Matrices with shadows

Question

I'm trying to draw the shadows and also the dots between the two matrices:

\documentclass{article}
\usepackage{amsmath}
\begin{document}


    \centering
        \begin{tabular}{|c|c|c|c|c|c|c|c|}
        \hline
        5   & 1 &  3&  &2  & 1 & &  \\ \hline
        4   & 2 &  &  3&  & 4 & 3 & 1 \\ \hline
        1   & 3 &  &  &  & 2 & 5 & \\ \hline
        2   & 5 &  1& 2 &  & & & 1 \\ \hline
        0   & 4 &  &  & 5 &3  &1 & \\ \hline
        & 2 &  &  4&  &4 & 1& 2 \\ \hline

        \hline
    \end{tabular}
    \begin{tabular}{|c|c|c|c|c|c|c|c|}
        \hline
     1  & 1 &  1&  &1  & 1 & &  \\ \hline
     0  & 1 &  &  &  & 0 & 0 & 1 \\ \hline
    0   &  &  &  &  & 0 & 0 & \\ \hline
    1   &  &  1& 1 &  & & & 1 \\ \hline
    0   & 1 &  &  & 1 &1  &1 & \\ \hline
        & 1 &  &  &  & & &  \\ \hline

        \hline
    \end{tabular}



\end{document}

I wouldn’t play that card! All I can tell you is that all the solutions are fantastic! — NaveganTeX, Jul 16 '19 at 15:17

score 11 · Accepted Answer · 2019-07-06T08:58:30.583

Shadows can be cast (a) by groundhogs, (b) with shadows, and (c) with shadows.blur. I use shadows.blur here because you seem not to be asking for a duck-shaped shadow. The styles for the dividing lines and the frame are from here. I made the cells quadratic in the update.

\documentclass[tikz,border=3.14mm]{standalone}
\usetikzlibrary{matrix,calc,positioning,fit,backgrounds,shadows.blur}
\makeatletter
\long\def\ifnodedefined#1#2#3{%
    \@ifundefined{pgf@sh@ns@#1}{#3}{#2}%
}
\tikzset{matrix vlines/.style={execute at end matrix={
\foreach \XX in {1,...,\the\pgf@matrix@numberofcolumns}
{\xdef\FitList{}
\foreach \YY in {1,...,\the\pgfmatrixcurrentrow}
{\ifnodedefined{\tikzmatrixname-\YY-\XX}{\xdef\FitList{\FitList (\tikzmatrixname-\YY-\XX)}}{}
}
\node[fit=\FitList,draw=none,fill=none,inner sep=0pt,draw=none] (\tikzmatrixname-col-\XX) {};
}
\foreach \XX in {2,...,\the\pgf@matrix@numberofcolumns}
{\draw[#1] ($(\tikzmatrixname-col-\XX.west)!0.5!(\tikzmatrixname-col-\the\numexpr\XX-1\relax.east)$)
coordinate (aux) (aux|-\tikzmatrixname.north)
 --  (aux|-\tikzmatrixname.south);
}
}},matrix hlines/.style={execute at end matrix={
\foreach \YY in {1,...,\the\pgfmatrixcurrentrow}
{\xdef\FitList{}
\foreach \XX in {1,...,\the\pgf@matrix@numberofcolumns}
{\ifnodedefined{\tikzmatrixname-\YY-\XX}{\xdef\FitList{\FitList (\tikzmatrixname-\YY-\XX)}}{}
}
\node[fit=\FitList,draw=none,fill=none,inner sep=0pt,draw=none] (\tikzmatrixname-row-\YY) {};
}
\foreach \XX in {2,...,\the\pgfmatrixcurrentrow}
{\draw[#1] ($(\tikzmatrixname-row-\XX)!0.5!(\tikzmatrixname-row-\the\numexpr\XX-1\relax)$)
coordinate (aux) (aux-|\tikzmatrixname.west)
 --  (aux-|\tikzmatrixname.east);
}
}},
matrix dividers/.style={matrix vlines=#1,matrix hlines=#1},
matrix frame/.style={execute at end matrix={
\draw[#1] (\tikz@fig@name.south west) rectangle (\tikz@fig@name.north east);
}}}
\makeatother
\begin{document}
\begin{tikzpicture}[font=\sffamily\itshape,node distance=0.3em,
s-matrix/.style={matrix dividers,matrix frame,matrix of math nodes,
nodes in empty cells,fill=white,blur shadow,inner sep=0pt,
cells={text height=0.8em,text depth=0.2em,minimum width=1.8em,minimum height=1.8em}}]
\matrix[s-matrix] (mat1) {
        5   & 1 &  3&  &2  & 1 & &  \\ 
        4   & 2 &  &  3&  & 4 & 3 & 1 \\ 
        1   & 3 &  &  &  & 2 & 5 & \\ 
        2   & 5 &  1& 2 &  & & & 1 \\ 
        0   & 4 &  &  & 5 &3  &1 & \\ 
        & 2 &  &  4&  &4 & 1& 2 \\ 
};
\node[right=of mat1] (users) {users};
\matrix[s-matrix,right=of users] (mat2) {
     1  & 1 &  1&  &1  & 1 & &  \\ 
     0  & 1 &  &  &  & 0 & 0 & 1 \\ 
    0   &  &  &  &  & 0 & 0 & \\ 
    1   &  &  1& 1 &  & & & 1 \\ 
    0   & 1 &  &  & 1 &1  &1 & \\ 
        & 1 &  &  &  & & &  \\ 
};
\foreach \X in {1,2}
{\node[above=of mat\X] (items-\X) {items};
\node at (mat\X-1-1|-items-\X) {i$_1$};
\node at (mat\X-1-8|-items-\X) {i$_8$};}
\foreach \X in {1,6}
{\draw[dotted] (mat1-\X-2-|mat1.east) -- (mat2-\X-2-|mat2.west)
 node[midway,fill=white,inner sep=0.5pt]{u$_\X$};}
\node[below=1em of mat1,font=\sffamily\bfseries]  {Explicit ratings};
\node[below=1em of mat2,font=\sffamily\bfseries]  {Implicit ratings};
\end{tikzpicture}
\end{document}

