3D box plot in Tikz LaTex

Question

How can we make these make kinds of 3D plots easily in LaTex? Is there any easy way available so we can draw this in that and in the end we can get tikz code? Any help would be highly appreciated

Answer 1

A solution with 3d library can be:

\documentclass[border=3mm]{standalone}
\usepackage{tikz}
\usetikzlibrary{calc,math,3d}
\newcommand{\myCube}[2]{%
%back face
\begin{scope}[canvas is yz plane at x=#1]
\foreach \Hpnt in {1,...,4} {
\coordinate (bfb-\Hpnt) at (\Hpnt-1,0);
\coordinate (bfu-\Hpnt) at (\Hpnt-1,3);
}
\foreach \Vpnt in {1,2} {
\coordinate (bfr-\Vpnt) at (3,\Vpnt);
\coordinate (bfl-\Vpnt) at (0,\Vpnt);
}
\end{scope}
%front face
\begin{scope}[canvas is yz plane at x=#1+1]
\foreach \Hpnt in {1,...,4} {
\coordinate (ffb-\Hpnt) at (\Hpnt-1,0);
\coordinate (ffu-\Hpnt) at (\Hpnt-1,3);
}
\foreach \Vpnt in {1,2} {
\coordinate (ffr-\Vpnt) at (3,\Vpnt);
\coordinate (ffl-\Vpnt) at (0,\Vpnt);
}
\end{scope}
\draw[fill=#2] (bfu-1) -- (ffu-1) -- (ffb-1) -- (ffb-4) -- (bfb-4) -- (bfu-4) -- cycle;
\foreach \jur in {2,3} {
\draw (bfu-\jur) -- (ffu-\jur) -- (ffb-\jur);
\tikzmath{
integer \jlr;
\jlr=\jur-1;
}
\draw (bfr-\jlr) -- (ffr-\jlr) -- (ffl-\jlr);
}
\draw (ffu-1) -- (ffu-4) -- (ffb-4) (bfu-4) -- (ffu-4);
}
\begin{document}
\begin{tikzpicture}[z={(90:1cm)},y={(-30:1cm)},x={(210:0.75cm)}]
\myCube{-3}{green!50!black}
\node at ($($(ffb-4)!0.5!(bfb-4)$)!2mm!90:(ffb-4)$) {$1$};
\myCube{-1.5}{red!80!black}
\node at ($($(ffb-4)!0.5!(bfb-4)$)!2mm!90:(ffb-4)$) {$1$};
\myCube{0}{orange}
\draw[orange,dashed,very thick] let
\p1=($(current bounding box.south west)!0.5!(current bounding box.north east)$),
\p2=($(current bounding box.south) - (current bounding box.north)$),
\n1={veclen(\x2,\y2)}
in (\p1) circle[radius=\n1/2+6.28mm]; 
\node at ($($(ffb-4)!0.5!(bfb-4)$)!2mm!90:(ffb-4)$) {$1$};
\node at ($($(ffb-1)!0.5!(ffb-4)$)!2mm!90:(ffb-1)$) {$3$};
\node at ($($(ffb-1)!0.5!(ffu-1)$)!2mm!90:(ffu-1)$) {$3$};
\end{tikzpicture}
\end{document}

Edit

For add more cubes you must add another `\myCube{position}{color} at the code. If you add it at the end of the code then the cubes will be on the foreground, if you add it at the beginning of the code it will be on the background. The position of a cube must be lesser than that of the cube in front of it. For example:

