\elev{Aj} & \elev{Ak} & \elev{Al} & \elev{Am} & \elev{An} & \elev{Ao} & \elev{Ap}   \\
\bottomrule
\end{tabular}

\bigskip

\begin{tabular}{*{8}{c}}
\toprule
\multicolumn{8}{c}{How many} \\
\midrule
0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\
\HowMany{0} &
\HowMany{1} &
\HowMany{2} &
\HowMany{3} &
\HowMany{4} &
\HowMany{5} &
\HowMany{6} &
\HowMany{7} \\
\bottomrule
\end{tabular}

\bigskip

\getmost{%
  The number that appeared the most is \mostnum;
  it appeared \mostcount\ times.}

\end{table}
\end{frame}

\end{document}

The macro \getmost takes as parameter the text you want to print. If you use \getmost{}, no text will be printed, but you'll have available \mostnum and \mostcount anyway. Note that this is local to the group where \countappearances appears. It's not difficult to make it global, if needed.

enter image description here

egreg
  • 1,121,712