Excellent! One small observation, how can we make the size of the cells in both matrices are the same? If you take a look at the screen shot, it does not matter if there is or there isn't numbers inside each cell, each one is a perfect square. — NaveganTeX, Jul 06 '19 at 08:37

score 6 · Answer 2 · edited Jul 06 '19 at 09:32

Less elegant than marmot's solution but maybe also interesting:

\documentclass{article}
\usepackage{amsmath,blkarray}
\usepackage{tikz}
\usetikzlibrary{shadows, shapes, positioning, matrix}
\makeatletter % from https://tex.stackexchange.com/a/129322/121799
\tikzset{no shadows/.code=\let\tikz@preactions\pgfutil@empty}
\makeatother
\begin{document}

   \begin{tikzpicture}[
    every node/.style = {draw, 
                         fill=white, 
                         drop shadow, 
                         anchor=center,
                         text height=2ex,
                         text depth=0.25ex},
    ]

    \matrix (leftmatrix) [no shadows,
                          draw=none, 
                          matrix of nodes, 
                          nodes in empty cells, 
                          row sep =-\pgflinewidth, 
                          column sep = -\pgflinewidth, 
                          minimum width=0.5cm, 
                          minimum height=0.5cm, 
                          ]
                          {
    5  & 1 & 3 &   & 2 & 1 &   &   \\  
    4  & 2 &   & 3 &   & 4 & 3 & 1 \\  
    1  & 3 &   &   &   & 2 & 5 &   \\  
    2  & 5 & 1 & 2 &   &   & 1 &   \\  
    0  & 4 &   &   & 5 & 3 & 1 &   \\  
       & 2 &   & 4 &   & 4 & 1 & 2 \\  
    };

    \node[right=0.25 cm of leftmatrix, draw=none, no shadows, font=\itshape] (users) {users};