\documentclass[border=3mm]{standalone}
\usepackage{tikz}
\usetikzlibrary{calc,math,3d}
\newcommand{\myCube}[2]{%
%back face
\begin{scope}[canvas is yz plane at x=#1]
\foreach \Hpnt in {1,...,4} {
\coordinate (bfb-\Hpnt) at (\Hpnt-1,0);
\coordinate (bfu-\Hpnt) at (\Hpnt-1,3);
}
\foreach \Vpnt in {1,2} {
\coordinate (bfr-\Vpnt) at (3,\Vpnt);
\coordinate (bfl-\Vpnt) at (0,\Vpnt);
}
\end{scope}
%front face
\begin{scope}[canvas is yz plane at x=#1+1]
\foreach \Hpnt in {1,...,4} {
\coordinate (ffb-\Hpnt) at (\Hpnt-1,0);
\coordinate (ffu-\Hpnt) at (\Hpnt-1,3);
}
\foreach \Vpnt in {1,2} {
\coordinate (ffr-\Vpnt) at (3,\Vpnt);
\coordinate (ffl-\Vpnt) at (0,\Vpnt);
}
\end{scope}
\draw[fill=#2] (bfu-1) -- (ffu-1) -- (ffb-1) -- (ffb-4) -- (bfb-4) -- (bfu-4) -- cycle;
\foreach \jur in {2,3} {
\draw (bfu-\jur) -- (ffu-\jur) -- (ffb-\jur);
\tikzmath{
integer \jlr;
\jlr=\jur-1;
}
\draw (bfr-\jlr) -- (ffr-\jlr) -- (ffl-\jlr);
}
\draw (ffu-1) -- (ffu-4) -- (ffb-4) (bfu-4) -- (ffu-4);
}
\begin{document}
\begin{tikzpicture}[z={(90:1cm)},y={(-30:1cm)},x={(210:0.75cm)}]
\myCube{-4.5}{blue}%<- another cube on background
\myCube{-3}{green!50!black}
\node at ($($(ffb-4)!0.5!(bfb-4)$)!2mm!90:(ffb-4)$) {$1$};
\myCube{-1.5}{red!80!black}
\node at ($($(ffb-4)!0.5!(bfb-4)$)!2mm!90:(ffb-4)$) {$1$};
\myCube{0}{orange}
\node at ($($(ffb-4)!0.5!(bfb-4)$)!2mm!90:(ffb-4)$) {$1$};
%\node at ($($(ffb-1)!0.5!(ffb-4)$)!2mm!90:(ffb-1)$) {$3$};%<- label not visible therefore commented
%\node at ($($(ffb-1)!0.5!(ffu-1)$)!2mm!90:(ffu-1)$) {$3$};%<- label not visible therefore commented
\myCube{1.5}{violet}%<- another cube on foreground
\draw[orange,dashed,very thick] let
\p1=($(current bounding box.south west)!0.5!(current bounding box.north east)$),
\p2=($(current bounding box.south) - (current bounding box.north)$),
\n1={veclen(\x2,\y2)}
in (\p1) circle[radius=\n1/2+6.28mm];%<- since the dashed circle is draw using the bounding box of the picture you must shift here this part of the code
\end{tikzpicture}
\end{document}

Edit slanted labels

Changing a little the macro \myCube (see the comment in the code) you can get the following:

\documentclass[border=3mm]{standalone}
\usepackage{tikz}
\usetikzlibrary{calc,math,3d}
\def\OffsetLabels{0.7mm}
\newcommand{\myCube}[5]{%
%back face
\begin{scope}[canvas is yz plane at x=#1]
\foreach \Hpnt in {1,...,4} {
\coordinate (bfb-\Hpnt) at (\Hpnt-1,0);
\coordinate (bfu-\Hpnt) at (\Hpnt-1,3);
}
\foreach \Vpnt in {1,2} {
\coordinate (bfr-\Vpnt) at (3,\Vpnt);
\coordinate (bfl-\Vpnt) at (0,\Vpnt);
}
\end{scope}
%front face
\begin{scope}[canvas is yz plane at x=#1+1]
\foreach \Hpnt in {1,...,4} {
\coordinate (ffb-\Hpnt) at (\Hpnt-1,0);
\coordinate (ffu-\Hpnt) at (\Hpnt-1,3);
}
\foreach \Vpnt in {1,2} {
\coordinate (ffr-\Vpnt) at (3,\Vpnt);
\coordinate (ffl-\Vpnt) at (0,\Vpnt);
}
\node[transform shape,anchor=north] at ($($(ffb-1)!0.5!(ffb-4)$)!\OffsetLabels!90:(ffb-1)$) {#3};
\node[transform shape,anchor=east] at ($($(ffb-1)!0.5!(ffu-1)$)!\OffsetLabels!90:(ffu-1)$) {#4};
\end{scope}
\begin{scope}[canvas is yz plane at x=#1+0.5,transform shape]
\node[anchor=west] at ($($(ffb-4)!0.5!(bfb-4)$)!\OffsetLabels!90:(ffb-4)$) {#5};
\end{scope}
\draw[fill=#2] (bfu-1) -- (ffu-1) -- (ffb-1) -- (ffb-4) -- (bfb-4) -- (bfu-4) -- cycle;
\foreach \jur in {2,3} {
\draw (bfu-\jur) -- (ffu-\jur) -- (ffb-\jur);
\tikzmath{
integer \jlr;
\jlr=\jur-1;
}
\draw (bfr-\jlr) -- (ffr-\jlr) -- (ffl-\jlr);
}
\draw (ffu-1) -- (ffu-4) -- (ffb-4) (bfu-4) -- (ffu-4);
}
\begin{document}
\begin{tikzpicture}[z={(90:1cm)},y={(-30:1cm)},x={(210:0.75cm)}]
%macro parameter meaning
%\myCube{position}{color}{bottom label}{left label}{right label}
\myCube{-4.5}{blue}{}{}{blue block}
\myCube{-3}{green!50!black}{}{}{green block}
\myCube{-1.5}{red!80!black}{}{}{red block}
\myCube{0}{orange}{}{}{orange block}
\myCube{1.5}{violet}{3}{3}{violet block}
\draw[orange,dashed,very thick] let
\p1=($(current bounding box.south west)!0.5!(current bounding box.north east)$),
\p2=($(current bounding box.south) - (current bounding box.north)$),
\n1={veclen(\x2,\y2)}
in (\p1) circle[radius=\n1/2+6.28mm];
\end{tikzpicture}
\end{document}