    \matrix (rightmatrix) [no shadows,
                          draw=none, 
                          matrix of nodes, 
                          nodes in empty cells, 
                          row sep =-\pgflinewidth, 
                          column sep = -\pgflinewidth, 
                          minimum width=0.5cm, 
                          minimum height=0.5cm,
                          right= 0.25cm of users]
                          {
    1  & 1 &  1 &   & 1 & 1 &   &   \\ 
    0  & 1 &    &   &   & 0 & 0 & 1 \\ 
    0  &   &    &   &   & 0 & 0 &   \\ 
    1  &   &  1 & 1 &   &   &   & 1 \\ 
    0  & 1 &    &   & 1 & 1 & 1 &   \\ 
       & 1 &    &   &   &   &   &   \\ 
    };
    \node[above= 0.1cm of leftmatrix, draw=none, no shadows, font=\itshape] (itemsleft) {items};
    \node[above= 0.1cm of rightmatrix, draw=none, no shadows, font=\itshape] (itemsright) {items};
    \node[below= 0.25cm of leftmatrix, draw=none, no shadows, font=\bfseries] (captionleft) {Explicit Rating};
    \node[below= 0.25cm of rightmatrix, draw=none, no shadows, font=\bfseries] (captionright) {Implicit Rating};
    \node[draw=none, no shadows] at (leftmatrix-1-1|-itemsleft) {$i_1$};
    \node[draw=none, no shadows] at (leftmatrix-1-8|-itemsleft) {$i_8$};
    \node[draw=none, no shadows] at (rightmatrix-1-1|-itemsleft) {$i_1$};
    \node[draw=none, no shadows] at (rightmatrix-1-8|-itemsleft) {$i_8$};
    \draw[dotted] (leftmatrix-1-2-|leftmatrix.east) -- (rightmatrix-1-2-|rightmatrix.west)
 node[midway,fill=white,inner sep=0.5pt, draw=none, no shadows]{$u_1$};
    \draw[dotted] (leftmatrix-6-2-|leftmatrix.east) -- (rightmatrix-6-2-|rightmatrix.west)
 node[midway,fill=white,inner sep=0.5pt, draw=none, no shadows]{$u_6$};

    \end{tikzpicture}

\end{document}

Or without no shadows.

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{shadows, shapes, positioning, matrix}
\begin{document}

   \begin{tikzpicture}[
    every node/.style = {draw, 
                         fill=white, 
                         anchor=center,
                         text height=2ex,
                         text depth=0.25ex},
    ]

    \matrix (leftmatrix) [fill=white,drop shadow,inner sep=0pt,
                          draw=none, 
                          matrix of nodes, 
                          nodes in empty cells, 
                          row sep =-\pgflinewidth, 
                          column sep = -\pgflinewidth, 
                          cells={minimum width=0.5cm, 
                          minimum height=0.5cm}
                          ]
                          {
    5  & 1 & 3 &   & 2 & 1 &   &   \\  
    4  & 2 &   & 3 &   & 4 & 3 & 1 \\  
    1  & 3 &   &   &   & 2 & 5 &   \\  
    2  & 5 & 1 & 2 &   &   & 1 &   \\  
    0  & 4 &   &   & 5 & 3 & 1 &   \\  
       & 2 &   & 4 &   & 4 & 1 & 2 \\  
    };

    \node[right=0.25 cm of leftmatrix, draw=none, font=\itshape] (users) {users};

    \matrix (rightmatrix) [fill=white,drop shadow,inner sep=0pt,
                          draw=none, 
                          matrix of nodes, 
                          nodes in empty cells, 
                          row sep =-\pgflinewidth, 
                          column sep = -\pgflinewidth, 
                          cells={minimum width=0.5cm, 
                          minimum height=0.5cm},
                          right= 0.25cm of users]
                          {
    1  & 1 &  1 &   & 1 & 1 &   &   \\ 
    0  & 1 &    &   &   & 0 & 0 & 1 \\ 
    0  &   &    &   &   & 0 & 0 &   \\ 
    1  &   &  1 & 1 &   &   &   & 1 \\ 
    0  & 1 &    &   & 1 & 1 & 1 &   \\ 
       & 1 &    &   &   &   &   &   \\ 
    };
    \node[above= 0.1cm of leftmatrix, draw=none, font=\itshape] (itemsleft) {items};
    \node[above= 0.1cm of rightmatrix, draw=none, font=\itshape] (itemsright) {items};
    \node[below= 0.25cm of leftmatrix, draw=none, font=\bfseries] (captionleft) {Explicit Rating};
    \node[below= 0.25cm of rightmatrix, draw=none, font=\bfseries] (captionright) {Implicit Rating};
    \node[draw=none] at (leftmatrix-1-1|-itemsleft) {$i_1$};
    \node[draw=none] at (leftmatrix-1-8|-itemsleft) {$i_8$};
    \node[draw=none] at (rightmatrix-1-1|-itemsleft) {$i_1$};
    \node[draw=none] at (rightmatrix-1-8|-itemsleft) {$i_8$};
    \draw[dotted] (leftmatrix-1-2-|leftmatrix.east) -- (rightmatrix-1-2-|rightmatrix.west)
 node[midway,fill=white,inner sep=0.5pt, draw=none]{$u_1$};
    \draw[dotted] (leftmatrix-6-2-|leftmatrix.east) -- (rightmatrix-6-2-|rightmatrix.west)
 node[midway,fill=white,inner sep=0.5pt, draw=none]{$u_6$};

    \end{tikzpicture}
\end{document}

It is, without a doubt, a very interesting solution. Thank you! — NaveganTeX, Jul 06 '19 at 08:58
@marmot: Thanks for the edit. You are right that adding the shadow where needed is definitely a lot easier than removing it where not needed. Somehow this idea did not come to my mind. — leandriis, Jul 06 '19 at 09:42

Keks Dose · Answer 3 · 2019-07-12T13:00:56.360

If you are not inclined to learn Tikz, then just make a tabular. It doesn't take much code, the most sophisticated thing was the »shadow«.

\documentclass{article}
% \usepackage{amsmath}
\usepackage{array}
\usepackage{booktabs, colortbl}
\begin{document}

\renewcommand{\arraystretch}{1.3}\sffamily{}

    \begin{tabular}{*{18}{|c} |c|}
      \multicolumn{3}{l}{$i_1$} & \multicolumn{2}{c}{\raisebox{2ex}{items}} &
     \multicolumn{3}{r}{$i_8$} &   \multicolumn{3}{c}{}  &
     \multicolumn{3}{l}{$i_8$} & \multicolumn{2}{c}{\raisebox{2ex}{items}} &
                 \multicolumn{3}{r}{$i_8$}\\\cmidrule{1-8} \cmidrule{12-19}
     5 & 1 &  3&  &2  & 1 & &   & \multicolumn{3}{c|}{\dots $u_1$ \dots} &  5   & 1 &  3&  &2  & 1 & &\\\cmidrule{1-8} \cmidrule{12-19} 
     4 & 2 &  &  3&  & 4 & 3 & 1  & \multicolumn{3}{c|}{} &  4   & 2 &  &  3&  & 4 & 3 & 1 \\\cmidrule{1-8} \cmidrule{12-19}  
     1 & 3 &  &  &  & 2 & 5 &  & \multicolumn{3}{c|}{} & 1   & 3 &  &  &  & 2 & 5 &\\\cmidrule{1-8} \cmidrule{12-19}  
     2 & 5 &  1& 2 &  & & & 1  & \multicolumn{3}{c|}{users} & 2   & 5 &  1& 2 &  & & & 1\\\cmidrule{1-8} \cmidrule{12-19}  
     0 & 4 &  &  & 5 &3  &1 &  & \multicolumn{3}{c|}{} &  0   & 4 &  &  & 5 &3  &1 & \\\cmidrule{1-8} \cmidrule{12-19}  
       & 2 &  &  4&  &4 & 1& 2  & \multicolumn{3}{c|}{\dots $u_6$ \dots} &  & 2 & &  4&  &4 & 1& 2\\\cmidrule{1-8}
      \cmidrule{12-19}\addlinespace[-4pt]
      \arrayrulecolor[gray]{0.5}
      \cmidrule[2pt](l{0.2ex}r{-.2ex}){1-8}
      \cmidrule[2pt](l{0.2ex}r{-.2ex}){12-19}\addlinespace[2ex]
      \multicolumn{8}{c}{\textbf{Explicit ratings}}& \multicolumn{3}{c}{} & \multicolumn{8}{c}{\textbf{Implicit ratings}}\\
    \end{tabular}


\end{document}

Result:

EDIT

As suggested, with \clines, instead of \cmidrules. We need the latter for the shadow (as far as I know you can't trim \clines).

And I took the opportunity to streamline the layout, code of version 2:

\documentclass{article}
% \usepackage{amsmath}
\usepackage{array}
\usepackage{booktabs, colortbl, ragged2e}
\begin{document}

\renewcommand{\arraystretch}{1.8}\sffamily{}

    \begin{tabular}{*{18}{|>{\Centering}p{1em}} |>{\Centering}p{1em}|}
      \multicolumn{3}{l}{$i_1$} & \multicolumn{2}{c}{\raisebox{2ex}{items}} &
     \multicolumn{3}{r}{$i_8$} &   \multicolumn{3}{c}{}  &
     \multicolumn{3}{l}{$i_8$} & \multicolumn{2}{c}{\raisebox{2ex}{items}} &
                 \multicolumn{3}{r}{$i_8$}\\\cline{1-8} \cline{12-19}
     5 & 1 &  3&  &2  & 1 & &   & \multicolumn{3}{c|}{\dots $u_1$ \dots} &  5   & 1 &  3&  &2  & 1 & &\\\cline{1-8} \cline{12-19} 
     4 & 2 &  &  3&  & 4 & 3 & 1  & \multicolumn{3}{c|}{} &  4   & 2 &  &  3&  & 4 & 3 & 1 \\\cline{1-8} \cline{12-19}  
     1 & 3 &  &  &  & 2 & 5 &  & \multicolumn{3}{c|}{} & 1   & 3 &  &  &  & 2 & 5 &\\\cline{1-8} \cline{12-19}  
     2 & 5 &  1& 2 &  & & & 1  & \multicolumn{3}{c|}{users} & 2   & 5 &  1& 2 &  & & & 1\\\cline{1-8} \cline{12-19}  
     0 & 4 &  &  & 5 &3  &1 &  & \multicolumn{3}{c|}{} &  0   & 4 &  &  & 5 &3  &1 & \\\cline{1-8} \cline{12-19}  
       & 2 &  &  4&  &4 & 1& 2  & \multicolumn{3}{c|}{\dots $u_6$ \dots} &  & 2 & &  4&  &4 & 1& 2\\\cline{1-8}
      \cline{12-19}\addlinespace[-1pt]
      \arrayrulecolor[gray]{0.4}
      \cmidrule[3pt](l{0.2ex}r{-.2ex}){1-8}
      \cmidrule[3pt](l{0.2ex}r{-.2ex}){12-19}\addlinespace[1ex]
      \multicolumn{8}{c}{\textbf{Explicit ratings}}& \multicolumn{3}{c}{} & \multicolumn{8}{c}{\textbf{Implicit ratings}}\\
    \end{tabular}
\end{document}

And the picture:

The idea is good but I wouldn't use booktabs with vertical lines. — CarLaTeX, Jul 12 '19 at 12:40
@CarLaTeX Yes, but I needed \cmidrules for the empty gap between the two parts. — Keks Dose, Jul 12 '19 at 12:41
@CarLaTeX I had forgotten that \clines exist. Thank you, I amended my answer! — Keks Dose, Jul 12 '19 at 13:05

Matrices with shadows

3 Answers3

EDIT